lers
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SoWe*
'J
Analytical
Geometry
(Co-ordinate Solid Geometry)
A. R. Vasishtha
D. C. Agarwal
tu\W
SoW®*
I Analytical Geometry
3D
(Co-ordinate Solid Geometry)
tons
(For Degree and Honours Students of Indian Universities &.for Van
Competitive Examinations like P.C.S.&. l.A.S. etc.)
Bv
D.C. Agarwal
M.Sc., Ph.D
A.R. Vasishtha
Retired Head,
Retired Principal &. Head,
Department of Mathematics
Meerut College, Meerut.
Department of Mathematics
S.S.V (P.G.) College, Hapur.
&
A.K. Vasishtha
M.Sc.. Ph.D
C.C.S. University, Meerut.
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Analytical Geometry 3D
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Book Code:450-18
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PREFACE TO THE REVISED EDITION
In thi4 edition the book has been thoroughly revised.
Many more questions asked in various university
examinations have been added to enhance the utility of
the book. Suggestions forthe further improvement of the
book will be §;ratefully received.
— The Authors
PREFACE TO THE FIRST EDITION
This book has been written to meet the requirements
of the B.A. and B.Sc. students of Indian Universities, and
is so written* ias to coyer the courses commonly prescribed
for these examinations.
The book has been written in a very simple and lucid
style so that both students and teachers will find it an
interesting study. Care has been taken to explain the
fundamental principles fully and rigorously. Each article
is illustrated by means of typical examples fully solved,
so as to give the students a clear idea of the types of
examples he has to solve.
Any suggestions for the improvement of the book
will be gratefully accepted.
Finally, our thanks afe dufe to the publishers and the
printers who have taken great pains in producing the
book in the present nice form
— The Authors
CONTENTS
Chapters
1.Systeufi of Co-ordinates
Pages
1 — 14
2.Direction Cosines and Projections
15 — 45
3.The Plane
46 — 87
4.The Straight Line
88 — 173
5.Shortest Distance
174 — 200
6.Volume of Tetrahedron
201 — 216
7.Skew Lines!
217—233
8.Change of Axes
234 — 241
9. The Sphere
242 — 309
10. The Cylinder
310 — 324
11. The Cone I
325 — 393
12. Cientral C^coids
394 — 464
/
*.* .
1
Systems of Co-ordinates
§ 1. IntrodactioD. Students know well that In co-ordinate
geometry of two dimensions {i.e. plane analytical geometry) the
portion of a point in a plane is referred to two intersecting lines
(in the plane of the point) called the axes of reference and their ^
point of intersection called the origion of co<Ordinates. The axes
are called rectangular axes if they are at right angles, otherwise
they are called oblique axes. Whatever the axes may be, they
divide the plane into four quadrants called the first, second, third
and fourth quadrants respectively.
But it is not always possible to determine the positions of all
the points we can imagine with reference to above co-ordinate
axes. For example, codsider the five corners of a rectangular
parallelopiped, they do not lie in one plane. Such points are
called points in space, A point in space can be demonstrated as
follows:
Consider your study room and let dimensions of the room be
that ofa rectangular parallelopiped. Now consider any particle In
alrt then this particle In air Is a point In space.
The geometry of such points in space is discussed in
**Analytical geometry of three dimensions** also called **Sblid
geometry.
§ 2. Definitions. Origin, axes and co-ordinate planes.
Draw two mutually perpendicular lines X'PX and TOZ in
the plane of the paper. Let these lines intersect at O, then
through O imagine a third line Y[OY perpendicular to both of
the above lines, so that OY is perpendicular to the plane of the
paper and is directed upwards.
2
Analytical Geometry 3-D
The point O is called the origin. The three mutually per*
pendicular lines namely X'OX^ TOY and Z'OZ are called the
axes of reference (rectangular), and are said to be x>axis, y-axis
and X-axis respectively. OX\s taken to be positive direction of
x-axis whereas OX* as negative direction of x-axis. In a similar
way OY and OZ are taken to be the +ve directions and OY'and
OZ* as ^ve.directions of y and z-axes respectively.
M
^JCPLALE
L
’SC
o
'A
^J/PLANe'
V
When these three axes are taken in pairs, they give us three
planes YOZ^ ZDATand 3rDyr-Tt^e4hree planes are called yx^ xx
and xy-planes respectively. The set of these three planes is called
the set of co-ordinate planes(rectangular).
The axes are oblique axes, if they are not rectangular.
Note. In the rest of the book the axes will be assumed to be
rectangular unless otherwise stated.
§ 3. Co-ordinates of a point In space. Consider a point?in
space; Through P draw a plane PNAM parallel to YOZ plane
i.e. perpendicnlar to x-axis meeting it in the point A \ if OA>=»a,
then a is called the x-co-ordinate of P. Similarly through P
draw planes PNBL parallel to the plane ZOX and Pil/CL parallel
to the plane JfOy meeting y and x axes In the points Pand O
respectively ;iTpB<=b and OCf=»c then b is called y-co-ordinate
of P and c is called x-co-ordinate of P. The three numbers a, b, c
are called the co-ordinates of the point P and are written by
3
Systems of Go-ordinates
fY
%
ordered triads of the form (a, b, c). These co-ordioates are
measured positive or negative in the sense explained in § 2.
Let i, j, k denote the unit vectors along OAT, OF, OZ respect
ively. Let r be the position vector of the point P whose co-ordi
nates are (a, b, c). Then we have
OP^ON+NP
te●$
^pA+AN+NP^OAi-OB+OC
r=fll+frj+ck.
The vector oi+6j+ck is more conveniently written as (a» b, c).
Hence we may write
Thus (a, b, e) are the co-ordinates of a point P if and only if
the position vector of the point P is the vector ai+^j+ck wA/cA Is
simply written as the vector (a, A, c).
Remark 1. If (a, b, e) are the co-ordinates of a point P in
space, then it is usually written as the point P (a» A, c).
Remark 2. The co-ordinates (a, A, c) of the point P deter
mined as above are called the cartesian co-ordinates of the point P,
' but for convenience we shall simply say that (a. A, c) are the
co-ordinates of the point P.
.
Thus we see that the distances with proper signs of the origin
from the points on the axes in which the planes through the given
point P drawn parallel to the co-ordinate planes meet, are'" called
the co-ordioates of the point P.
4
Aualyttcal Geometry 3-Z)
§ 4. _ Properties of co-ordinates of a point P.
(A) The co-ordinates of a point P are respectively the distances
with proper signs of the point Pfrom the three co-ordinate planes.
See figure of § 3. Through the point P draw planes parallel
to the co-ordinate planes cutting the axes in the points
B and
C respectively. These three planes together with, the co-ordinate
planes form a rectangular parallelepiped. We have,
the perpendicular distance of the point P from the ;>z-plane
«=£»P«= CAf<=
=fl
=x-co-ordinate of the point P,
the perpendicular distance of the point P from the zx-plane
«AfP=CL=C?P=A
=^-co-ordinate of the point P,
and the perpendicular distance of the point P from the x>>-plane
^NP^AM^OC^c
«=*z-co-ordinate of the point P.
(B) The co-ordinates ofa point P are the distances from the
origin O of thefeet of the perpendiculars from the point P to the
co-ordinates axes.
See figure of § 3. Since the plane PNAM is perpendicular
to OX,PA (a line in this plane and cutting OX) is perpendicular .
to OX. Similarly PB and PC are perpendiculars to OY and OZ
respectively. Thus
the x-co*ordinate of the point P=Oi4, A being the foot of
the perpendicular from the point P on the x-axis.
Similarly the ^-co-ordinate of P<= OP, the z-cb-prdinate of
PaOC,where the points B and Care the feet of the perpendiculars
from P on the and z-axes respectively.
1
V
; .
§ 5. Octants. The three co-ordinate planes namely ;^z-plane,
ZX-plane and xj'-piane divide the space into eight parts called the
octabts, and to which octant the point P belongs is determined by
the.signs of the corordinates pfthe point P. The following table
detemihes the signs in eight octants r
Systems of Co-ordinates
5
5^
p
p
o
X
+
y
+
z
+
N
o
o
o
+
+
+
o
o
+. +
+
+
k
o
+
+
Ex. 1. What are the positions of the following points ?
(0 (1,2.3),
(//) (1.-2.3). (i7/> (0. 0.-3).
(/V) (-1,-2.0).
(V) (2. 0.0).
(V/) (-,1.-2,-3).
Sol. (i) (1. 2, 3) is a point in the octant OXYZ and its dis
tances from the co-ordinate planes
zx and x;;. are 1. 2 and 3
respectively,
(ii)(1. —2.3)is a point in the octant OXY'ZnnA its distances
from the co-ordinate planes yz. zx and xy are 1, 2 and 3 res
pectively,
(iii) (0,0, —3)is a point on OZ' Le, on the —ve side of
the z-axis situated at a distance 3 from the origin O.
(iv) (—I, —2,0) is a point in the co-ordinate plane xy
since its z-co-ordinate is zero. It lies in the. octant OX*Y*Z and
its distances from the co-ordinate,planes yz and zx are 1 aind 2
respectively,
(v) (2,0. 0) is a point on the positive side of the x-axis
situated at a distance 2 from the origin O.
(vi) (— 1, —2, —3)is a point in the octant OJfT'Z' hnd
its distances from the co-ordinate planes yz. zx and xy are 1. 2
and 3 respectively.
§ 6. Change of origioi Let OX, OY,OZ ba a rectangular
set of axes. Referred to these axes let the co-ordinates of two
points P and Q be (Xi, yi. Zi) and (Xa. ya, ^a) respectively; Suppose we want to shift the origin from O to the point P. i,e» we
want to find the co-ordinates of Q referred to P as origin.
6
Analytical Geometry 3~t>
Draw the new axis PXu PYi and PZi parallel to the original
ax«s OX» OYand OZ respec
tively.
The position vectors of
the points P and Q with res
pect to O as origin are given
by
X,
oJdL
OQ=>Xii+yil+Zik.
Also the position vector
ofthe point Q with respect to /
t
P as origin is PQ, Now we have
/^=5a-S>=(Xai+)^aj+Zak)-(*ii+J'ij+2tk)
i+(7s-;'i)j-h(^a-^i) k
Mxt-Xu yi-yi, zr-zi).
Therefore the co-ordinates of the point Q with respect to the
new origin P are (Xa—;fi. yt-^yu xt—zih
§ 7. The distance between two given points.
Let P and Q be two given
points in space.
Let the co-ordinates of
iZ
the points P and Q be (Xi, yu
Zi) and (xa, y^. Zi) with res
pect to a set OX, OY, OZ of
rectangular axes. The posi
/
tion vectors of the points P
V
and Q are given by
X
and O^aXai+^'aj+Zak
Now we have PQ^OQ^OP,
=W+J'ai+Zak)-(Xii-|-;»,j+Xik)
*=>(Xa-Xi) i+(>'a->»i)j+Ut-Xi)k.
Thus the distance PQ beween two points P(xu yuZi) ond
Q iXt, y», z») is given by
7
Systems of Co-ordinates
Corollary. The distance between the points (0, QrO) and
zi) is=V'(Xi*+V+^i*)*
Jt 4
f
§ 8. Division'of a line. To determine the co-ordinates of a
point R which divides the join of the line Joining the two points
P {Xu J'l, Zi) and Q (Xa, yu Za) internally in the ratio mi: m».
Let OX, OY, OZ be a set of rectangular axes.
The position vectors of the two
p
given points. P (Xi, yu z\) and
Q (Xa, yt, Za) are given by
i
^=Xii^‘;;ij+2ik ...(1)
and
5g=Xal+)'2j+Z8k' ...(2)
Also if the co-ordinates of the
point R are (x,>,z), then
I
/
/
...(3)
o
Now the point R divides the
y
join of P and Q in the ratio mi : ma,
so that
rni ^ or mi (RQ)-mi (PR),
OR=x\+yii-zk.
«●
X
mi^RQ
Hence
m^PR’=*miRQ
or
ma (5^-0?)==mi(5e-C^)
or
(mi-j-ma) OR—m\ OQ-l-ma OP
mi OQ+mj OP
mi-f-ma
or
or
xi 1 jj I
i-Kmij>a+mayi) j-f-(miZa+maZi) k
[using (1), (2) and (3)]
. Comparing the coefficients of i, j, k, we get
miy^+miyi
miZ^+rntZi
miXa+maXi
rni+rna * “ mi-f-ma
mi+ma
Cor. 1. The middle point of the segment P|2 is obtained by
putting mi=3ma. Hence the co-ordinates of the middle point of
Pfi are (i (x,-hxa), i (Pi+M i (^i+^a)).
Cor. 2. If mi: ma=p : 1, then the co-ordinates of the point
R are Xi-|-/*Xa yi±pyt Zi + /AZa
( ju-Hl * /*+! ’ ./*+! )●
s
Analytical Geometry 3-i)
These are called the general co«ordinates of a point on the
UnePQ.
Cor. 3. If the ratio (ffii/ws) is +ve then the point R divides
PQ internally and if it is —ve then externally.
For direct applications, the co-ordinates of the poidt R which
divides externally the join of P and Q in the ratio mi: m% are
miXg—miXi miyi—m^yi m\Z%^m%Zi \
( mi—mg »
mi—m» V mi—ma /
§9.(A) Centroid of a triangle. Let
be a triangle.
Let the co-ordinates of the vertices
^ and C be (Xi, yi, Zi),
end (X3, yg, Zs) respectively. Let AD be a median of the
AABC. Thus D is the mid. point of BC.
The co-ordinates of Z) are
/Xg+Xa yg-fys Zi+Za
\ 2 ● 2 * — )●
Now if G is the centroid
■ 2
(/.«., centre of gravity) of AABC,
s
7
then G divides AD in the ratio
2:1. Let the co-ordinates of G ^ S ^7 7? I
c
be(X,y,2). Then
Xx
■
-,otx
2+1
Similarly
3
J (yi+y.+y,), 2=iJ (*,+^+z,).
(B)
Centroid of a tetrahedron.
Let ABCD be a tetrahedron,
the co-ordinates of whose verti
ces are (x„ y,, Zr), r=* 1. 2, 3, 4.
Let Gi be the centroid of the
face ABC of the tetrahedron. Then
the co-ordinates of Gx are
/Xi+^fg-t-xg yi+yg-l-ya
I
3
»
3
*
The fourth vertex D of the tetrahedron does not lie in the
plane of A-dBC. We know from statics that the centroid of the
tetrahedron divides the line DGi in the ratio 3 ; 1. Let G be the
centroid of the tetrahedron and if (je, y, 2) are its co-ordinates,
then
9
Systems of Co-ordinates
Xi+Xa-fXa+Xj
> or X«=>
3+1
Similarly j>=*i (;>i+;^a+y8+J'4)»
§ 10.(A) Spherical polar co-ordinates.
LetX'OX, rOY and Z\OZ
be the set of rectangular axes.
Let P be a point in space.
Draw FN perpendicular from
P to the xy-plaoe. The posi
tion ofP is determined if the
length OP, angles ZOP and
XON are known. Suppose
OP=r, ZZOP=6 and f_XON
measured positively in the
directions shown by arrows
in the figure. The quantities
r, d, 0, defined as above, are
jc=i
4
(xi+xa+xa+ra).
called the spherical polar co-ordinates of P and are written as
Now we shall find relations between these co-ordinates and
cartesian co-ordinates. Let(x, y,z) be the cartesian co-ordinates
of P. Hence we have
z^PN=OP cos iAOPN)=»r cos(AZOP)^r cos 6,
...(1)
Also Oi\f= OP sin ZOPiV=r Sind
[v Z.O^P«90®]
x«=OiVcos
cos^ Sind,
.(2)
and
;>e=OiVsin ^=r sin tf> sin 6,
...(3)
Thus relations (2),(3) and (1) give the relations between
z and r, d,
Now squaring the relations (2) and (3) and adding, we get
x*-fy><=»OiV*, or u*c3x2+y® where ties* O^
or
‘V/(x*+P*)*=>M«r.sin d.
...(4)
Dividing (4) by (1), we get tan d*=^(x^+y^)lz.
Dividing (3) by (2), we get tan 4>^ylx.
Squaring (1) and (4) and adding, we get
x*+/+z*c=r*.
Thus the relations between spherical polar co-ordinates and
cartesian co-ordinates are
. xe=*r cos ^ sin d, year sin ^ sin d, z«=rcosd
x*+y®+2*«r®, tan d=>V(x*+y®)/z, tan <f>=yjx.
X,
10
Analytical Geometry i-D
(B) Cylindrical co-ordinates. See figure of § 10(A). Let P
be a point in space. The position of P can also be determined if
the measure of OJV, ^XON and NP are known. Suppose
AXON^4>, NP=z, The quantities «, z are called the cyclindrical co-ordihates of P and are written as (m, z).
Let(x, yt z) be the cartesian co-ordinates of P, then N has
the co-ordinates (;c, y, 0). Hence, we have
x=ON cos
cos y=>u sin 2=z.
Also
tan 4>=>yjx,
Solved Examples
Ex, 1. P is a variable point and the co-ordinates of two points
A and Bare (—2,2,3) and (13,-3,13) respectively. Find the
locus of P if 3P^=2PB.
Sol. Let the co-ordinates of P be (x, y, z).
...(1)
P^=-^{(x4-2)H(y-2)H(z-3)a),
and
...(2)
PB=V{(*-13)H()'+3)»+(z-13)»}.
...(3)
Now it is given that 3PA=>2PB i.e.y 9P^®=>4jP-B*.
Putting the values of PA and PB from (1) and (2)in (3), we
get
9{(x-f2)>-K>-2)»-i-(z-3)2}=4{(x-13)H(;'+3)«+(^-13)*}
or 9{x*-H>*-|-z*+4x—4;v—6z-)-17}=a4
26x-|-6>'
-26Z-I-347}
or 5xa-}-5;;2^5z*+140x-60>^-h50z-1235=0
or x*-|-/+z*+28x-12;v+10z-24=0.
This is the required locus of P,
Ex. 2. A, B, C are three points on the dxes of x^yandz
respectively at distances o, cfrom the origin 0\ find the co-ordt^
nates of the point which is equidistantfrom A, B,C and O.
Sol. Let the required point be P(x, y, z).
The co-ordinates of the points A, B, C and O are (o,0, 0),
(0,6, 0),(0, 0. c) and (0,0,0) respectively.
PA==»PB=PC^PO.
It is given that
Taking
PA^PO, or PA^=PO*, we have
(x—fl)*-l-y“*+z^=x*+y*-|-z*
or
—2ox-|-o*=.0, or x»a/2.
...(I)
Now taking
PB=»PO, or PB^t=>po^, we have
x'^+iy—b)^+z^=x^-[-y*-^z^
or
-2by-{-b^=0, or y«=»ft/2
...(2)
Again taking PC^PO, or PC*=PO>, we getz=»c/2. ...(3)
11
System of Co-ordtnates
Heace the co-ordinates of the required point P are (fl/2, 6/2,
c/2)
Ex. 3. Find the centre of the sphere which passes through the
points O (0,0, 0). A (a, 0, 0), B(0. 6.0)and C(0,0, c).
Sol. Let the centre of the sphere be.P(x,;>.z).
Since the sphere passes through A, B,C and O,so P(x, y, z)
is equidistant from A^ P,C and O.
PA=>PB=PC*=»PO.
Now proceeding as in Ex. 2 above, the centre P of the sphere
is given by
(Ja,
Jc).
Ex. 4. Show that (0, 7, 10),(-1,6, 6), (-4,9,6)'/orm an
isosceles right angled triangle:
Sol. Let ABC be a given triangle and let the co-ordinates
of the vertices A, B and C be (0,7,10),(— 1, 6, 6) and(—4,9, 6)
respectively. .*. i4P=»V{(0+l)*+(7—6)*+(l6—6)*}=*V'(1®)
BC= y{(-1+4)H(6 -9)*+(6-6)*}=^(18)
C^=V{(-4-0)>+(9-7)«-K6-10)*}=.v'(36).
We have AB=BC, hence A ABC is an isosceles triangle.
Again ^P>-fPC*=18-M8=36=C^». .*. f_ABC-=>W,
Hence A
is also right angled triangle. Therefore, the
given triangle is an isosceles right angled triangle.
Ex. 5. Find the co-ordinaies of the point which divides the join
o/(2, 3,4)and(3, -4,7) in the ratio 2:-4.
Sol. Let the co-ordinates of the required point be (x, y^ z),
then by § 8. we have
2(-4)-4(3) -20
10;
x«2(3)~4(2)^6~8
-2
2—4
2-4 “-2
— -
■
. ●
. ■
C3
^
z=»2(7)-4(4),14-.16„-2_
2-4
-2 -2
Hence the required point is (1, 10, 1).
Ex.6. A point P lies on the line whose end points are
A(l,2y 3)and B(2, 10, 1). Ifz co-ordinate ofP is 7,find Its other
co-ordinates.
Sol. Let the co-ordinates of the point P be(x, y,z) and let
it divide the join of A (1, 2, 3) and B(2, 10, 1) in the ratio /x: 1.
. /tt(l)-H(3)
Then [by cor. 2, § 8]
/^+1
But it is given that the z-co-ordinate of the point P is 7.
7ju-f7«=/a-1-3 or. 6/i=3—4 or7*=-2/3.
12
and
Analytical Geometry 3-D
x=/^(2)+l (1) _(-4/3)-M -1
M+1
(-2/3)+l“
ti(\0)+\(2)_-(20/3)+2
y^
14.
AiH-1
-(2/3)+r
£x. 7. F/firf the rath in which the xy-piane divides the join
(5. —6,4).
find the point of intersection
. o/ the line with the plane.
^'
Sol. Let the x;^-plane (/.«., 2=0 plane) divide the line
joining the points(-3,4. -8)and (5, -6.4)in the ratio p ; 1,
in the point R. Therefore, the co-ordinates of the point R are
4/i-8\
[See cor. 2 6 81 l^tzl
Vm+1
...(1)
But on xj'-plane, the z co-ordinate of R is zero.
(4/*—8)/(/i+l)=0, or p=>2.
Hence
p\\=>2: 1. Thus the required ratio is 2: 1.
Again putting /a=2 in (I), the co-ordinates of the point R
become (7/3,—8/3, 0).
Ex. 8. Find the ratios in which the sphere
-}->'*+Z*™ 504,
divides the line Joining,the points (12, -4,8) n«d(27, -9, 18).
Sol. Let the sphere x‘4’J'®+.z*«=>504 meet the line joining the
given points in the point (xi, yi, z{).
Then Xi*+j;i*+zi>c=504.
...(1)
Now suppose that the point (Xuyu Zi) divides the join of
the points (12, -4,8) and (27, -9, 18)in the ratio fi: 1.
27p+l2
-9#t-4
18/i+8
Then Xi
*yi^
● Zi-a
P+l
P+1
^ P+1
Putting the values of Xi, yu Zi in (1), we get
(27/t+12)«^(-9/i-4)8 . (18m-1-8)»
504
(/*+!)*
t/4-1-1)*
(/*+!)»
or
9(9/«+4)*+(9/*+4)*+4(9/x+4)»«504 (/t+1)®
or
14 (9/t+4)*=504(p+lf, or (9/i+4)*«36(p+i)K
Taking the square root, we get 9/«+4=±6(p+l).
Taking +ve sign, 3^1=2, or /n/1=2/3, or pi 1=2:3.
Again taking -ve sign, 15/t=—10, or p/l=> -2/3,
or
P : 1=2: -3.
Ex. 9. From the point (1, -2^3) lines are drawn to meet
the sphere x*+y^+z^-^4 and they are divided, in the ratio 2:3.
Prove that the points of section lie on the sphere
5(x*-i-j»Hz“)-6(x-2j;+3z)+22=0.
13
Systems of Co-ordinates
Sol. Suppose any line through the given pomt (1, ^2,3)
meets the sphere
in the point (xi, yu zi). Then
...(1)
Now le^ the co-ordinates of the point which divides the join
of(1, —2,3) and (xi, yu Zi) in the ratio 2:3 be (xa, y%t Za). Then
we have
2.x,4-3:1
2+3 .
or
2.;y,+3.(-2)
or
Xa=3
2a
2+3
2.2t+3.3
2+3
or
Xi
5xa-3^
2
...(2)
Zl
4?1
2 J
Putting the values of Xu yu zi, from (2) in (1), we have
(5xa-3)a+(5ya+6)»+(52a-9)>=4x4
or
25 (Xa*+;>a*+Za*)—30x2+60>>a—902a+110«0
or
5(xaHV+Z8*)-6(Xa-2;;a+?Za)+22=0.
the locus of(Xa, yu 2a) is
5(x*+J'®+z®)-6(x-2j;+3r)+22=*0,
which is the equation of a sphere.
Ex. 10. Prove that the three points A, B and.C whose co-ordi
nates are
—2,4),(1, 1, 1) and (—1,4, -2)respectively^ are
colllnear.
Solution. The general co-ordinates of a point R which divides
the line joining A (3, —2,4) and B (1, 1, 1) in the ratio /u : 1 are
ft+3
/it+4
...(1)
(/*+!’ /*+i* /*+1)
If C(—1,4, —2)also lies on the line AB, then for some value
of p the co-ordinates of the point R will be the same as those
ofC.
let the x-co-ordinate of the point /?=the x-co-ordinate
of the point C.
Then (/a+3)/(/li+1)= —1, or p=—2.
Putting /*«—2 in (1), the co-ordinates of R are (—1,4, —2)
which are also the co-ordinates of C. Hence the points A, B and
C are collinear.
Also we note that C divides AB in the ratio
:1,i.e.f —2: 1.
14
Analytical Geometry 3-Z)
Exercises
1. Find the locus of a point P which moves in such a way that
its distance from the point A (u, v, w) is always equal to a»
Ans.
2t/x—2vy—
2. The axes are rectangular and
P are the points (3, 4, 5),
(—1,3, —7)» A variable point P moves such that (i)PA=PB
and (ii) PA*—PP‘=2fc*. Find the locus of P in each of the
above cases.
Ans. (i) 8x+2;;+24z+9=0. (ii) 8x+2;;+24z+9+2fe*«0.
3. Show that the points(1, 2, 3),(2, 3, 1) and (3,1, 2) form an
equilateral triangle.
Hint. Show that the length of each side of the triangle is y/6:
4. Prove that the three points A, B and C whose coordinates are
(3,2, —4), (5, 4, —6) and (9, 8, —10) respectively are
collinear.
2
Direction Cosines and Projections
§ 1. Angle between two non*coplanar (i.e. non>intersecting lines).
Let PQ and MN be two non-coplanar lines. The angle
between two non-coplanar lines PQ and MN is equal to the angle
between two straight lines OA and O^B drawn from any point O
parallel to PQ and MN respectively. Thus the angle between
p
the lines PQ and MN is equal to the angle AOB,
§ 2. Direction cosines of a line.
Definition, i/’a, j8, y are the angles which a given directed line
with the positive directions of the axes of x.y andz respec●jpti
t
then cos ct, cos J8. cos y are called the direction cosines
{briefly written as d.c:s) of the line. These d.c:s are usually denoted
by If m,n.
Let AB be a given line. Draw a line OP parallel to the line
AB and passing,through the
origin O. Measure angles a, y
as shown by arrows in the
figure, then cos ct, cos p, cos y
are the d.c.*s of the line AB. It
can be easily seen that lfm, n
are the direction cosines of a
line if emd only if li+mj-{-nh Is
X
'A
a unit vector In the direction of
p*
that line.
4
● Clearly OP' {i.e., the line
through O and parallel to BA) makes angles
Analytical Geometry 3-D
16
180°-y with OX,07 and OZ respectively. Hence d.c.’s of the
line BA are cos (180®- a), cos(180®-^), cos (180®-y)
are
—cos a, —cos j3, —cos y.
Remark. Since the angles a, p, y are not coplanar,
/. a+i34-y#3.60®.
D.c.*s of the coordinate axes.
Since the axis of x makes angles 0®, 90°, 90® With the axes of
Xf z respectively, therefore by definition, its d.c.’s are cos 0®,
cos 90°, cos 90® /.c., 1,0, 0.
Hence the d.c.’s of the x^axis are 1, 0,0.
Similarly the d.c.’s of the y-axis are 0,1,0
and the d.c.’s of the z-axis are 0, 0, 1.
§ 3. If the length of a line OP through the origin O be r, then
the co-ordinates ofP ore (Ir, mr, nr) where /, m,n, are the d c.’j
of OP.
Draw PM perpendicular
z
p
from P to OX meeting it at M.
y\
Let (X, y, z) be the co-ordi
nates of P, then OM=»x. From
the right angled l^OMP^yte
,
.
have OMaCOS Qt^l
OP
nM
X
or
xfr=l or x=>lr.
Similarly y=mr,z=>nr.
A P is the point (/r, mr, nr).
/
§ 4. If /, w, n are directlon cosines of any line AB, then
(Kanpnr 1983)
to prove that
-f =1.
Through the origin O draw a line.DP parallel to the given
line AB so that the d.c.’s of OP are I, m, n. Suppose OP is of
length r. If the co-ordinates of P are (x, y, z), then we have
...d)
x=/r, y=smr, z=>nr.
(see § 3)
Now r®«DP2 or r*=(x—0)2-}-(y—0)^-l-(z—0)®
or
r*s>x2-l-y*-l-z2 or r^^Pr^+m^r^-^n^r^
using (1),
or
12-f-m2-l-n2=l.
Remark. We have DP*=xl-l-yj-l-zk=/ri-|-mrj-l-«rk..
a unit vector in the direction of OP
OP
— e=/i-|iwj l-;ik
[Y lOPHrJ
...(2)
Direction Cosines and Projections
17
The relation (2) shows that the direction cosines of the line OP
are the coefficients of\^ j. k in the rectangular resolution of the unit
vector in the direction of OP.
Thus if/, m, n are the d.c.’s of a line, then a unit vector
along that line is /i-|-mj+nk.
l /Hwj+nkl=l
or
§ 5. Direction ratios.
Definition. If the direction cosines /, m,n of a given line be
proportional to any three numbers a^ b,c respectively, then the
numbers a, b, c are called direction ratios (briefly written as d.r *s)
of the given line.
Relation between direction cosines aud direction ratios.
Let a, b, c be the direction ratios of a line whose d.c.’s are
/, m, n. From the definition of d.r.’s, we have
l/a=^mlb-nlc=k (say). Then l=‘kai m=»kb, n=kc,
But
Jfca
or
lKa^+b*+c^)
or
Taking the positive value of k, we get
c
b
a
/=
m«=>
»
V(aHb*+c»)
Again taking the negative value of k, we get
—c
-b
Remark. Direction cosines of a line are unique. But the
direction ratios of a line are by no means unique. If a, b, c are
direction ratios of a line, then ka, kb, kc are also direction ratios
of that line where k is any non-zero real number. Moreover if
o, b, c are direction ratios of a line, then ai+bj+ck is a vector
parallel to that line.
Rule. Let a, b. c be the d r.’s of a given line, then^ to find
actual direction cosines ofthis line, divide each of a, b, c by
SOLVED EXAMPLES 2(A)
Ex. l. Find the d.c.*s of a line whose direction ratios are
2, 3, -6.
Solution. We haveV((2)*f(3)H(-6)*}=V^4^-9-|-36)-7.
2 2
^
Hence (by § 5) the d.c.*s of the given line are p
-y
18
Analytical Geometry 3-D
Ex. 2. Prove that sin* d-^-sln^
sin* y=>2, where Pt y ore
the angles which the given Urie makes with positive directions of the
(Agra 1978)
axes.
Sol, We have (See § 4) cos* a+cos® /5+cos* y=» l..
Changing cosines into sines, we get
(l-sin* a)+(l“Sin* j8)+(l-sin» y)=I
or
sin* a+sin* jS+sin® y=2.
Ex. 3. Find the direction cosines of the line which ts equally
(Agra 1979)
inclined to the axes,
Sol. Let the direction cosines of“ the line be /, m, n.
Since the line is equally inclined to the co-ordinate axes, therefore
/9e3/|}®G=iR®.
But /*+m*-l-»*«l.
3/®=l or /®=l/3 or /=±l/v'3.
Similarly
m=*±l/V3i
Hence the direction cosines of the line are ±l/i/3, ±l/\/3,
±l/\/3. There will be eight such lines, one lying in each octant.
Ex. 4. Find the direction cosines /, m, n of two lines which are
connected by the relations /—5m+3n=0 and 7/*-|-5m*—3n*e»0.
Sol. The given relations are
...(1)
/—5m+3n=0 or /«=5m—3n
...(2)
and
7/*+5/M*-3n®=0.
Putting the value of / from (I) in (2), we get
7(5OT-3n)®+5m®-3n®=»0
or
180m®—210mn-i-60h*=0- or 6m®—7mn+2n*s=0
or
6m*—4mn—3m/i-f-2n®«=i0 or (2m—n)(3m—2n)=0,
and
m/R=2/3.
.*. m/rt=i.
Now when m//ieaI/2 i.e. n«=»2m, we have from the relation (1)
1
/
/c=.5m—6m or /=»—m or — ”1 *
m
m 1
. /
1 ..
I m n
Thus -=2-and
giving -i=T=2
1
n
m
/
^(P+nfW) _
or
-I“T= 2“ V{(-l)*+l*+2®}“V6
●. The d.c.’s of one line are —1/^6, l/i/6, 2f^6.
Again when
, .
9m
l==5m—j
, m
or
or
/
m
Direction Cosines and Projections
19
m
Thus —
giving
n Jjand
1
I m
n
T=T“T‘“V(1’*+2»+3»)”V(WJ
The d.c.*s of the other line are l/VO^), 2/i/(14), 3/V0^)»
Ex. 5. Find the direction cosines /, m, n of the two lines which
are connected by the relations /+wi+n=0 and
2/m>=0.
(Meerot 1985)
. Sol. The given relations are
...(1)
/+m+n=»0 or l^-m-^n
...(2)
mn~-2nl—2lm=0.
and
Putting the value of / from (1) in the relation (2), we get
«n_2»(-./»-n)-2(-m-n)m=0 or 2m»+5mn+2n*«0
or
(2w+«)(m+2/i)=0.
m
i-and-2.
n ““""S
m
/ —m—n
From (1), we have — =a n
...(3)
n
Now when m/n=—i,(3) gives//n=J—l=—i.
m/l=«/—2 and //l=n/—2
1
/ m
n
l.e..
1 “1 “-2- v/{l*+l*+(-2)*}
The d.c.*s of one line are
1/a/6, —21^6,
Again when m/2c=i —2» (3) gives ///i«2—1«»1.
I
m n
ViP+m^+n*)
J_
●● r“-2“r“v'{T*+(-2)»+l*}“V6‘
/. The d.c.*s Of the other line are
—2/^/6, l/\/6»
§ 6. Projection of a point on a given line.
Let P be a given point and ifP the
P
given straight line. Draw perpendicular
PM from P to AB» meeting AB in M,
Then the,foot of the perpendicular M is
. called the projection of the given point
Af
3
P on the given line AB.
The point M Is the point in which the
plane through P and perpendicular to
meets the line AB.
A
§ 7. Projection of a segment of a line on another line (in the
same plane or another)^
Suppose we are to find the projection of a segment CD of a
20
Analytical Geometry 3-D
line on another given line AB. Let the points C', D' be the
projections of the points C, D on the line AB, then the segment
c
<9
/9
'P
C'
C*D* is the required projection of the segment CD. The projec
tion of the segment CD on the line AB may also be defined as the
intercept CD* made on the line AB by the planes through the
points C and D each perpendicular Xo AB.
To find the length of the projection C'D'. Through D draw a
line DP parallel to D'C’ and meeting the plane through C per
pendicular to AB in P. Thus DP^D'C.
.:.(!)
Let 9 be the angle between the lines AB and CD. Also AB
is parallel to PD and hence A_CDP^9. Again the line CP lies
in the plane which Is perpendicular to
and hence CP is per
pendicular to DP.
Therefore
DP=DC cos $.
...(2)
From (1) and (2), we have D'C'=DC cos 0
or CD*=CD cos 6. ^
Remark*. Let DC=a and let b be a unit vector along BA.
Then
a«b=|a 1 1 b 1 cos d, by definition of dot product of two vectors
=DC cos $
[V, |b|=l andla|«DCj
=projection of DC on BA.
Thus to get the projection of DC on a line BA, we take the
dot product of the vector DC with a unit vector along BA.
§ 8. Projection of a broken line an a given line. Or Given n
points Cl, Cs,..., Cfl (say) in space, to find the projec
tion of CiCn on a given line.
Suppose AB is a given line. Let the projections of the n
points Cu Ctt ...«Cji on the given line AB be the points
21
Direction Cosines and Projections
il/i, Ms,
Affl respectively. Thus the points Mu A/2,..., M„ lie
on the same straight line AB. Hence in view of § 7, we have
projection of CiCz=MxMu
projection of
and projection of
C„“Affl_i Af„.
sum of the projections of C1C2, CaCs,..., C«-i C„ on the
line AB—MxM^A^M^M^-^ . ■ AtMn^\ Mn
=A/iAfft=projection of CiCi on AB.
Hence we conclude that :
,
Projection of CiC„ on a line /4.8=Sum of the projections of
^i^2»
C71 on the line AB.
§ 9. Direction cosines of a line joining two points P (Xi, yi, Zi)
and Q (Xa, ja, Zg).
Let the line PQ make angles
a, /3, y with X, y, z axes respec
tively. If /, m, n be the d c.’s
of this line, then
z
P/LC^-
/=cos a, m-cos ]8, «=cos y.
Now if L and M are the
projections of the points P and
Q respectively on the x-axis,
then we have
Af
X
V
OL^Xu OM—Xu so that
LM=^OM-OL^Xi-Xu
But LM is the projection of PQ on the axis of x. Hence by
§ 7, we have
LM=PQ cos a or Xg—
or
{x^--x,)H^PQ.
...(I)
In a similar way by projecting PQ on y and z axes, we have
(ya—yx^|m =^PQ and (Za—ziVn^PQ.
...(2)
So from (1) and (2), we have
/
m
B
Pfi.
Hence we conclude that the actual direction cosines /, m,n of
the line Joining two points P-(xi, yi, z,) and Q(Xa, ya, Zg) are
Xa -X, ya-yi
Zg-Zj
PQ * 7PQ ● PQ
reipectively, where P0== v^{(Xa-Xi)H(y2~.yi)*+(Zf~Zi)*}.
Again the direction cosines of the line PQ are proportional to
Anilyiical Geometry 3-D
22
Xt—Xi, ya—yii *a—which are therefore the direction ratios of the
line PQ.
Vector method. Since the co-ordinates of the points P and Q
are(Xu yi, Zi) and (Xa, ya. Za) respectively, therefore
DP*=*the position vector of P=Xii+yij-l-Zik
and Dfi^Xai+yaj+Zak.
PQ=^OQ-~dp^(Xi-Xi)i+(y3“yi)j+(za-zi) k*
Also 1 PQ HPe=v^{(Xj-x,)>+(l’»-;’i)’+(rs-Zj)*)A a unit vector along
Hence the d.c.*s of the line PQ are
Xa-Xi ya-yi Za-Zi
PQ * PQ * PQ'
Also the d.r.*8 of the line PQ are Xa—Xi, ya—yi» Za-Zi§ 10. If O and P are two points (0,0.0) and (Xi, yi, Zi), then to
prove that the projection of OP on a line whose direction
cosines are 1. m, n is Ixi-fmyi-i-nzi.
Construct a rectangular parallelopiped with diagonal as OP
and faces parallel to the co-ordinate planes. [See figure of § 3,
chapter 1, page 3]. Then clearly we have
OA^Xu OB=^yx=ANt CO=Zi=NP.
Now considering OP as a broken line consisting of the parts
04
and iVP, we have
the projection of OP on a line with d.c.*s l,m»n
«sum of the projections of OA,AN and NP on the line with
d.c.’s /, m,n
[by§7J
1xi+ihyi-l-nzi.
Vector method. We have OPc=xii+yiJ+Zik. .
Also a unit vector along the line whose d.c.*s are /, m, n
c^ti-i-wj-l-wk.
A projection of OP on the line whose d.c.*s are /, m,n
=(Xii+yjj-hZik)●(/i-1-mjHh nk)=/xiH-myi-h nzi.
§ 11. To find the projection of the line joining two points P(Xi,yi,Zi)
and Q(X3y.ya. Zs) on another line whose d.c.*s are 1, m, n.
Let O be the origin. Then
C7P»Xii-l-yij-i-Zik and Ofi^Xai+yaJ-j-Zak.
Direction Cosines and Projections
23
^=*00—0?*=*(xa—Xi) i+(>'8—
j+(2a—2i)
Now the unit vector along the line whose d.c.’s are /, m, h
=/i+mj+nk.
projection of PQ on the line whose d.c.’s are Um^n
■=[(Xa-A:») i+(;;a-.Vi) j+(Z2-Zx) k].(/i+mj+nk)
=*/ (Xa~-Xi)+/M (.Va—3'i)+« (za—^i)*
Remark, In the articles 10 and 11, l^m^n are the actual
direction cosines and not direction ratios.
SOLVED EXAMPLES 2 (B)
Ex. 1. If P (2. 3. -6) and Q (3, -4. 5) are two points, find
the d.c:s of OP, PO, OQ and PQ where O is the origin,
Sol. By § 9, the direction ratios of OP are 2—0, 3—0, —6—0,
i.e,, are 2, 3, —6.
Also C?P=V(2*+3a+(-6)>}«V(49)=7.
Hence d.c.*s of OP are
2
are 2= 3 -;6
Ans.
Tv 7
OP* OP* OP ^ '
The d.r.’s of PO are 0—2, 0—3, 0—(—6) /.e., —2, —3, 6.
Also
V{(-2)>+(-3)>+6»}=7.
Ans.
/. d.c.’s of PO &TQ —2/7, —3/7, 6/7.
the d.r.’s of OQ are 3-0, -4-0, 5-0 i.e., 3, -4, 5,
Also 00= V{(3)>+(-4)>+(5)»}“V(50)=»5i/2.
d.c.*sofO0are3/(5V2),-4/<5>/2),5/(5v'2).
Ana.
The d.r.’s of PO are 3-2, -4-3, 5-(-6) f.c., 1, —7,11.
Also P0 = V{(3 -2)*+(-4 - 3)»H- (5+6)*}=V( 171).
/. d.c.’s of PQ are l/v'(171), -7/V(171), 11/V(171). Ans.
Ex. 2. Find the iength of a segment of a line whose projections
on the axes are 2,3, 6.
Sol. Let CD be a segment of a line whose direction cosines
are /, m, n. Then, we have
● ●● (1)
2=the projection of CZ) on the x-axis=/.CZ>
●●● (2)
3cathe projection of CD on the y-axis=m.CZ)
● ●● (3)
6=>the projection of CD on the z-axis=n.C/)*
-1
Squaring (1). (2), (3) and adding, we have
4+9+36=(i>-i-m*+n*) CD*
49=1.CD*
[y /*+m*-l-rt*=i3
or
Ans.
CD=7.
or .
Ex. 3. If P, Q, R* S are four points with co-ordinates (3, 4, 5),
i4
Analytical Geometry 3-/)
(4, 6i 3),(—1,2, 4),(I, 0, 5)respectively, then find the projection
of P.Q on RS. Also find the projection of RS on PQ.
Sol. To iBod the projection of PQ on RS, we should find the
d.c.’s of RS.
The direction ratios of RS are
l-(-l),0-2,5-4/^., 2, -2, 1.
Also /?5=v'[{I-(-l)P+(0-2)a+(5-4)a]=V(4+4+l)=3.
d.c.’s of RS are 2/3, -2/3, 1/3.
Hence the projection of PQ on RS i$ (See § 11)
Ans.
=f(4-3)-|(6-4)+J (3-5)=f-|-|=-4
8‘
Again to find, the projection of RS on PQ, we should find the
d.c.*soff»g.
The direction ratios of PQ are
4—3,6—4. 3-5/.e., I, 2,—2,
Also PQ -- V{(4-3)2+(6 7 4)2 -|-(3-5)2}= V(1 +4+4)=3.
the d.c.’s of PQ are 1/3, 2/3, -2/3,
projection of.iJS* on PQ is (.See § 11]
Ans.
■=J {l-(-l)} + f (0-2)-f {5-4)=f-|-|=-t.
Ex. 4.
V Pt Qi R, S are four points with co-ordinates
(2,3,-1), (3,5, “3)» (1,2,3), (3,5,7) respectively, prove by
projections that PQ is at right angles to RS.
Sol.
In order that PQ is at right angles to RS, the projection of PQ on RS should be zero.
The direction ratios of i?,Sare 3-1, 5-2, 7-3, i.e., 2, 3, 4.
Also /?S=-v/{(3-1)2 4-(5 -2,2+(7-3)2}
.
‘=V'(4+9+16)-V(29).
d.c.’s of RS are 2/v'(29), 3,V(29), 4/v/(29).
Hence the projection oPPQ on RS\s [See §11]
V(29)
.(5-3)-f
V(29)^
3-(-l)}
^2+6-8 =0.
V(29)
Therefore Pg is at right angles to RS.
§ 12. Angle between two linesi.
^ To show that the angle $ between any two lines whose direction
cosines are /,, «j,,
and U, m^, n^ is given by
cos ^=lila-fniini2+nina.
(Meerut 1980, 85S: Kanpur 82)
25
Direction Cosines and Projections
Let AB and CD be two given lines whose d.c.’s are /i, mi, «i;
and /a, ma, respectively. Through the origin O draw lines OP
and OQ parallel to AB and CD
respectively so that direction
cosines of OP and OQ are
/i, mi, nil and /a, ma. n^ respec>
tively. Take a point P{xi, yu zi)
on OP such that OP—ti. Since
the d.c.'s of OP are li,muHu
therefore the co-ordinates of P
may be written as
(/iri. mir„ «ir,).
iG
P
y.
o
...(1)
Zs) on OQ such that
Zi=.«iri.
Similarly if we lake a point Q (jca,
OQ=Ti, then
●C2)
Xi - /gfa, yz=nizrg, Za=n^rzLet 6 be the angle between the lines AB and-CZ); then B is the
angle between O/-and Og.
Now the projection of OQ on OP
~li (■^2—0)+mi O’a -0)+rti (Z2—O)
...(3)
/lA'a-f wi>'2-f/J,Za
...(4)
But the projection of OQ on OP-^^OQ cos 0=ra cos 9,
or
or
From (3) and (4), wc have ra cos ^—Axa+mi^a+WiZa
using (2)
racos 0=/,/ara-l-mimar2+WiWara
...(5)
cos d = Ilia-hmimj-l-niQa-
Remark. If 9 is the acute angle between the two lines, then
cos 0 is -f ive and so we have
cos 0=1 /i/a+mima f «ina !●
Corresponding formula when direction ratios of the lines are
given. Let Oi,
Ci and a^, 6a, Ci be the direction ratios of the
two given line-.
Then their actual direction cosines are given by
b
6a
ViOz^f bfyc2^) *
02
+
Cl
and
Cz
6a^ + Ca«) *
Using these values of d.c.’s in (5), the angle 9 between these
two lines is given by
COS 0«=—
Viai' -h
Oiaa^ biha-f-CiCa
Ci*7v va2^ + baHca")
...(6;
26
Analytical Geometry 3-2)
Cor. 1. Tofind sin 6 and tan 9 In terms of d.c's and d.r.*s
of the two given lines.
First of all we state the Lagrange^s identity which is as
follows :
If/i, nil, r>i and /g, m^, «a are two sets of real numbers, then
.●●(7)
(m,na—wi2«i)*+(Wik—WsAT+(A^2— /aWi)*.
[Remember]
Now, we have
sin* 0=1 —cos® 0=-l —(/i/2-t-WiHi2+«i«2 .2
= (/l* + Wi* + «,*) (/22 + /M22-t-W22)-(/i/2 + WIiW2 + «ina)®
=(m,«2-m2«i)* + («i/24-ff2/i)*+(A/M2-/2mi)*, using (7).
...(8)
sin 9=^/{S (minz—mzni)^}.
The value of sin* 9 may be conveniently remembered in terms
of determinants as follows :
We write the corresponding direction cosines of the two lines
above and below in two rows as follows ;
Then sin® 9=
A
/Ml
/Ji
I2
/Ma
//2
nil
ni
Wa
/la
2
+
1
A
A
.. (A)
2
Ma
+
A
/2
nil
2
/Ma
To get the first determinant we suppress the first column in
(A), to get the second determinant we suppress the second column
and to get the third determinant we suppress the third column.
In terms of direction ratios,;the formula (8) is given by
sin* 9= {bid
Again, tan
- bjCiY+(Cifla-C8gi)^+ (gifea-fl2bi)*
.(9)
sin 6 y/{S (/Mi/l2~/MaM,)*>
0 = COS0'
AA+^i^2+Wi/ia
■■●(10)
in terms of direction cosines, while, in terms of direction ratios,
we have
tan 0= y/{2
(^iCa~baCt)*}
diaa+biba+Ci^'a
...(H)
Note that the formula for tan 0 is the same whether we are
given direction cosines or direction ratios;
Cor. 2. Condition for perpendicularity. If the lines are per
pendicular to each other, then 0=90® i.e, cos 0=cos 90®=0 and
therefore from (5), the required condition is
IJa+mima+D1D2=0.
...(12)
27
Direction Cosines and Projections
In terms of direction ratios the required condition of perpen¬
dicularity [from (6)] is
aiSa i-b,ba+CiC2=0.
...(13)
Note that the condition for. perpendicularity is she same
whether we have direction cosines or direction ratios.
Cor. 3. Condition for parallelism : If the lines are parallel to
each other then 0=0/.e. sin 5=0 and from (8), we have
+(«J2—Wa/j)**+(A Wa“/2^1)® ●= 0.
(miWa“
The L.H.S. being sum of three squares of real quantities will
be zero if each of them is separately zero and hence . we have
m/na ” wja«i=0, Oi/a—«2/i--0, /ifWa—/a»*i=0
/,
or
Wi
«i
)
V(^)_i
...(14)
li~la» mi=ma» ni=na.
This shows that the two iines will be parallel if their direction
cosines are the same.
In *erniis of direction ratios the required condition of parallelism is
aa^ba”* Ca
This shows that the two lines are parallei if their direction
ratios are proportional.
Vector approach for § 12. (i) To find the angle between two
lines whose d c.*s are /i, nii,
and /a, Wa, na.
We have a=the unit vector along the line whose d.c.’s are
/j[,/«!, «i=/ii-|"tWij+Wik,
and
b=theunit vector along the line whose
/2, twa, na=/2i+7M2j+«2k.
d.c.’s are
If 0 be the angle between the two lines, then
a»b=l a I 1 b 1 cos 0, by definition of dot product of two vectors
=> (/ii4-mij+nik)«(/2i+W2j+nak)
=(l) (l)cos0
=> cos 0=/i/a+^iW2+nina*
(V |a|=lHbl]
Also by definition of cross product of two vectors, we have
a xb = | a I I b I sin 0 N, where N is a unit vector perpendi
cular to both a and b
r- sin 0 N, because | a |=il=il b |.
iXb)®=sin>0.
[V N»=N.N=1]
28
Analytical Geometry 3-2)
But axb=i i
/1
3
m,
i k
m2
»2
(WiWa-W2«i) i+(«i/2
j+C/iWa-Zam,) k.
sin2 0=faxb)3=--(axb)*(axb)
=(m,Wa-Wafi,)»+(«,/a-w2/i)2+(/ima- /am,)*.
Condition for perpendicularity. The two lines are perpendi
cular if and only if a«b=0
/.e., iff (/ji+mij+n,k)«(/ai-|-m2j*l"W2k)=0
/.e., iff/,/2+mim2-fn,ff2*=0.
Condition for parallelism. The two lines are parallel if and
only if the vectors a and b are collinear
r.e., iff a=Ab, where A is some scalar /.e., real number
/e., iff/ii-hm,j-|-«,k=A (/2i+m2j-i-/iak)
i.e., iff /i=A/2, mi«=»Ama, «i=Ana
,●
/I,
re., 1117=—c=a—●
/a m2 «2
(ii) To find the angle between two lines whose dr*s are a^ 6,,
Cl and aa,
c^.
We have
A=a vector along the line whose d.r.’s are a,, o,, c,
=-«7ii + />jj+rik,
B=a vector along the line whose d.r.’s are a^, 62, Ca
=02i+/>2j+cak.
Now proceed as in case (i). Here | A ]=
§ 3.
and
To fi nd the perpendicular distance of a point P (x'» y', z*')
from a line through A (a, b, c) and whose direction cosines
are 1, m, n
Let/4B be a line through A (a, b^ c) and whose d.c.’s are
/, m, «. Let FNbe the perpendicular from P to AB.
Now
projection of the
line segment joining A (a, />, c)
and P {x\ ; z) on the line AB
-a)/-}-(/-/>) m
4{2'-c)w,
and
distance between the
points A and P
L
CCr,6,C)
/>/■
29
Direction Cosines and Projections
or
or
We have. PN^=-AP^-AN^
PA^*={(x'-a)2+(y-{(x'-fl) /+(/-h) m+(z-c)n}*
PA^a-{(*'~a)>+(y-^0‘+C2'-c)*){/* + m*^n*} .
.-{(x'-fl) /+(/-h) m-t-(z'-c) n}*
=27{(y-6) n-(z'-c) m}*
[by Lagrange’s identity]
/. PN^ y/{D {{y-b) n-(z'-c) m}%
AUter. Let Z.PAN=d.
PJV*=y4P* sin* 0.
We have,
Now 6 is the angle between the lines AP and AB. Here
thed.c.*s of^Pare {x'-a)jAP,{f-b)}AP,(z’-c.)/^P
n
and
m.
I,
the d.c.’s of i4P.are
a
(/^b)/AP, (z’-^cUAP
sin* 0=
n
m
a
+
(x'-a)/AP {z*-c)IAP * \{x’--a)IAP
{y'-bjiAP
i
+
ra
/.
1
AP*
I
n
i
y-b
m
z’-c
n
a
+ x'—a
/
z
n
m
x'—a y b *1
^ *4m
^ /
\y'—b z'—c *
x'—a z'—c *
4
PJV»=AP> sin* 0=1 .
/
n
a
■ !
m
y-a y-h *
(1)
4/
ai
Remark. in the formula (1), /, m, a are the d.c.’s of the line
AB. If however, ]8, y are the d r.’s of the line AP, then to get
PiV* we should divide the R.H.S. of(1) by a*4-^*4y*.
SOLVED EXAMPLES 2(C)
Ex. 1. If points P, Q are (2, 3. -6),(3. -4, 5), then find the
angle between OP and OQ where O is origin.
Sol. The direction ratios of OP are 2—0,3—0,-6-0 i.e.,
are 2, 3, —6.
.
. Also OP-V((2-0)*4-(3-0)*4-(-6-0)*1=v'(49)=7.
/. the d.c.’s of OP are 2/7, 3/7, -6/7.
The direction ratios of OQ are 3—0,—4—0,5—0,/.e., 3,
-4, 5.
Also Ofi=y{(3-0)*4-(-4-0)*+(5-0)*}=V(50)=5-^2..-. the d.c.’s of OQ are 3/(5 v/2), ^4/(5v'2), 5/(5y2).
30
Analytical Geometry 3*2)
If d is the angle between OP and OQ, then
[See eqn. (5), § 12]
cos 0=/i/2+Wim2+ni«a*
-36
cos 6=
35x/2
or
cos 9=
■.
(=^j.
Ans.
35
Ex. 2 (a). If points P, Q are (2, 3, 4), (1. -2, 1). then prove
that OP is perpendicular to OQ where O is (0, 0, 0). [Magadh 68]
Sol. The direction ratios of OP are 2—0, 3—0, 4—0, i.e.
2, 3,4.
The direction ratios of OQ are 1-0, -2^-0, 1^ 0 /.c. 1,
-2.1.
If OP is perpen.dicular to OQ, then we must have
^1^2 "I"
0.
[See eqn. (13), § 12]
Now.fl,aa+M2+^?xr2=(2).(l)+(3).(-2)+(4).(l)
=2-6+4=0,
which shows that OP is perpendicular to OQ.
Ex. 2 (b). Show that the line joining the points (0, 1,2) and
(3,4,6) is parallel to the line joining the points (—4, 3, —6)
(5, 12,6).
Sol. The direction ratios of the line joining (0, 1, 2) and
(3, 4, 6) are 3-0, 4-1, 6-2 i.e., 3, 3, 4.
The direction ratios of the line joining (—4, 3, —6) and
(5, 12. 6) are 5-(-4), 12-3, 6-(-6) i.e , 9, 9,12.
We see that the direction ratios of the two lines are propor*
tional because we have
3/9=3/9=4/12.
Hence the two lines are parallel.
Ex. 3. If the vertices ?, Q and R of a triangle have coordinates
(2, 3, 5), (—1, 3, 2) and (3, 5, —2) respectively, find the angles of
the triangle PQR.
Sol. The direction ratios of PQ are — 1 —2, 3—3, 2~5 ie ,
-3,0,-3.
We have V'{(-3)H0*+(-3)2}=<v/(i8)=3V2.
3 0
-3 .
, I.e., the d c.’s of PQ are
3\/2* 3^2’ 3V2
Similarly d.c.’s of QR are
4
2
4
V(36)* V(36)’ V(36)
2 1 -2
i.e. ;
3»3* 3
and d.c.’s of P/? are
1
2
-7
3a/6’ 3V6’ 3^/6
31
Direction Cosines and Projections
Now using the formiila [eqn. 5, § 12] cos 0=/i/a+mima+ni»a>
we shall find the angles P, Q, R of APQR as follows':
We have, cos P=cos of angle between PQ and PR
6
6
=-f2-3V6+‘’-3^6+(-:;^)(3^) 3V(12) 6V3
“l/\/3.
Z.P=cos-'(l/\/3).
Again cos ^^^cos of angle between QP and QR
(-\y.
1
1
Note that d.c.’s of QP are^,0,
Finally cos R=cos of angle between RP and RQ
18
2
9V6 - V
P=cos-V(2/3).
Ex.4, Prove that the straight lines whose direction cosines are
given by the relations fl/+6m-Hcn-0 and fmn-\-gnl \-hlm^0 are
perpendicular if//a+gjb -{-hfc=0
and parallel if ^(af)±y/ibg)±y/{ch)=0.
(Meerut 1984. 85 P,87, 89)
Sol. From the first relation, we have »= —{ali-bm)lc.
Putting this value of n in the second relation, we have
or
afmi-\-bfm^-\-agP-\-bglm-chlm~0
or
og ^2+-^ {af-\-bg-ch)-{-bf=0.
...(1)
Now if /i, mi, and /a, ma. na are the direction cosines of the
two lines, then the roots of(1) are /i/mi and /a/ma.
/i .—=-^
h bf or hh^mimj
product of the roots
mi ma ag
fia gib
.
hU _mima_/iina
by symmetry.
if/ar(glb)°{hlcy
We know that the lines are perpendicular if
/i/a+mjma+«ina=0
●
i.e, if
flo+g!b+hlc=0t
which proves the first part.
If the lines are parallel then the direction cosines are the
same. This shows that the roots of(1) are equal, for which the
condition is *B^=4AC*
i.e.
(of-\-bg-chy f=»4ag,bf.
32
or
or
Analytical Geometry 3-D
Taking square root, af\-bg -ch— ±2y/{afbg)
af±2V(afbg)-\-bg=-ch
{VW)±\/ibg)Y=(ch).
Taking square root, V(of)±Vih)= ±V(ch)
or
V(af)±V(bg)±V(chh-0.
Proved.
which proves the second result.
Ex. 5. Show that the lines whose direction cosines are given by
the equations 2/+2/w-««=0, and mn+nl+lm—0 are at right angles.
(Meerut 1986 S)
..(1)
Sol. From 2/+2m—/i=0, we have «==2/+2m.
Putting this value of n in m«+n/+/m—0, we get
m (21-1-2m)+(21+2m)l+lm^O
or
2/?+5/m+2w8=0 or (/+2m)(2/+m)=0.
. /=—2m and 2/-|-m—0.
When /=—2m,from (1) n=—2m;
.
/_ m
_1
2 -I .n-:v{(2/+(-i>H(2n 3‘
The d.c.*8 of one line are 2/3, —1/3, 2/3.
Again when 2/= —m,from (1), n=m.
.
V(l^+m‘+n^)
_1
●● -1 2 2 V{(-1)*+(2)M(2)»}''3
.% the d.c.’s of other line are - 1/3, 2/3, 2/3.
The lines will be at right angles if Ijli+mim^-i ni«2=0.
We have
Hence the lines are at right angles.
Ex. 6. Show that the straight lines whose direction cosines are
given by the relations at+bm+cn^O and ut^+vm^+wn^=0 are
perpendicular or parallel according as
a* (v+w)+b^(u+w)+c^ (i/4-v)=0
or
a^lu+b^jv+d^fw^l).
(Meerut 1984, 86, 89S;
Sol. From the first relation, we have n=—(al+bm)lc.
Putting this value of n in the second relation, we have’
ul*+vm*+w
or
or
(c^u+ahv) l*+2abwlm+(b^w+c^v) m^~- 0
(c^M+a^iv)(//m)2 +2abw (lim)+(b^w +ch>)^0.
...tl)
Let /i, niu n\ and /^, mg, na be the d.c.’s oi iLc two lines.
33
Directions Cosines and Projections
Then the roots of(1) are /j/iwi and /a/m*.
. . .. .
\ h i» b*w4cH
●
/j/g
f*hmz
W|«a
by symmetry.
...(2)
’* b*w+c*u'~c^u+a^w'~a*v-\rb*u*.
The lines will be perpendicular if /i/a+mima+nina=0
i.e.
(6*w+c*v)+(c*ii+o*w)+(a*v+ft*tt)®*0
or
a*(v+w)+h*(«+w)+c*(v+«)=0.
Proved I.
The lines will be parallel if the d.c.*s of the lines are the same
/.e. if the roots of.(l) are equal, for which the condition is
/.e.
or
or
4a*6*w*=4 (c*ii+o*w)(h»w+c*v)
0*C*WV+b*C*UW+
=0,
Proved II.
flVw+^Vv+^Vw-0.
Ex. 7. Show that the lines whose d.cjs are given by /+m+n=0
(Meeratl983)
ani2mif+3/n—5/m»0 are at right angles.
/=a
...(1)
Sol. From the first relation, we have
Putting this value of / in the second relation, we have
2m«+3(—m—li)»—5(—m—«)m«=0
or
5m*+2mn-3rt*=0 or 5(m/«)*+2(m/n)-3-0.
...(2)
Let A, mi, rti and /*, m*, «a be the d.c.*s of the two lines.
Then the roots of(2) are milni and ma/na.
mi m*
3
mim* Hi«a
—.
product of the roots — ,—1=> —— or
3 ^-5
Hi »a
^
...(3)
Again from (I), n=>—/-m and putting this value of n in the
second given relation, we have
2m (-/-m)+3/(-/-m)-5/m-0,
or
3(//m)*+10(//m)+2=0.
/*/* mim*
. /r
./*_ 2
...(4)
ntx nt^ 3 ' 2
3
From (3) and (4), we have
— fc (say).
/i/a+mima+nina=*(2+3—5) A:=0.fc«=>0.
Proved.
The lines,are at right angles.
Ex.Is. Lines OA, OB are drawnfrom O with direction cosines
proportional to 1, —2, 1; 3, —2, 3. Find the direction cosines of
the normal to the piane AOB,
Sol. Since Oil and 0.8 lie in the plane AOB» therefore
these lines are perpendicular to any normal to the plane AOB.
34
Analytical Geometry 3-D
Let Ou 02, 08 be the direction ratios of such a.normal. Using the
con'ditioD of perpendicularity 01^1+0262+03^8=0,
we have
..(1)
01 (i)+08(“2)+fl3(—1)=0
and
..(2)
0i(3)+0a(-2)+03(3)=O.
Solving (1) and (2),
01
02
2.3-(-l)(-2)“-1.3-1.3
03
l.(-2)-(-2)3
01
or
02
08
0|
02
03
4"3 “ -2
Therefore the d.c.’s of the normal are
-2
4
3
\/.{(4)>+(3)>+(-2ja}» V{(4)*+(3r+(-2)2}’ V{(4)* +(3)*+(-2)»}
4
3
or
V(29/V(29)'V(2V)*
Ex. 9. Jf li, mu ni and
rt3 are direction cosines of the
two lines show that the direction cosines of the line perpendicular to
both are proportional /o W|«2—
01/2—02/1- l\m%—hmx.
Prove further if the given lint s are at right angles to each other
then these direction ratios are the actual direction cosines.
Sol. Suppose that the required direction cosines of the line
are /, m, n. Since the line is perpendicular to the given lines, we
have.
and //.2+
//i+wjmi+««!=() ...(I)
Solving (1)and (2), we have
/
m
.mi«2—
. (2)
+002=0.
n
■
Wa/, “/,/»2-72Wi
This shows that the required d r.*s are
.. (3)
—'0s0i. tij-y n«/,. .
/1/U3 —/aWi.
Now suppose 6 is the angle between the two given lines
'se d.c.’s are /i, /;ii, Wj and />, m., /ig.
Then sin 6-\/{2!(/UjUa—uFaU,)*} [see cor. 1, § 12J
...(4)
If f)=90° /.p.,.thc lines are perpendicular, then (4) gives ●
\/{£(m,«2—w-jw,)*}=!.
in this case from (3) the d.c.’s /, m, n of the line are given
by
I
Wi/i2“/Wa0i
m
f''. —
n
/, ■ /,u/2 - l.,mi
Vi2.\Win.,—rfun,)“'}
1 [Using (5) and/■‘*+m®+0®=Ij
35
Direction Cosines and Projections
“1.
Hence in this case the actual direction cosines /, m, n are
Proved.
ntin^—m%n\y n\l^—n%li^ /iWa—'/awii.
Ex. 10. Show that the direction cosines of a line perpendicular
to a pair of mutually perpendicular lines with direction cosines as
lu Wi, »i and k, nts, n^ respectively are
(Bbagalpor 1966)
m\n% — mzni^ n\l%^ndij l\rn%—lim\m
Sol. See example 9. This is other way of stating the same
problem.
Ex. 11. Prove that three concurrent lines with direction cosines
it, mi, nil /a, /Mg, Wa;(^nd /g, ma, «8 are coplanar if
h mx wi
I2
m%
Wg
h
ma
na
=0.
(Meerat 1983 S)
Sol. Let/, m, nbe the d.c.*s of the normal to the plane in
which the two concurrent lines with d.c.’s h, mi, ni and U, ma, na
lie. Then the line whose d.c.’s are /, m, n is perpendicular to the
lines whose d.c.’s are /i, Wi, «i and 1%, ma, na. Therefore
/i/+mim+ni«=0 ...(1)
and
lat-\-mam-]rnan=0. .. (2)
Again if the third concurrent line whose d.c.*s are Iz, ma, »z
also lies in this plane, then it is also perpendicular to the normal
to this plane.
...(3)
Ial+mam+nan*=*0,
Eliminating l,m,n between (1), (2) and (3), we get the
h mi ni
required condition as
la
ma
na
la
nta .na
=0.
Ex. 12. Prove that the three lines drawn from a point with
direction cosines proportional to 1,-1, 1; 2, —3,0, and I, 0,3.are
coplcmar.
Sol. Let a, b, c be the direction ratios of the normal to the
plane in which the two concurrent lines with direction ratios
1,-1,1 and 2,-3,0 lie.
Clearly these lines will be perpendicular to this normal. Hence
applying the condition for perpendicularity of two lines, we have
...(1)
l.a+(-l).b+l.c=0.
'h
36
and
Analytical Geometry 3-/)
2.fl+(-3).6+0.c«=0.
Solving these,
...(2)
_ or 14=^.
Again the third concurrent line with d.r.’s 1, 0, 3 will lie in
this plane if the normal with d.r.*s 3, 2, —1 is also perpendicular
to this third line.
We have(3)(I)+(2)(0)4-(—l)(3)=0, showing that the lines
with d.r.*s3, 2, —I and 1,0,3 are perpendicular. Hence the
three given lines are coplanar.
Ex. 13. The dx*s of two Intersecting lines are /i, W|, ni^and
Show that all lines through the intersection of these two
whose dc's are proportional to A+Zc/a,
«i+A:«a are cop¬
lanar with them.
Sol. Let /, m, n be the d.c.’s of the normal to the plane in
which two intersecting, lines whose d c.’s are A, mi, nj and
hi nJit »a lie. Clearly these lines will be perpendicul if to this nor
mal. Therefore
/i/-fmim-[-nin=0,
...(1)
and
kl+mini+nin—6.
...(2)
Now any line through the point of intersection of these lines
with d.c.’s proportional to A+Wa, mi-\-km»y ni+kn^ will lie in
this plane if the normal to this plane is also perpendicular to this
line.
We have /(/i-|-A:/a)-l-m(mi-f^m2)-hrt(n|-|-A;na)
=»=//i-f mmi-f««!+fc(//2+mma+nwa)
=0+fc.0, using (1) and (2)
0.
This shows that the line whose d.c.’s are /, m, n is perpendfcular to a line whosed.r.’s are /i+A:/a, Wi+Arm2, nx-\-knz. .
Hence all lines through the intersection of the given lines and
with d.c.’s proportional to A+fcA, mi-fA:ma, ni-f-/rrfa are coplanar
With' them.
Ex. 14. Ifa variable line in two adjacent positions has direction
cosines /, m,n and /4-S/, m-f8m, n-\-8n, show that the small angle
8b between the two positions is given by (8(9)*=a(8/)®4-(8m)*-i-(8B)*.
(Meerut 1984 P,87 P,88)
Sol. Since /, m, n and (/+S/>,(m+Sm), («-|-8n) are the
actual direction cosines, we have /‘+m*+R^=>l
...(I)
and (/+8/)*-Km-i-8m)»-Kn-|-8M)*=l
or
(/*+m*-hn*)+2/8/-l-2m8m-|-2/f8fi-|-(8/)»-i-(8m)*-i-(8b)*= 1
37
Direction Cosines and Projections
or
or
[Using (1)1
H-2(/8/+/m8w + «8/i)+(8/)2+(8w)*+(8/i)®= I
...(2)
2(/S/+mhm+n8«)=-{(8/f+.(8m)>+(8«)*}.
Now it is given that Sd is the angle between two adjacent
positions ot the line. Therefore
...(3)
cos S9—i.{i-^8l)-^m.(m-\-Sm)-{-n.(n-\-8n),
Now cos8dal
/.
.2!
4
if 80 is small, we have cos 86= 1
W
2!
Then from (3), we have
1—
or
1
«=(P++M®)+(/8/+ m8m+n8n)
(80)2
2
1-i {(8/)H(MH(8/i*)}
(using (1) a- J (2)]
or
Proved.
(80)2=(8/)2+(8m)2+(8n)».
Ex. 15. Prove that the acute angie between the lines whose
direction cosines are given by the relations
/-l-w4«=0 and P’\-ni*—n^=Q is tc/3. (Meerut 1986 P)
Sol. The relations giving the d.c.’s of the two lines are
/+m4n=0,
...(1) and P+w*—n*=0.
...(2)
From (i), /i=-(/+m). Putting this value ofn in (2), we get
/*+/m2“(/+/m)*=0, or 2/m=0, or /m=0.
i=0
or
m=0.
When /=»0, wc have from (1), m=—n.
n
m
0“ 1 “-1
-
.(3)
Again when m=0, we have from (1), l=—n,
I m
n
...(4)
. From (3) and (4), we observe that the d.r.’s of the two lines
are 0, 1, — 1; and 1, 0, —1.
,
their d.c.’s are 0, 1/V2,
and 1/V2» 0. -l/\/2.
If 0 is the acute angle between the two lines, we have
1
cos 0
«(-*)■ (-7.) 2
A
e=nl3.
Ex. 16. Show that Jhe area of .a triangle whose vertices are
the origin and the points (x,, > i, Zj) and (Xi, y%, z^) is
3^
Analytical Geometry 3-2)
Sol.
The direction ratios of OA are Xi, yu Zi and those of
OS are Af», ja.
and
Also O^= v'{(Jfi-0)»+()^i-0)»+(zi-0)*}=VW+>’i*+Zi“)
OS=V{(^a-0)a+(;^a-0)H(2a-0)»)= V(X2>*+V+^2*).
.*. the d c.*s of OA are
yi
gi
and the d.c.*s of OB are
x»
y%
Zi
V(Xi*-i-yi^-\-Zi*) * VM+yi^-^Zi^) * VU?+jvT^'
Hence if $ is the angle between the lines OA and OS, then
\/{S(j>|Z8-J?2Zl)*}
V{S(yiZ2-yiZi)*}
sin 0=
OA.OB
Vl^»*+>l®+V)V(Jf2*+)'2”+Z8*)
Hence the area of A0/4S=.i,Oi4.OS sin
[V
Z_AOB=^6]
UA.OB
Proved.
^W{S {yiZi-yiZx)^}.
Ex. 17. If lit iMi, «! and /a, /Wa, na ore the d.c.*s of two concurrent lines, show that the d.c:s of two lines bisecting the angles
between them are proportional to /i±/a, w,±ma. ni±/ia.
Sol. Let O be the origin.
Draw 0^ and OS parallel to
the given concurrent lines. Let
A. mu fti be the d.c/s of OA
and lu ma, na the d.c.*s of.OS.
Cut off OA=^OB=r. Then the
co-ordinates ■ of A and S are
(Ar, wiir,n,r) and (Ar, mar, war)
respectively.
Again take a point C on SO produced such that OS=OC- r.
Thus the co-ordinates of C are(—Ar, —mar, — »ar).
Suppose M and N are the middle points of yfS and CA. Then
the co-ordinates of Mand Nate
jl\r+/ar m,r4 mgr -nir+n^r\
2
and
(Ar—Zar
mir- War» “2~|
Wir—War \
2 ’"I
respectively. Clearly OAf and O/V are the internal and external
bisectors of the angle AOB.
Direction Cosines and Projections
39
Hence the direction ratios of OM and ON are
i (/i+/a)^ i(mi+ma) r, J (ni+«») r
and
i (/i -/») r. 1
r, ^ (n,-«a) r respectively,
/.e ,the d.c.’s of 0/lif and OJV are proportional to U+lt, mi+m%.
Proved.
and /1-/2, Wi-ma, ni—«a respectively.
Ex. 18. Find the direction cosines of the ilnes bisecting the
angles between the lines whose direction cosines are /i, mi, ni and
i»x Wa* «2 and the angle between these lines is 0.
Sol. Proceeding as in Ex. 17 above the direction ratios of
the internal bisector OM of the angle AOB are
...(1)
l\Arh* mi-4*m2, Wi+Wa.
Since 0 is the angle between the lines whose d c.’s are
/is Wi» and /a, ma, «2*
cos 0=/i/2+mima+WiWa“^ hh*
Also 2/i*=/i*+mi*+«i*=« 1 and Z /a*=/a* +«a*+«a*=* 1.
We have
/i*+^/a‘H-22? /i/a}.
=-v/(l+H-2 cos 0)=»V{2(l+cos 5)}=2 cos
Dividing the direction ratios (1) by
V{(/i+/8)H(mi+m2)H(ni+«a)’}
i.e. by 2 cos 0/2, the d.c.’s of the internal bisector OM are
mi+ma «i+”a
/i+/a
Ans.
2 cos 0/2 * 2 cos 0/2’2 cos 0/2
Similarly the d.c.*s of the external bisector ON are
mi—ma Wi-Wa
l\.~h
Ans.
2iiT^2’2 sin 0/2"* 2 sin 0/2
Note that in this case, we have
V((/j“^a)®+(w»i-W2)*+(«i~na)*}=>\/{/?/i*+^/a*—
/lia}
= V(i -f I“2 cos 0)=2 sin ^0.
Ex. 19. The vertices. of a triangle PQR are the points
(—1, 2, —3),(5, 0, -6)and(0, 4, —1) in order. Find the direction
ratios of the bisectors'of the angle. QPR.
Sol. Suppose the internal bisector of the LQPR meets the
side QR in L. Then we know that
...(1)
QL:LR=PQ : PR.
Now />g- V{(5+l)H(0-2)*+(-6+3)>}=V(49)=7,
and
P/j=V{(0+l)«+(4-2)H(-l+3)»}= v'(9)=»3.
Putting the values of PQ and PR in (1), we have
QL:LR=>1:3.
40
Analytical Geometry 3-Z)
This shows that L divides QR internally in the ratio 7:3 and
hence the co*ordinates of L are
/.e..
7.0+3.5 7.4+3.0 7.f-l)+3.(-6)
7+3
( 7+3' 7+3 »
)
/3 H
\2 *5 *
the direction ratios of the internal bisector PL are
14
● 5 4 I
'5 -2.
l.e., are
^
are 25, 8, 5.
4)-
Again the external bisector of the £QPR will meet the side
QR in Af, where M divides QR externally in the ratio 7:3 and
hence the co*ordinates of M are
7.0-3.5 7.4 3.0 7.f - I)-3.(-6)
7-i
)
( 7-3 ♦ 1-3 »
i.e●$
(44)-
the direction ratios of the external bisector PM are
11 - 23
(-^+!,7-2,n+3 ji.e.. are- 4 ♦ 4
f.e., are
-11.20, 23.
Ex. 20. If the edges of a rectangular parallelopiped be a, b, c
show that the angles between the four diagonals are given by
cos -1
i±a*±b^±cn
l(a*+ba + c»j/
(Kanpnr 1982, Meerut 84 S)
Sol. Let O, one of the vertices of the rectangular parallelbptped, be taken as origin and the three coterminous edges
OB and OC as the co-ordinate axes.
f
/n
4
\
\
>c-
●z
a
..."
\
2^ 4:
41
Direction Cosines and Projections
Let the edges OA=a, OB=b, OC»c. The four-diagonals are
OF, AG, BE
DC.
The co-ordinates of the vertices of the parallelepiped are
o (0, 0, 0) ;^ {a,0,0); 5(0, h, 0); C(0. 0. c);
given by
F(a, b, c); D {a. b,0)\E(a,0, c); G (0, b, c).
Note that the vertex D lies in the xy-plane* the vertex E in
the 2x-plane and the vertex G in the j>z-plane.
The d.r.’s of the diagonal OF are a—0,b—0, c—0 i.e..
fl,b, c.
the d.c.’s of the diagonal OF are
c
a
b
Similarly the d.c.’s of the diagonal AG are
b
—a
c
the d.c.’s of the diagonal BE are
a
^b
c
and the d.c.’s of the diagonal CD are
b
a
c
the angle 0 between the diagonals OF and /40 is given by
ax(—a)-l-^xF-l-cXc
cos 6
i.e.,
[using cos 0=/i/24'^i^2"i"nina]
-a--\-b‘^+c^
a^+b^-\-c‘‘
-{
0=»cos
[
The total number of pairs of the diagonals are *Cz i.e., 6. In
a similar way the angles between the rest five pairs of the diago
nals are determined and all of these six angles are given by
cos
\
J
The above expression will give only six valid values because
the ambiguous signs cannot be either all -f ive or —ive for in that
case 0^cos-i 1 or cos"*(-1) ie., 0^0 or 180® which is impossi
ble as no two of lhe diagonals are parallel.
Ex. 21. A line makes angles ol, B, y, S with the four diagonals
of d cube. Prove that cos* ct-^-cos- p+cos^ y-fcoi* 5=4/3.
(.Gorakhpur 1982; Meerut 81, 83, 85; Lucknow 81; Kanpur 83)
42
Analytical Geometry 3-D
Sol. lo a cube all the edges are equal. Proceediug as above
in Ex. 20 the d.c.*s of the diagonal OF(putting b=c=>a) are
a
a
a
i.e..
1/V3, l/v'a, 1/V3.
Similarly the d.c.’s of the diagonal AG are
-1/V3, 1/V3, 1/V3.
the d.c.’s of the diagonal BE art l/\/3, — 1/V3, l/\/3,
and the d.c.’s of the diagonal CD are l/\/3, l/v^3, —1/V3.
Now let /, m, n be the d.c.’s of the line which makes angles
a. )3, y, S with the four diagonals of the cube. Then we have
or
or
cos a=/.(l/V'3)+/n.(l/'v/3)+n (l/V'3)=(/+m+n)/\/3
...(1)
cos* a=(/+m-|-n)*/3,
cosi3=/.(- I/v/3)-f-m.(l/v'3)-|-/i.(l/V3)=(-/+«i+n)/v'3,
cos* j8=(--/-|-m+n)*/3.
Similarly
and
cos* y=(/-w+«)*/3
cos* S==(/-f-m—n)*/3.
...(3)
..(4)
Adding the relations (1),(2),(3) and (4h we get
cos* a4-cos* jS+cos* y-j-cos* 8
"J {(/+m-l-«)*-f(-/+w-hn)*-f-(/-m+«)*-h(/+m-«)*}
[V
«*=!].
«=i {4 (/*+m^-f-n*)}=*t
Ex. 22. Find the angle between two diagonals of a cube.
Sol. Proceeding as in Ex. 21 above, the d.c.’s of the two
diagonals OF and AG of the cube are J/\/3, l/\/3» 1/V3 and
— Ify/Zt 1/V3, I/V3 respectively. Thus i( 0 is the acute angle
between the diagonals OF and AG, then we have
cos
(1/V3).(-l/V3;+(l/v'3).(l/v/3)-i-(I/V3).(l/V3)|=|
i.e„
d==co&-^ (i). .
This is the required angle.
Ex. 23. If Oi, mi, ni),(/2, n/*, n^), (/a, ms. «a) are the direction
cosines of three mutually perpendicular lines, then find the direction
cosines of a line whose direction cosines are proportional to
li^h+h, mi+ma+mz, Wi-j-na+Ws. and prove that this line is
equally iriclined to the given lines.
Sol. Since li, mi, «i ; k, m2, «a; k* m^, /I3 are the d.c.’s ’ of
three mutually perpendicular lines, we have
Also /i*
/i/s-f- mims+Wina=0, /a/a+ms»M3 UzUs*=0,
|
/a/i-hmami-i-nani^O. ● ...(1)
mi*-1- »!*=/a* ma*-t-/la*=/a*-f-m3*+/la*=1.3
Direction Cosines and Projections
4^
Now we have \/((/i+/2-|-/3)*+(mi+/W2+m3)®+(«i+«9+«a)*]
=
+(/a*+W2=+»2*)+(/3=*+W8*+«3*)
+2 {(/,/2+mim2+«in2)+(/2/8+»»2W8+W2»8)
+(/a/i+WaWi+ns«i)}]
[using the relations (1)]
^[1+1 4-1+2(0+0+0)],
-V3.
Thus the required d.c.*s of the line are given by
/i+/a+k mi+ma + nh ni+«2_+»8
...(2)
y/y '
V3
*
V3 " ●
Let d be the angle between the lines whose d.c.’s are lu
m
and those given by (2). Then we have
cos ^=/i.(/i+fa+4)/\/3+mi .(mi+ma+^7if»)'\/3
+/ii.(Wi+na+W8)/V3
=(1/V3)[(/i*+mi*+«!»)+(/1/2+m.ma+ihn^)
+(/s^+WaWi+Ma^i)]
[using the relations (1)].
=(1/V3)[1+0+0],
/. tf=cos-^ (1/V3).
Similarly the angle between each of the lines with d.c.’s
hf ma, na; is. ms, ns and the line with d.c.’s given by(2) is
cos“* (1/V3).
Ex. 24. If two pairs of opposite edges of a tetrahedron are
perpendiculart then prove that the third pair is also perpendicular.
Sol. Let OABC be a tetrahedron. Let O be chosen as origin.
Let the co-ordinates of the vertices
and C be (Xi,
2i),
(Xa, Pi, 2a) and (Xg, yz, 23) respectively.
The d.r.’s of the edges OA^ OB and OC are Xi,
2i;
Xa,
2a; and Xs, yz, Zz respectively. Also the d.r.’s of the edges
5C,CA and AB are X3—Xa,;^s—J'a. 23—.23 ; Xi—Xa,J'x—J'a, 21—23
and Xa-^Xi.^'a-;/!, 2a-2i respectively.
Now suppose the edge OA is perpendicular to the opposite
edge 5C. Then using the condition for the perpendicularity of
two lines, we have
...(1)
X,(xs-xal+yi {yz—y2)+zi (zs-22)=0.
Also if the edge OB is perpendicular to the opposite edge
CA^ then we have
...(2)
Xa (X,—Xs)+.r2 (Vi-.V8)+2a (2i-2s)=0.
Adding (1) and (2), we have
...(3)
Xs (Xi-Xa)+;-3 {yx-yz)-Vzz(2j-Za)=0.
The relation (3) shows that the third pair of opposite edges
OC and
is also perpendicular.
.●
44
Analytical Geometry 3-D
Ex, 25. If a pair of opposite edges of a tetrahedron be perpen
dicular, then show that the distances between the middle points of
the other two pairs of opposite edges are equal.
Sol. Let OABC be a tetrahedron. Proceeding as in Ex. 24
above, if the pair OA and BC of opposite edges be perpendicular,
then we have
X,(Xa-Xsl+J'i {yz-yi)-¥zt (Z8-Za)=0.
...(1)
Let Ml and
be the middle points of OB and CA, so that
their co-ordinates are given by Mi(^Xa.
i^a)and MafJfXi-l-Xa),
Now
MxA/a» =(i (xx-l-Xa-XalP+a O'l
U (Zi+Za-^a)}*
Similarly if Afs and Af* are the middle points of OC and AB,
we have
MzM^—\ {(Xi-}-Xa-T-X8)*-f-(.yiH-J^a—7a)“+(zi+^a—Zs)*)- ●●●(3)
Now we want to prove that MiMi^MzM^.
Subtracting (3) from (2), we get
MiMi-M^M^
*=*i tK-^i + Xa—^a)*—(■^i+'^a —^a)®} + {(J'l+^’a—J'a)*—(j'l+.l'a -j'a)*}
+{(^1 + Z3 - Za)* -(Zi -1- Za—Za)-}]
(2.V, (2x3 -2xa)+2>'i (2>’3-2>;a)+2zi (2za-2za)],
using the formula fl‘‘*—i>*c=(a-{-d) (a—
=Xi (X8~Xa)+>i (.ya-3'a)+^i (^a-Za)
e=0,
using (1).
or MiM^^MzMi.
Ex. 26. If in a tetrahedron OABC, OA^+BC^<=^OB^-\‘CA^
=tOC^-\-AB*, then show that its pairs of opposite edges are at right
angles.
Sol. OABC is a tetrahedron. Let O be chosen' as origin
and let the co-ordinates of the vertices A, B and C be (xi, yi, Zi),
iXi, ys, Zi) and (X3, y^, zs) respectively.
Now D^*+^C»={(xi-0)H(:i'i-0)2-l-(zi-0|2}
+ {(Af3-X8)M-0'3-3'a)*+(Z3-Z8)2}
=W-¥yi^-\-Zx^)+W+yi^+Zi^)
+ {Xi* i-yi*+Za») - 2 (XaXa +;'ay3+^2^8)
Xx»-}-D Xa^+i; ;c8*-2i;xxrX3,
...(1)
Proceeding similarly, we have
OB*-\-CA^c=2: Xi^+JSXi*+i: Xi^-2S x^Xi
...(2)
and
x,»-l-2: Xa>4-r Xi^-2Z XiXi.
...(3)
Direction Cosines and Projections
45
Now OA*+BC^=OB*-\-CA* gives
Z
X2H2? Xz^—2Z Xi-Xs-Z Xi^-\-Z Xi*+Z x^—2Z x^Xi
or
or
or
Z XtXi--Z x^X\~Q
(X2X9-^y^ya+ZiZ;i) ~(x3<i+j’8>'i+2:3Zi)=’0
^3(^2-.«i)+;'8 (;'a->'i)4-^8 (ra~^i)=-0.
This shows that the edge OC is perpendicular to the opposite
edge AB /.e., an opposite pair of edges of the tetrahedron is
perpendicular.
Similarly by taking OB^-\-CA*=^OC^+AB* and OAHBC*
«=OC*+-<45*, the other two opposite pairs of edges of the tetra
hedron will be at right angles.
EXERCISES
1. The direction cosines of two straight lines, inclined at an angle
6 are A, mi, rii and /g, mj, n^. Show that direction cosines of
the bisector of the angle between them are
/1+/2
mi-f-ma
ni4*«a
2 cos tf/2'2 cos <?/r 2 cos y/i'
(Meerut 19845, 85S)
2. Prove that the line joining the points (1, 2, 3) and (—1, —2,
3) is perpendicular to the line joining the points(—2, 1, 5)
(Meernt 1985 S)
and (3. 3, 2).
3
The Plane
Plane.
Definition. A plane is a surface such that all the points of a
straight line joining any two points on the surface lie on it. Or in
other words// we take any two points on the surface, the straight
line joining these two points wholly lies on the surface.
§ 2. (A) The equation of a plane. (Normal form).
Tofind the equation of a plane in terms ofp l.e., the length of
the perpendicularfrom the origin to the plane, and direction cosines
/, m, n of this perpendicular.
Let OX, OY, OZ be a set
of rectangular axes with 0 as
origin. Let p be the length
of the perpendicular ON
from the origin O to the given :
plane ABC. We shall take
p always positive. The d.c.*s
of the perpendicular ON are
/, m,n. The direction of
the perpendicular ON is from
origin towards the plane.
A line perpendicular to the
plane is called a normal to the
plane. Thus /, m. n are the
d.c.’s of the normal to the plane, the direction of the normal
being that from the origin to the plane. If n is
along the perpendicular ON, then
n=/i+»»j+nk
and hence
a unit vector
CW=pn=>/7(/i+mj+nk)
...(I)
47
The Plane
Let P with co-ordinates (x, y, z) be any point on the plane,
so that
OP>=‘Xi+y\+zk.
...(2)
We have, NP^OP-ON^ix-pl)i+iy-pm)IH^-pn)k.
So long as P lies on the plane, NP is always parallel to the
plane and consequently perpendicular to ON and so also perpen
dicular to D which is a unit vector in the direction of ON.
NP*n^0.
...(3)
Putting the values of NP and n in (3), we get
{(x—pl) i-\-(y-pm)j-Hz—pn) k}»0i+ml+nk)<=»0
or
(x—pl)l+iy-pm) m+(z—pn) w«0,
or
Ix-\‘my -{-nz—p
or
...(4)
lx+my-)rnz=p
[V
Jhe equation (4)is satisfied by the co-ordinates of every point
on the plane, but by no point off the plane. Hence this is the
equation of the plane and is known as the normalform of Xht
equation of the plane.
Remark. In the equation (4), p is positive and l*-\-m^+n**=l.
Only then p is the length of the perpendicular from origin to the
plane and /, m, n are the d.c.’s of the normal to the plane, the
direction of the normal being from origin towards the plane.
Cor. If the perpendicular ON makes angles a, jS, y with the
co-ordinate axes then clearly we have /=cos a, w=cos n=cos y,
and the equation (4) becomes
...(5)
x cos cc+y cos /3+z cos y«=»p.
(B) The general equation of a plane.
The equation (4) may be written as
...(6)
(xi-l-:Fj+zk).(/i-fmJ+»k)«p,
where /, m, n are the d c.*s of the perpendicular ON.
Now suppose o, 6, c are the d.r.*s of ON, so that /, m, n in
terms of o, b, c are given by
(/, m, n)=(a, 6, c)/-v/(fl»+6»+c*).
Putting these values of /, m,n in the equation (6), we get
(xi+yj+zk).(fli-l-^»j+ck)«pV(flH**+c>)
or
...(7)
(xi -l-pj+zk)●(fli-f -1-ck)«=* q.
48
Analytical Geometry 3-D
where ^=/;v'(fl*+6®+c®) and ai+6J+ck is a vector perpendicular
to the planed Hence the equation of the plane can be written in the
formfl).
Taking qr=^-d, the equation (7) of the plane may be written
as
...(8)
ax-\-by-\-cz^-d=0.
The equation (8) is the general equation of a plane, where
the numbers a,.b, c arc the d.r.’s of the normal to the plane t.e., a
line perpendicular to the plane and the length of the perpendicular
from origin to the plane is —dfy/{a'^+t^-^c^)t the number ^ being
negative.
§ 3. To prove that the general equation of the first degree in
Xi z nam‘^ly ax by+cz+d=^0 represents a plane and that the
coefficients a, b, c of x, y, z in this equation are d.r*s of the normal
to this plane.
The general equation of the first degree in x, y* z is given by
...(1)
ax+by-{-cz^-d=0.
Let i4(xi, yi, z,) and JB (Xa,.Ma. Z2) be any two points on the
surface (!}, so that we have
..(2)
ax,+hyi+ezi.+£/=0
. (3)
and
flXa+&ya+cza+^=0.
Multiplying the relation (3) by V and adding to (2), we get
a (Xi+/iXa)+b (yi+/tya>+c (Zi+/tZal +d{l+p)~0.
Dividing both the sides by (1 +m)» we get
a (xt+iiXi). b (yi±PZ^ , c(zi-\-fiZj)+rf-0.
l+p
l-\-n
1+/*
...(4)
The relation (4) shows that for every value of/x96 - 1, the
„„ ,he surface (1). But
point ^
these are the general co-ordinates of a point which divides the
join of i4(xi, yu Zi) and BfXa, yz, za)in the ratio p: 1. Since p may
take any real value other than — I, every point of the straight line
AB lies on the surface (t). Hence the equation (1) represents a
plane.
Subtracting (2) from (3), we have,
...(5)
a (Xa-Xi)-h^> (.F8-.)'i)+c (za-z,)=0.
The relation (5) shows that the two lines whose d.r.’s are
fl,b, c and Xa—Xi, yz—yu
are perpendicular. But X2—xi,
49
The Plane
z^-zi are d.r.’s of the line AB which is any line In the
plane (1). Therefore a line whose d.r.’s are <i, b,c is perpendicular
to every line lying in the plane (1) and so it is perpendicular to
the plane (1). Hence fl, fr,c are d.r.*s of the normal to the
plane (1).
Remark. In the general co-ordinates of a point on'the line
we cannot have
—1 because there can be no point on the
line AB which divides it in the ratio —1: 1 i.e., which divides it
externally in the ratio 1 : 1.
Note^ The number of arbitrary constants in the general equa
tion of the plane.
The general equation of the plane is
6t {ald)xf(bld) y+(cl.d) 2——
This equation shows that there are three arbitrary constants
namely a/d^ bid, cjd in the equation of a plane. Therefore the
equation of a plane can be determined to satisfy the three condi
tions, each condition giving us the value of a constant.
§4. To reduce the general-equation of the planie to the normal
form.
The general equation of the plane is
dX’\-by-VcZ’¥d=^a.
If /, m, n are the d.d.’s of the normal to the plane, then the
equation of the plane in the normal form is
●●(2)
' lx-\-my ^' nz=p.
If(1) and (2) represent the same plane, then
1
/ m
n
p
.
w*-!-«*)■ .
'a~T"‘ c
*
where the same sign either -five or —ive is to be chosen throughout.
.-.
/=±a/V(o*+b*+c**), m=±b/v/(fl*+b*+c*).
n = ±f/V(a®+b“-f c**). and p=±rf/v^(a3=fbHc*).
Substituting the'se values in (2), the normal form of the plane
(1) is given by
ax
.
-
.
4.
cs
d
=*= Vio*+ «● + «●)■
The sign in equation (3) is so chosen that pie.,
±dfy/(a^ + bH c'-*) is always positive.
-(3)
50
Analytical Geometry 3-Z)
Working role fo.rednce the equation of a plane in normal fornii.
Transpose the constant term in the equation of the plane to
the R.H.S. and adjust the equation in such a way that this const
ant-term on the R.H.S. is positive. Now divide the equation by
where a, b, c are the coefficients of x^y^ z\n the
equation of the plane. The resulting equation will be the equation
of the plane in the normal form
Ix+my+nztstp,
Here p will be the length of the perpendicular from the origin
to the t>lane and I, m, n will be the d.c;*s of the normal to the
plane.
§5. .Intercepts form:
Tofind the equation to the plah6 in terms of the intercepts
u, bt c which the plane cuts on the coordinates axes. (Meerut 1985)
Let the general equation,of the plane be
Ax+By+Cz-{-D—0.
...(I)
Since the plane (1) makes an intercept a on the x-axis, . the
point (a, 0,0) lies on the plane (I), so that we have
Aa+D^O i e.f A=3 — Pla.
Similarly the points (0,;^, 0) and (0, 0, c) lie on the plane (1),
so that
or
JBb-\‘D>==0 and Cc-^^0, giving B=—Dlbt C= -Djc.
Putting the values of.
Cin (1), we get
i-‘Dla)x+(^Dtlky+i-D/c)z-^D=0
xla+y/b+zfc=\.
The equation (2) is (he required equation of the plane in
terms of the intercepts a, b^und c made by the plane *on the axes
of X, y and z respectively.
§ 6. Plane through a given point and perpendicular to a given line.
Tofind the equation ofa plane through a given point A(Xit yu Z\)
and perpendicular to a line whose direction ratios are a, b, c.
Let (x, y, z) be the coordinates of any current point P on the
plane. Since the plane passes through the point A (xi, >i, Zi), the
line AP lies in the plane.
The d.r.’s of the line AP uic x—Xu y-^yu z—Zi. Also the
d.r.*s of the normal to the plane i.e,, of a line perpendicular to
the plane are a, b, c.
Now the normal to the plane is perpendicular to every line
lying in the plane and therefore the lines [whose d.r.*s are a, b^ c
and x—Xi,y—yu
are perpendicular.'
51-
The Plane
a (x—Xi)-\-b iy-yi)+c {z-zi)=0t
which is the equation of the required plane.
Remark. The equation of any plane passing through the
point (xi,
zi) is
a(x-x,)+h (y-yi)+c(z-zt)=0.
In this equation a, h, c are d.r.*s of normal to the plane.
As a particular case» the equation of any plane passing
through the origin is
ax+h>'+cz=(V
in which the coefficients of x, y,z i.e., a, h c are
of the nprmai tqihe plane.
§ 7. Equation of a plane through three points.
Tofind the equation of a plane which passes through three points
. whose co-ordinates are (Xi, yu Zi), (Xg,}'2» Za) and (Xa, J'a, Za).
(Kanpur 1983)
Let the general equation of the plane be
ax-lrby+cz-^d=Or
CD
If the equation (1) of the plane passes through the given
points (Xi, yu Zi),(xa, y^ Za) and (xa, ys* Za), the coordinates of
these points will satisfy the equation (1), so that we have
● ●●(2)
V,
-T
x 4^
</=0,
ox^-\-byz-\-cz3+d=(T, ' —
Eliminating a, h, c and d from the above equatioiT~
(3) and (4), the equation of the required plane is given by
X
1
y
z
and
Zi
1
Za
1
...(3)
...(4)
...(5)
Xi
y2
1
Xi
yi
Za
Cor.Condition forfour points (Xj, yu Zi),(Xa, yz, z*),(Xa. yz^ Za)
(Xa, yu z«) to be coplanar.
The equation of the plane passing through first three points
is given by equation (S). If the fourth point namely (Xuyir^*)
also lies on this plane, then the co-ordinates of this point will
satisfy the equation (5), so that we have
1
1
*1
yi
zi
Xa
yi
Za
Xi
yi
Zi
1
Za
yz
Za
1
=0.
=0, i.e,.
Xa
^'a
Za
1
Za
yz
Zs
1
Za
yz
Za
I
Xa
yt
Za
1
...(6)
52
Analytical Geometry 3~D
The condition (6) is the required condition for four gi^en
points to be coplaoar.
Note. For solving numerical-examples an easier method can
however be follpwed as^xplained in Ex. 7 and Ex. 8 below.
Solved Examples 3(A)
Ex. 1. Reduce the equation of the plane x+2y-'2z~9=Q to
the normalform and hence find the length of the perpendicular
orawnform the origin ta the given plane.
X^®-Cqualion of the given plane is
'
JC-f-2;;—2z -9^0.
. ^
~
Bringing the constant term to the R.H.S., the equation
becomes
x-\-2y-2z^9,
0)
[Note that in the equation (1) the constant term 9 is positive.'
If it were negative, we would have changed the sign throughout
to TIrex^^p05itiv^_
Now the square root of the sum of the squares of the coeffi>
cients of x, y, z in (1)= V{(i)»+(2)8+( -2)*}== V9=3.
Dividing both sides of(1) by 3, we have
1
/
(2)
'^i^equation
of the plane is in the normal form
lx-Jfmy+nz=p.
T 0
Hence the d c *s /, m,n of the normal to the plane are ^
*9 *
—s
ahd the length p of the perpendicular from the origin to the plane
is 3.
Ex. 2. The co-ordinates ofa point A are (2, 3, —5). Determine
the equation to the plane through A at right angles to the line OA^
where O is the origin.
(Meerut 1986S)
Sol. Here the plane passes through the point A(2,3,—5)
and is perpendicular to the line OA i e , the line OA is normal to
The d.r,*s of OA are 2—0,3—0, —5-0 /.e., 2, 3, -5.
Thus the plane passes through the point (2, 3, —5)and 2, 3,
—5 are d.r.*s of normal to the plane. Therefore the equation of
the plane is
or
2(x-2)+3(:F-3)-5(z-(-5)}=0
2(x-2)+3(j;-3)-5(z+5)=0
[Refer §.(6)1
The Plane
53
or
or
2x+3j>—5z—38«=ap
2x+.3>'—5z«38.
Ex. 3. O is the origin and A is the point(o, b, c), Findtbe
d.c*s of the^join OA and deduce the equation of the plane through A
at right angles to OA.
Sol. The co-ordinates of the points O and A are (0, 0, 0)and
(fli, b, c) respectively. Thus the direction ratios of the join 0/1 are
fl—0,^-0,c—0 i.e., a, 6, c.
Hence the d.c.*s of the join OA are
e/v'(fl*+fr*+ c>).
Now we are to find the equation of the plane passing through
the point A {a. b, c) and perpendicular to the line OA. Here (he
line OA is normal to the plane and its d.r.*s are a, b, c. Therefore
Diane is
a{x-a)+b(y-b)+c(z~c}=^0. or ax+by
Ex. 4. Find the equation of the plane perpendicular to the line
segment from(-3, 3, 2) to C9, 5, 4) at the middle point of the
segment.
'
Sol. The end points of the given line segment are(—3,3, 2)
and (9, 5, 4). The d.r.’s of the line segment are 9—(—3),5—3,
r^., 12, 2, 2. The co-ordinates of the middle point of the
line segment are {i(9-3), i(5+3j. i(4+2» i.e.,(3,4, 3).
Thus the plane is to pass through the point (3, 4, 3) and d.r.’s
of normal to the plane are 12. 2. 2. Therefore the required equa
tion of the plane is
12(x-3)-|-2a-4)+2(2-3)«0
or
6(x-3)-ff:v-4)+(z-^3)=0. or 6x-f;;-|-z=s25. '
Ex. 5. Find the intercepts made on the co-ordinate axes by the
plane
x~3y+2z=9.
Sol. The equation of the givqn plane is
x~3y+2z==9.
...ID
Dividing both sides^by 9, the equation (1) may be written as
x . Z. L ^
1
1
9 +-3+(9/2l“^‘
(2)
Comparing'the equation (2) with the equation
xla+ylb-\‘Zlc=»U
the intercepts on the co-ordinate axes are given by
a=the intercept on the x-axis<=>9
6=the intercept on the y-axiso.—3
and c^the intercept on the z-axis«a9/2.
54
Analyttcai Geometry 3-D
Ex.6. A plane meets the eo-ordinate axes in A, B. C such that
the^centroid of the triangle ABC is the point(p, g, r)i show that the
equation of the plane is xlp+y/q-\-zlr=3.
(Kanpur 1983: Meerut 85S}
Sol. Let the equation of the plane be
xla-\-ylb+zlc=l.
...(!)
The plane (1).meets the x-axis in the point A,so putting y=0
and z=0 in (1) we get x=a. Thus the co-ordinates of the point
A are
0,0). Similarly (he plane (1) meets y and z axes in the
points B and C whose co-ordinates are given by B (0, b, 0)and
C(O.O.c).
Thus the co-ordinates of the centroid of the triangle ABC are
given by (K/H-O-f0), J(0-|-6+0), J(O-i-O-l-c)) /.e.. (Ja. ib, Jc).
But it is given that the co-ordinates of the centroid of the
triangle ABC are (p^ r), so that we have
■
Substituting these values of o, 5, c in the equation (1), the
equation of the required plane is given by
^/(3p)+W(3^)+^/(3r)=l or x/p-\-ylq-\-zlr=3.
Ex. 7. Find the equation to the plane through the three points
(0,-1, -1),(4, 5/1) W (3, 9,4).
Sol. The equation of any plane passing through the point
(0, —I, —1)is given by
fl(X_0)+h{;;-(-l)}+c{z_(-l)}=0
or
...(1)
flx-l-h(;>4-l)+c(z-|-l)=6.
If the plane (1) passes through the point (4, S, 1), we have
4a-|-66-l-2c=0.
If the plane (1) passes through the point (3, 9, 4), we have
3u-|-10h-|-5<;=Q.
...(3)
Now solving the equations(2) and (3), we have
a
b
c
sr:::M“6::S3'=4(n8“^(“5').
A fl=10A,6=-I4A.e=22A.
Putting these values of a, b,c in (1), the equation of the
required plane is given by
A[l0x-I4(y-1-1)4-22(2-/1)]=0
or
l0x-14(>;-l-l)-}-22(z-/l)=0
V
or
5x—7>-/llz-i-4=0.
Ex. 8. Show that the four points
—1),(4, 5, 1), (3,
9^4) and(—4,4, 4) are coplanar.
(Meerut 1982)
The Plane
55
Sol. Proceeding as in Ex. 7 above the equation of the plane
passing through the three points(0, —1, —1),(4, 5, l)and (3,9,4)
...(1)
is given by 5x:-7y+112+4=0.
if the fourth point(—4,4, 4) also lies on the plane (1) then
the co-ordinates of this point must satisfy the equation (1).
Putting jc=—4,
2=4, the L.H.S. of(1)
e: 5.(-4)-7.4+ll.4+4=0=the R.H.S. of(I).
Hence the equation (1) is satisfied by the point(—4,4, 4).
Therefore the given four points are coplanar.
§ 8. Equations of the co-ordinate planes,
(i) The equation to yz-plane. The x-coofdinate of each point
lying on the yz-plane is zero, and hence the equation to ^z-plane
is given by jc=0.
(ii) The equation to zx-plane. It is gixen by >>=0.
(ili) The equation to xy-plane. It is given by z=0.
§ 9.(A) The equations to the planes parallel to tibe co-ordinate
planes.
The equation of the plane parallel to the yz-plane and at a distauce ^a*from It. The x-coordinate of every point on this plane
is equal to 'a*. Hence the equation of the required plane is given
by x=fl.
Similarly, the equation of the plane parallel to the xz-plane and
at a distance *b* from it is given by yi=»b.
Also the equation of the plane parallel to the xj'rplane and at
a distance
from it is given by z=c.
(B) The eqnacion of the planes perpendicular to the co-ordinate
axes.
The equation of the plane perpendicular to the x-axis. This
plane is obviously parallel to the j;z-plane and hence its equation
is given by x=a.[See § 9(A) above];
Similarly the equations of the planes perpendicular to y and
z axis are respectively given by >>=6 and z=c.
§ 10. Angle between two planes.
Definition. The angle between two planes is defined as the angle
between their normals drawnfrom any point to the planes.
Let the equations to the two planes be
OiX+hiy+ciz+di=0.
...or
...(2;
and
aaX+hay-i-.C22+da=0.
56
Analytical Geometry 3-jD
The' d r.’s of the normal to the plane (1) are at, bu Ci and the:
d r.’s of the normal to the plane (2) are oa, ha. Ca.
If^ is the angle between the planes (1) and (2), then $ is the
angle between the lines whose d r.’s are at, bu Ci and-fla.
Ca*
.*.
cos'0=gifla+Ma + C|Ca
...(3)
V(«i*-trV + Ci*; V(aa“+ha*i-Ca»)
For the-acute angle between the two planes, cos $ is p'ositive
and for the obtuse angle it is negative. The numerical value of cos 0
in both these cases is the,same because cos(n~0)=i — cos 0,
Condition of perpendicularity of two planes.
Two planes are perpendicular if their normals are perpendi
cular. Therefore the planes (1) and (2) are perpendicular if the
lines whose d.r.’s are Oj, hi, ci and Oa, ha, Ca are perpendicular the
condition for which is
flifla+hiha+CiCa =-0.
...(4)
Condition of parallelism of two planes.
Two planes are parallel it their normals are parallel. There
fore the planes (1) and (2) are parailei if the lines whose d.r.’s are
®ii bi, Cj, and aa, ha, Ca are parallel the condition for which is
flj/fla=^i/ha--=Ci/Ca
...(5)
l.e., the coefficients of y, z in the equations.of the two-planes
should,be proportional.
is
Remember. The equation of any plane parallel to the plane
ojc+hj'+cjt+*=»0
0JC+hj> cz-f-A «=a0.
§ 11. The two sides of a plane.
Two points P(Xu yu z^) and Q(Xa, y^,zj lie on the same or
opposite sides of the plane axrfby>^cz-^d^Q according as
0Xi+byi+czi+dandaXi-\-byi-{-cZi+d
are of the same or opposite signs, ■
The equation of the plane is. nx4-hj'-fcz+rf=0
Suppose the line Fg meets the.given plane (1) at the point R
such that PR: RQ.^ nn: mi. Thou tfie co-ordinates of the point
R are
/hiiXa-HWa^i WiyaH-Way, .iWiZa-f-maZt \
\
Mi . * mi+ma * mi-tmt J
Since the point /? lies oh the plane (1), therefore
a w«-^2+WaXi )4.*
g!*Vi \ , /m«^«+ w,za+d=0
V '"i4-/«a;
. .hJ A- ^,rMh P \ mi^
I
Off
mi(ax2-jrbyi ^czi -\- d) f«Ma-r,4-h>»i^-czi+d)~o
$7
The Plane
4- byI +£?,-f«</
ma” dxi^byz+cZi-j-/
Now the ratio nti/ntz is positive or negative according as PQ
is divided at i? internally or externally i.e., the points A Q lie on
the opposite or same side of the plane (I).
Hence from (2) if ax\-\-byx-\-czx-\-d and axg+dyaH-c^a+rf are
of the same sign, then mi/ma is negative /.e., the points P and Q
lie on the same side of the plane (1) If axi-\-byi+czi-td and
aXi-^byi-\-cz^-\-d are of the opposite signs, /Wi/ma is positive f.e.,
the points P and Q lie on the opposite sides of the plane (1).
§ 12. To find the length of the perpendicular from the point
.(Xi, yi, Zi) to a given plane.
Let the equation of the given plane be
ax-i-by+cz=dz=,o.
(1)
To find the length of the perpendicular from the point
(^i» J'l, ^i) to the plane (1).
or
Shifting the origin to the point (Xi, yx, Zx), the equation (1)
becomes
a (x-+Xx)-i-b (y+yx)+c (z+Zi)-hd>=-0
or
ax+by+cz+axx+byx-hczx+d=0.
...(2)
Dividing both sides of (2) by V(a‘+b^+c^), we get
a
b
c
x -\-
+ +
, flXj+6^1
...(3)
The equation (3)of the plane is in the normal form with a
proper adjustment of sign throughout the equation.
The length of the perpendicular from the new origin to
_ , flfi+^Vi+£Zi+^
thu plane (3)
Via^+bHc^)'
Hence the length of the perpendicular from the point
(xuyu zi) to the plane (I)
=±oxx-^byi-{-cZf{-d
V{a*+b^-\-c*)
Since the perpendicular distance of a point from the plane is
always positive, therefore a positive or negative sign is to be
attached before the radical according as
-axi -i^b^i^czi^rd
IS positiveor negative Le., according as (Ai,2^1, zO lies on the
same side or On opposite side of the plane as the origin, provided
d is positive.
Working rule.
To find the length p of the perpendicular
.
58
Analytical Geometry 3-D
from the point ixuyi^zi) to the plane ax+by-\-cz+dc=»0, we
substitute the co-ordinates of the given point in the left hand side
of the equation"^of the plane and then divide this expression by.
\/[(coeff. of A)“+(coefF of j)*-t-(coeflF. of z)%
_ axH-byH-czH-d
Thus
y(a»+b»i-c*)
If the value of p obtained from this formula is negative, we
can ignore the sign and give the positive value in answer, unless
there is some special reason.
If the equation of the plane is in the normal form
Ix-^my+nz—p—Of the length Pi of the perpendicular from the
point (xi, >’i, rO to the plane is given by
pi^lxi+myi-\-nzi—p, for in this case
+'«*+«*)= !●
§ 13. To find the distance between two parallel planes.
Find the lengths of perpendicular distances of each plane
from the origin and retain their signs. The algebraic difference
of these two perpendicular distances is the distance between the ,
givdh parallel planes. But while applying this^ meihod-we' should
be careful that the coefficients of X in the two equations of the
planes are of the .same sign.
Alternative method. Take a point on one of the two given
planes, then the required distance is the length of the perpendicu
lar drawn from this point to the other plane.
SOLVED EXAMPLES 3 (B)
Ex. 1. Write the equations of the planes in the following cases:
(/) parallel to the xy-plane and 5 units below it^
{ii) parallel to the yz-plane and having x intercept 3,
{Hi) perpendicular to the z-axis at the point (0, U, 4),
and {iv) parallel to the zx-plane and 6 units behind it.
Sol, in view of § 8, the equations of the planes in the difife- .
rent cases are as giyeq belovi^ ; ■ . ;
V
^
":a ^
and (iV)
-6,
pT
Ex. 2. I^nd^ i^ eqi^tiph i^ the plane which is horizontal and:
pasies
^
.
■ v', - - ^ ^
. ,;
JUelt us'choose .the eq-qrdmaip: ix^.^so 'thitfTKthe’ ^xes; \
^;xfpcp^iie ih a hPrizohtal plana (the plane of the ’
and; 4|U^tS bfi-is pefipendiculat to this plaie^
required plane is perpendicular:to the ^-axis -and*
$0
The Plane
the point (1, -2, -5), therefore, the equation of the required
z — 5 or z-i-5=0.
plane is given by
Remark. If the axes of x and z be taken in a horizontalplane
(the; plane of the paper, say) then >/-axis is perpendicular to this
plane and hence the equation of the required horizontal plane wi'l
be ^=-2.
Ex. 3. Find the equation of the plane through the point
(— 1,2,4)and parailel to the plane 2x+3>^—5z+6=0,
Sol. The equation of the given plane is
..(1)
2x+3;;--52+6-0.
Since the equations of the parallel planes differ only in the
constant term, therefore the equation of any plane parallel to the
plane (1) is given by
...(2)
*
~ 2x.-j-3ji—5?^
0.
If the plane (2) passes through the point(—1,2, 4),^ we have
. 2(-l)+3(2)-5(4)+k=0 or k=16..
Substituting this value of k in the equation (2), the equation
2x |-3;;-5z+ 16 =0.
of the required plane is
Ex. 4. Find the equations of the planes parallel to the plane
3jc—6y—2z 4=0 at a distance 3from the origin.
Sol The equation of any plane parallel to the plane
...(1)
3x-6;;-2z-4=0 is 3x-6>;-2z+fc=0.
Let p be the length of the perpendicular from the origin to
the plane (1). Then
k
[See § 12]
VU3)“+(-6)» +(-2)*}
or
p-±fc/7.
But according to the given condition p is 3. Hence
kp=±3 => k= ±2\.
Putting the values of /: in (1), the equations of the required
3x-6x-2z±.:i=0.
planes are given by
Ex. 5. Find the equation of the plane parallel to the plane
2x-3;'--5z+l=0 and distant 5 unitsfrom the point(—1,3, 1).
Sol. The equation of any plane parallel to the plane
...(1)
2x--3;’—5z+I--0 is 2x~3y — Sz-\-k-=0.
Let ^ be the length of the perpendicular from the point
(—1, 3, 1) to the plane (1), then [see § 12]
-164-^
● ^=±2f-l)--3(3)-5(!)-^A:
Vt3s>
±-
60
Analytical Geometry 3-Z)
But according to the question, q=^5.
5=±(-I6+A:)/V(38) or Jk=16±5«v/(38).
Substituting the values of k in succession in (1). the equations
or the required planes are given by
2jc-3y-5z+16db5V(38)=0.
Ex. 6. Find the equation of the plane through the point (oi, B, y)
and parallel to the plane flx-f by+c2=0.
or
Sol. The equation of any plane parallel to the plane
ax-\-by-\-C2^0 is ax -{-by -)-cz+k=0.
If(l)l3asses through the point (a, jS, y), we have
<2a+ -fcy+A:=0
k=»^aoL—b^—CY^
...(1)
tuo
cz—aoL—bfi—cy=0
fljc+by4-C2e= fla 4-Aj8+ey.
or
Ex. 7. Find the equation of the plane through (I 0, —2) and
■ pstpendicular to each of the planes
2x+y~z—2=^0 and x---y—z—3=0,
(Gorakhpur 1978; Meerut 85S)
Sol. The equation of any plane through the point (1, 0, -2)
isu(x-l)4-b(y_0)+c(2+2)='0.
Ifthe plane (I) is perpendicular to the planes 2;c+y—z~2-0
andx-y-z-3==0,wehave
[See§7oi
o (2)4bn)4c(-l)=0/.e., 2fl4b-c=0
●●●(2)
and
a (l)4-b(-D4.C(- I)=0 /.e„ fl ^-cc»o!
..(3)
Adding the equations (2) and (3). we have
f«.
Subtracting (3) from (2), we have b= -*a.
Putting the values of b and c in (1), the equation of the
required plane is given by
or
c(jc-l)-j fly4_|a(^^2)«0
2a:—2—y4-3z4-6=0, or 2a:—y+32-}-4^0.
ri
poUits
1 , —2) and perpendicular to the plane
^4-2y~3z=5.
(Gorakhpur 1988)
Sol. The equation of any plane
passing through the point
(1.-2. 2) is
A ^)% t—
a (Ac -l)+h fy f 2)+c (2_2)=o.
.. (I)
If the plane (1) passes through the point (-3, I,● —'2l. wc have
The Plane
or
61
a(-3-l)+6U4-2)-i-cf-2-2)=0
-4fl+3!>-4c-=0, or 4a-364-4c=0.
...(2)
If the plane (1) is perpendicular to the plane x+2;>—3^=5,
or
we have a.I+6.2+c.( - 3)=0,
fl.+2h-3c=0. ...(3)
Solving the equations(2) and (3) for <i, b, c, we have
a
b
c
(—3)(-3j-2.(4) lx4-4.(-3)“4x2-l.(-3)~
. or
<3=*A, 6*=16A,
r”T6”n”*’
Putting the values of a,b^c in (I), the equation of the
required plane is
A [(X - I)+16 (3^+2)f 11 (2-2)1-0.
or
JC+16>-+I1z+9=0.
Ex. 9. Find_the, angle between the planes 2x —y-^z=6
■_^otidX'\-y'^Tzf=*‘7.
SoK The d.r.’s of -the normal to the pWne 2x—3>”|-z=6
are 2.-1, 1.
The d r/s of the normal to the plane .v4-;>>+2z=7 are I, 1, 2.
Now the angle between two planes is equal to the angle
between their normals. Therefore if d be the angle between the
given planes, we have
cos d -
2.1-K-l).1 + 1.2
v/{(2)2+(-i)H(i)“*)V{(i)*+(ir+m
$=klT.
=3/6=f.
Ex. 10. Find whether the two points (2, 0, 1) and {3, —3, 4)
lie on the same side or opposite sides of the plane x -:2>>+z»6.
Sol. Taking all terms to the left hand side, the equation of
. (I)
X—2v+z —6=0
the plane may be written as
Substituting the co-ordinatc.s of the point (2, 0, P, the value
of the left hand side of the equation (1) of the plane
-2-0+l-6^^-3.
Again substituting the co-ordinates of the point (3,—3, 4),
the value of the left hand side of the equation (i) of the plane
«=3+6+4-6=7.
Since the values —3 and 7 are of opposite signs, the given
two points lie on the opposite sides of the given plane. [See § 11]
Ex. 11. Find the equation of the locus of a point P whose
distance from the plane tx — 2y-\-3z-\-4=0 is equal to its distance
from the point ( — 1, 1,2).
62
Analytical Geometry 3-D
Sol. Let ihe co-ordinates of a point P be (xi, yu «i). It is
required to find the locus of the point P.
Let p be the perpendicular distance of the point P {Xu yu ^i)
from the plane 6x—2;^H-3z+4=0.
Then p
6jCi —27]+3z,+4 6xi-2;>i+3zi-t-4
7
\/{3o+4i- 9}
...(1)
Again let d be the distance of the point P (Xj» y^ Zx) from
tl^e point(— 1, 1, 2).
__
Then d=‘>/{(Xx^\)^+[yi-\?+{zx-2n
...(2)
According to the given condition, we have p^d i.e., p^=d*
i.e.
or
or
(6xi-2;;i4-3zi-F4)V49=(x,+l)H(>'i-I.H(Zi-2)*
36xi*-l-4;;i*+9z,*-M6—24xi >',4;36zix,-f48xi—12;>iZi
-16;;i+24zi«=.49 {V+2xi-H
4zi+4}
13xi*+45y,H40zi*-b24jCi:Ki-36z,Xi+12j;,Zj+50xi-82j;,
-220zi+278=0.
Hence the locus of the point
<Xj, yu Zi) is
13xa+4Sj^*+4Qza+24xx-36z.r+I2;;z+50x-82;;-220z-|-278=0.
Ex. 12. Find the perpendicular distance between the parallel
phfftes 2x-3y-6z-2\^0 and 2x-3;;-6z+14-0.
Sol. The equations of the given planes are
2x~3;»-6z-.21=0, ...(1) and 2x-3;^-6z-|-l4«=0. ...(2)
[Note that the coefficients of X in the equation H)ond (2)
are of the same sign].
Let Pi and pz be the lengths of the perpendicular distances of
the.planes (I) and (2)from the origin. Then we have
2.0-3.0-6.0-21 -21
Pi
-3..
V(4+9-f-J6)
“● 7
and
pz=
2.0-3.0-6.0+14
14
V(4+9+36)
“7
Hence the perpendicular distance between ihe given planes ■
“Pa—p,=2—(—3)= 5.
Alternative method, The co-crdinates of a point on the first
plane 2x-3p-6z-2I = 0 are (0, -7. 0).
The perpendicular distance between the given planes
“the perpendicular distance of the point ^0, — 7, 0) from
the second plane 2x- 3p- 6z+14=0
●
'
0+21-0 + 14
● ” \/(4+9f36] “7
63
The Plane
Ex. 13. Find the hern ofa point, the sum of the squares of
whose distancesfrom the planes
x-\-y-\-z=0, x—y=0, x+y>—2z=»0 isl.
Sol. Let P{Xu yu 2i) be a point whose locus is required.
Let pi=the distance of jP from the plane x+^+r=0
^(xi+yi+Zi)IVi^)f
p-,=»tbe distance of P from the plane x—
dnd
Ps<=>the distance of P from the plane x+y—2xs=0
{xi+yi-2zi)fVi6h
Now according to the given condition, we have
S3
...(1)
Substituting the values of pi, pz, Pa in (I), we get
(xi+:f,—2zi)*=>7
or 2(V+;;ia+2:i*4-2x,:i;i+2z,X|+2;>iZi)+3 (x,*-;^i*-2Xiyi)
+W
+4zi*+2x,j;i-4ziXi-4;?iZi)«=42
or 6x,*+6;;iH6zi*^=42, or Xi*+7i*+Zi*«=»7.
the required locus of P(xj, yi, z{) is
§ 14. A plane through the intersection of two given planes.
Tofind the equation ofany plane through the Ihse of intersection
of the two given planes.
Let the equations of the two given planes be
..-(1)
i»sflix+y+CiZ-1-=0,
and
...(2)
QsaMX-\-biy-\-CzZ-\-di=0.
Then the equation P-f-A Q=0,/.e., the equation
...(3)
(aiX+biy-\-CiZ+di)+\ (aaX+hj;>+caZ+</2)=0
is the required equation of any plane through the line of inter*
section of the planes(I) and (2).
First we observe that the equation (I) is of first degree in
X, y and z and so it represents a plane. Again whatever A may be
if a point satisfies both the equations (1) and (2), it defihitely
satisfies the equation (3). Thus ail the points of the line ofinter
section of the planes(1) and (2) also lie on the plane(3). Hence
(3) is the equation of any plane passing through the line of inter
section of the planes (1) and (2).
;§ IS. Tofind the condition that a line whose d.r.*s are /,iw, n
may be parallel or be perpendicular to d given plane.
Let the equation of the given plane be
...(1)
ax+by+cz-{-d=0.
Thus the d.r.’s of the normal to the plane (1) are a, b, c. The
d.r.*s of the given line are /, m, n.
64
Analytical Geometry Z-D
Tbe line is parallel to the plane. ir the given line is parallel
to the line (1), it is purpendicular to the normal tp the plane (1
the condition for which is
...(2)
al+ bmfCD"0.
The line is perpendicular to the plane.- If tbe given line is
perpendicular to the plane (1), it is parallel to the normal to the
plane (j), the condition for which is
a/l=»b/m = c/n
§ 16. The angle between a line and a plane.
Definition. The angle between a line and a plane is defined to
be the complement of the angle between the line and the normal to
the plane,
Clearly this angle can be determined by the methods explained
earlier.
Solved Examples 3(C)
Ex. 1. Find the equation of the plane passing through the line
ofintersection of the planes 2x—7y+42=3, 3x —5y+4z+ll=0,
and the point (~-2, 3),
Sol. The equation of anv plane through the line of interSTOtion of the g'P^h planes is [See § 14]
.(1)
(2x-7y-f42-3)+A (3x-5y+42+11)=0.
If the plane (I} passe.s through the point (-●2» 1, 3), then
substituting the co-ordinates of this point in the equation (1),
we have
{2 {-2)-7(l)+4(3)~3)+A {3 (-2)~5(l)+4(3)+1}=0
(●-2)fA (12)=0, or A-1/6.
Putting this value of A in (1), the equation of the required
plane is (2x-7;'+4z-3)+(l/6) (3x-5y+42+ll)=^0
or
15x-47y+28z==7.
Ex. 2. Find the equation of the plane through the line of irttersection of the planes x-\-2y—3z— 6<= 0 and 4x+ 3y— 2z+2«=a0 and.
passing through the origin.
.
Sol. the equation of any plane through the line of inter
section of tbe given planes is
..(1)
(x+2>»-3z-6)+A (4x+3y-2z+2)=0.
If the plane (1) passe.s through the origin /.e., through the
point (0, 0, 0), then we have
● '
.(0+0-0~6)+A f0+0-0-|-2)«0, or-6+2A=i0, or A*3,
.
Substituting this value of A in the equation (1), the equation
of the required plane is
(.r + 2i —3z - 6)-1-3 (4.v+3y—2r+2)=0,
or
i.rv-i l l
or
The Plane
65
Ex. 3. Find the equation of the plane through the line of inter- .
section of the planes x+2y-^3z"4—0 and 2x+y^z+5—0 and
perpendicular to the plane 5x-^3y-\-6z+^'=0.
(Luckaow 1982, Meerut 84S)
Sol. The equation of any plane through the line of inter
section of the planes .r+2>>+3z—4=0 and 2a:+>--z4-5=0 is
-4)4 A f2x+,v-z4 5)-0
or
-0)
X (l+2A)+y (2+A)+z(3-A)-(4 - 5A)---0.
If the plane (1) is perpendicular to the plane 5x+3j;.+ 6z-f 8=0,
we have
(1+2A).5+(2+A).3+(3-A).6=0
or 5+10A+6+3A+l8-6A=0, or 7A+29=0, or A= 29/7.
Substituting the value of A in (1), the equation of the required
plane is
X (l-58/7)+y'(2-29/7)+z(3+29/7)-(4+145/7)=0
or
—51x~15j;+50z-173=0,
or
51x+l5y-50z+173=0.
Ex. 4. Find the equation of the plane through the line of inter
section of the planes flx+6y+cz+rf=0 and ctx-{-fiy-\-'yz-\-8=0 and
perpendicular to the xy-plane.
Sol. The equation of any plane through the line of inter
section of the planes ax+i>;'+cz+t/=0 and ax+(3>'+)'Z+S=0 is
...(1)
{ax-\-by-^cz-\-d)-^\ («x+y?;^+yz+8)=0
or
...(2)
X (fl+aA)+>'(fef/?A)+z (c+yA)+(</+8A)=0.
Now the equation of the xj'-plane is given by z=0
or
...(3)
0.x+0.>»+l.z= 0
If the planes (2) and (3) are perpendicular to each other.
we have
0.(fl+aA)+0.(6+^A)+l.^c+yA)=0, or A^-c/y.
Putting; this value of A in the equation (!), the equation of the
required plane is
{ax-{-by-\-cz+d)-idy)(«x+j3>'+7z+S)=0
or
y (ax+8v+cz+f/)“c(ax+.'3y+yz+8)=0
or
{ay—col) x-\-(by—cp)j'+(cy cy)z+(dy-cB)=0
or
.{ay—ccx)x-^(hy—cP)y-\-{dy cS)=0.
Ex. 5. Find the equation of the plane through the line of inter
section of the planes flx+fty+rz+ </-—-0 and ax+/53’+yz+8=0 and
parallel to x-dxis.
Sol. The equation of any plane through the line of inter
section of the given planes is
66
Analytical Geometry 3-D
...(1)
ax+by+cz+d+X («xi-fiy-\-yz+8)^0
...(2)
X {a+a\)-{-y (A+j?A)+z(c-f yA)+(rf+aA)=0.
Now the d.c.’s of the x-axis are 1,0,0 and the d.r.’s of the
normal to the plane (2) are n-l-aA, ^+/5A, c+yA. The plane (2) is
parallel to the x-axis if the normal to the plane (2) is perpendicu iar to the x-axis, the condition for which is.
1.(arf-oA-f0.(ft-|-/5A)+0.(c-f yA)=0,
giving
A=—a/a.
Putting this value of A in the equation (l),.the required equa
tion of the plane is given by
ct(ax+by-\-cz-\-d)-a (oLX+p\-{-yz+8)=0
or
(ba—ap)y-\-(cx-ay)z+(da-a8)—0.
Ex. 6. Find the equation of the plane through the point
(1, —2,0) and normal to the line joining the points (2, 3,-2)and
(1.-2,4).
Sol. The d.r.*s of the line joining the. points (2, 3, — 2)
and (1, —2,4) are 1 -2, -2-3,4-(-2) z.e., -1, -5,
These
are the d.r.’s of the normal to the plane. Since the plane is to
pass through the point (1, —2,0), its equation is
-l(x-l)-5 (>;-i-2)-|.6(x-0)=0,
or
—X—5A-h6r—9=0 or x-f-5;;—6z-l-9=0,
or
x-h5y—6z=—9.
Ex. 7. Find the equation of the plane through the points
(1, —2,4) and(3, —4, 5).and parallal to the x-axis (i.e. perpedicular to the yz-plane),
Sol. The equation of any plane through the point (1, -2,4)
is
...(1)
a{x-\)-^b{y^l)^lHc{z-4)=0.
If the plane (1) also passes through the point(3,—4, 5),
we have
a (3-l)-l-i> (-4-1.2)-l-c f5~4)=0,
or
2a—2h+c=0.
...(2)
Now the plane (1) is to be parallel to the x-axis Le. perpen
dicular to the x,v-plane whose equation is
x=0 !>., l.x-l-0.;;-l-0.z=0.
Hence we have
a.l-t-6.0-bc.0=0, or a=0.
...(3)
Putting the value of q from (3) in (2), we get c=2b.
Substituting the values of a and c in the equation (1), the
equation of the required palane is
0-1-6(y+2)-f26(z-4)=0, or j;-f-2r-6=0.
or
67
The Plane
Some Important Solved Examp les
Ex. 1. A variable plane at a constant distance pfrom the origin
meets the axes in Ay B and C. Through A, B,C planes are drawn
parallel to the co-ordinate planes. Show that the locus oftheir point
(Meeratl984P)
ofintersection ir
Sol. Let the equation of the variable plane be
...(1)
xla-hylb-\-z/c^\y
where a, c are variables.
The plane (1) meets tb''^ co-ordinate axes in the points Ay B
and C. whose co-ordinates are (o, 0, 0), (0, by 0) aiid g,0, c)
respectively.
It is given that the length of the perpendicular frbin (0, 0,0)
to the plane (1) isp.
1
1
or
A p=
V{(l/fl)*+(iW+(i/^^)«}
...(2)
Now we shall find the equation of the plane through the point
A (dy 0, 0), and parallel to the pr-plane.
The equation of the yz-plane is x=0.
Any plane parallel to the plane x=0 is given by x=\.
If it passes through the point 4(ay 0. 0), we have a=^.
Hence the equation of the plane through A and parallel to the
x=a.
...(3)
yz-plane is
Similarly the equations of the planes through the points B
and C and parallel respectively to the co-ordinate planes y—0 and
z*=0 are
...(5)
y=:b
.^.(4) and^Zjsac.
The locus of the point of intersection of the plaherT3)T1i4)
and (5) is obtained by eliminating Oy by c between the equations
(2),(3).(4) and (5).
Putting the values of a, by c from (3),(4) and (5) in (2), the
required locus is given by
l/p8=l/x*-l-l/y»+l/z2 or
Ex. 2. A variable plane passes through afixed point (a, y)
and meets the axes of reference in A, By C. Show that the locus of
the point of intersection of the planes through Ay By C parallel to
(Meerut 1984)
the co-ordinate plartes is
.'
Sol. Let the equation of the variable plane be
...(1)
x/a-lfP/6+z/c=l,
where a, by c are parameters /.c., variables.
68
Analytical Geometry 3-D
The plane (1) passes through the point (a, p, y).
oi!a+filb-h7lc=].
...(2)
The plane (1) meets the co-ordinate axes in the points A, B
and C whose co-ordinates are respectively given by (a,0, 0),
(0,6,0) and (0, 6, c). The equation of the planes through A, B
and C and parallel to the co-ordinate planes are
...(3)
3C=fl, y=by z=c respectively.
[See Ex.. 1 above]
The locus of the point of intersection of these planes [given
by the equations (3)] is obtained^by eliminating the parameters
OybyC between the equations(2) and (3). Putting the values of
a. 6, c-froirt (3) in (2), the required locus is given by
djx-^PIyA-yfz^X or
Ex. 3. A point P moves on the plane x(a+y!b+zlc= 1 which
isfixed. . The plane through P perpendicular to OP meets the coordinate axes in Ay B and C. The planes through Ay B and C parallel
to theyzyzx and xy-planes intersect in Q. Prove that if the axes
be rectangulary the locus of Q is
JL4.14.J._L..L. J.
ax^by'^cz
(Kanpur 1983)
Sol. The equation of the plane is
(1)
xla-^y}b’lrzlc=\.
Let the co-ordinates of the point P be (a, p, y). ^ince the
point P(a, fiy y) lies on the plane (1), we have
(t!a-\-pfb-\-y!c=^].
...(2)
The direction ratios of OP are a—0,fi —Oy y-0 i.e., a, p, y.
Hence the equation of the plane passing through the point
P(«, fiy y) and perpendicular to OP is
a(x-a)+A (jF”/?)4y (z-y)=0,
or
..(3)
ax-t-py+yz=a2-f pH
The plane (3) meets the axes in the points Ay B and C whose
co-ordinates are"respectively given by
PHy’^Va. 0, 0),(0.{aH^Hy='}/P, P>
and
(0, 0.{dH^Hy2}/y).
Again the equation of the plane through ^4 and parallel to the
yr-plane i.e., the plane .\:=0 is
x=(aHpHy®)/a.
...(4)
Similarly the equations of the other two planes are
y=(«*+i»*+y«)/p ’ ...(S) and z=(aHPHy*)/y.
...(6)
69
The Plane
Now Q is the point of intersection of the planes (4), (5) and
(6). The locus of the point Q is obtained by eliminating a, y
between the equations (2),(4),(5). and (6).
From (4)» (5) and (6), we have
1
1
l
a®
.
7* .
a'-*+/?2+y2
“(a*+j8‘‘+y*)V
i
...(7)
[using (2)]
a2-i-^a+y2
From (7) and (8), the required locus of Q is given by
1/jcH l/^^H l/z"= l/(a^)+l/{6v)+l/(cz)..
...(8)
Ex. 4. A variable plane is at a constant distance 3p from the
origin and meets the axes in A, B and C. Prove that the locus of the
centroid of the triangle ABC is x~^-ry-^ArZ~'^=p’~K
(Meerut 1983, 85, S9S ; Lucknow 81)
Sol. Let the equation of the variable plane be
xlaA-ylb-\-z/c^=l.
...(1)
It is given that the length ofthe perpendicular from the origin
to the plane (I) is 3p.
1
1
1
1
1
/. 3p^-.
V(l/ttM-i/6"+l7c2)
+^2+C»**
...(2)
The plane (1) meets the coordinate axes in the points
3
and C whose co-ordinates are respectively given by (n, 0,0),
(0, b, 0) and (0,0, c). Let (x, y, z) be the co-ordinates of the
centroid of the triangle ABC. Then
x=(a-|-(H-0)/3,7=(0+6+0)/3,z«(0-1-0+c)/3
i.e*»
x=^ia,y=^b,z^y.
fl=3x, b=3y, c=3z.
.(3)
. The locus of the centroid of the triangle ABC is obtained by
eliminating a, b, c between the equations (2) and (3). Putting the
values of a, b, c from (3) in (2), the required locus is^ given by
1
_L-i_ -L _
9p^~~'9x^'^9y^'^9z'^
Ex. 5(a). A variable plane is at a constant distance pfrom the^
origin and meets the axes in A^ B and C. Show that the locus of the
centroid of the triangle ABC is
x-^~\-y-^-{r:r^=9p~^.
Sol. Proceed as in Ex. 4 above.
(Lucknow 1981)
70
Analytical Geometry 3-D
Ex.5(b). A plane meets the co-ordinate axes at A^ BiC such
that the centroid of the triangle ABC is the point(a, c). Find the
locus to the plane ABC.
(Meerut 1989)
Sol. Let the equation of the plane be xja'-^yfb*-\-zlc'
the plane (1) meets the co-ordinates axes, in Ay fl , C. Hence we
have (o', 0, 0), jB (0, bS 0)aud G (0, 0, c').
Since centroid of /^AIK^ is (a, 6,c)
(a'+O-l-O), b=i
C=i(0-1-0 -1-c')
or
a'=3fl, b'=36, c'=3c.
Substituting values in (1), the equation of the required plane
is
xla-^ylb-^zjc=3.
Ex.6. A variable plane is.at a constant distance pfrom the
origin O and meets the axes in Ay B and G. Show that the locus of
the centroid of the tetrahedron OABC is
(Lucknow 1982; Meerut 83)
Sol. Let the equation of the variable plane be
xlaJrylb-^zlc=X.
...(I)
It is given that the length of the perpendicular from the origin
O to the plane (1) is p.
1
●
, .
1
1
1
1
...(2)
The plane (1) meets the axes in the points Ay B and G whose
co-ordinates are (a, 0,0),(0, b,0)and (0,0, c) respectively. Let
(^*» >'» z) be the co-ordinates of the centroid of the tetrahedron
OABC. Then
x=(0+o.|.0+0)/4,p=:(0-l-0+b+0)/4, z=(0-i-0-i-0+c)/4. ,
.*. a=4x, 6=4p, c=4z,
-..(3)
The locus ofthe centroid of the tetrahedron OABC is obtained
by eliminating a, b, c between the equations (2) and (3). Putting
the values of a, by c from (3) in (2), the equation of the required
locus is given by
Ex. 7. Two systems of rectangular axes have the same origin.
Ifa plane cuts them at distances a, by c and a\b\c\ respectively
from the originy show that
1
1
1
-^4—
a* b^ c* fl'*'^b'®“^c'**(Lucknow 1978;Kanpur 82)
Sol. Let O be the origin. Let OAT, OYy OZ be one set of
The Plane
71
rectangular axes and let the equation of the plane with respect to .
...(1)
this set of axes be
x(a-\‘ylb+zlc=\.
Let the second set of rectangular axes be chosen as
0% 01
and let the equation of the same plane with respect to this set of
axes be
|/a'+>l/6'+C/c'=l.
v(2)
Now we know that from a point outside the plane only one
perpendicular can be drawn to the plane. .The origin being the
same for both the systems, the length of the perpendicular from
the origin to the plane in both the cases will be the same. Hence
we have
1
1
V{(iW+UOT+(i/<?“)}
or
l/a«+1/*«+ l/c2=l/a'a+
\\c^K
Ex. 8. A plane meets a set of three mutually perpendicular
planes in the sides ofa triangle whose angles are A, B and C respecti
vely, Saow that thefirst plane makes with the other planes angles,
the squares of whose cosines are
cot B cot C, cot C cot A, cot A cot B.
Sol, Let the three mutually perpendicular planes be chosen
as the co-ordinate planes, and let
-.(1)
x/<j+j^/h+z/c=l
be the equation of any palne which meets the axes of x, y and z
at the points A, B and C respectively.
Clearly the co-ordinates of the vertices.^4, -.5, C of the A
are (fl, 0,0),(0, 6,0),(0,0, c) respectively.
The direction ratios of the side ABzxt a, —6,0 and the direc
tion ratios of i4Care.a,0, —c. Hence the angle A between the
sides AB and AC is given by
tan Ass V[S(biCt-btC,)^]
Oia^'j-biba+CxCi
* [See § 12, chapter 2]
._VC(^c-0)2+(0+flc)»+(0+fl6)*]
or
tan^
—
52+0+0
.% cot A=a^ly/(b^c^+a^c^-{-a^lP),
Siniilarly cot B=lPly/(b^<^-\-c^a^+a^b'^)
and
cot C==c*/y(fl*6*+6*c*+c*«®).
Now suppose a is the angle between the plaqe (1) and one of
the co-ordinate planes say the plane x=0. The d.r.’s of the nor-:
mals to these planes are 1/a, 1/6, 1/c and 1, 0, 0 respectively.
n/a).l+n/6).0+(l/c).0
●
● e.
cos a
~ V(l/a»+l/6=*+l/c»).y(ia+0*+0*)
72
or
Analytical Geometry 6-D
cos* a«3)
cot B cot c
Similarly if p and y are the angles which the plane (1) makes
with the coordinate planes y=a and z=0 respectively, then we
have cos* ^=cot C cot^ and cos* y=cot A cot B,
Ex. 9. Find the equation of the plane which bisects the jtoin of
^(*i» yu 2i)
y^y
perpendicularly.
Sol. The required plane passes through the middle point of
the segment PQ and is perpendicular to PQ. The co-ordinates of
the tniddle point of PQ are
(i
^
i C^i+Za)*
Also the direction ratios oi PQ are {xx-x^, yx-yt. zx-z^). Thus
the direction ratios of the normal to the plane are Xj—JCj,
Vjj
Zx~ Zg and hence the e:quation of the required plane is given by
ixx-Xt){x-i (Xi+x^n^+CFi-y,)
(yx+yi)} .
: ^izi-Zx){z-i {Zx-\rZi))=0
0Tx(xx~xx)’hy(yx-y2)’hzizx-zx)
Ex 10. From any point P are drown PM and PN perpendiculars
zx and xy~planes. O is the origin and «, y and S are the angles
which OP makes with the co-ordinate planes and with the plane
OMN. Prove that if the co-ordinates of the point P are (a, b, c),
then.
to
(i) the equation of the plane OMN is xld-ylb^zjc=Q
abc
(«) B-sm
^{b^c^-fc*a*+a*b^)
and {Hi) cosec^ B=cosec^ «+cojec* p+cojec*y.
Sol. The co-ordinates of the point P are given to be (a, b^c).
The equation of the zx-plane is j;=G. Now since PM is drawn
perpendicular from P to the zx- plane, therefore M is the foot of
:
the perpendicular from P(h c) to the zjc-plane and hence the
co-ordinates of the point M are (a, 0, c). Similarly the co-ordinates of the foot N (lying in the xj'. plane) are (a, by 0).
Now we shall fiitii.the equatidn of the plane OMN.
the equation of any plane through the origin O (0, 0. 0) is
iix4-Py+Cz=G.
(I)
O
If the plane (I) passes through the points Af (a, 0, c) and
. . ^(fl* by 0), we have i4.n-|-Pi0-f-C.c=0 and A.a-^B.b-\-C.O
-
v>
Solving these for 4 B, C, we get —
—be
ca ab'
73
The Plane
Putting the values ot 4, 5, C in (1) the equation of the plane
OMN is given by ,
...(2)
- bcx-\-cay-\-abz=0 or xla ~yfb-zfc=0.
This proves the result (i).
To find the angle S. Since S is the angle between the line
OP and the pl^ane OMN^ therefore 90° -- B is the angle between the
line OP and the normal to the plane OMN.
The direction ratios of the line OP are a—0,£>-0, c-0 i.c.
fl,bt c. The direction ratios of the normal to the plane OMN are
[See equation (2)]
1/fl, -1/6, -1/c.
a({/a)+bi-\lb)+c{-Mc) _
Hence cos (90°-S)=
abc
neglecting the — ive sign because 90° - 8 is the acute angle between
the line OP and the normal to the plane OMN
abc
or
sin 8
...(3)
abc
^
8=sin j
*’
6H c*)' V(^®c'a+
a^bl^j
This proves the result (ii).
Again let a be the angle between the line OP and the co
ordinate plane ;c=0, so that 90°—a is the angle between the line
OP and the normal to the plane x=0 whose d.c.’s are 1,0, 0.
Hence we have
fl.l-f^O-f c.O
cos(90°-a)=
or
or
sin a=
a
, or cosec? a
<2*
Similarly, we have
cosec*
6*
and cosec* y=:
cosec* /r-fcosec* //-l-cosec*
(6*c*-f c*fl*-f
aWc'^
=cosec*8.
(«*-l-6*+c*)
c®)'
[using the relation (3)J
This proves the result (lii).
§
Eqaations of the planes bisecting the angles between
two given planes.
Let the equations of the two jiven planes be
14
Analytical Geometry 3-Z)
..(1)
...(2)
agX+&a>'+c,z+</2=0.
The equations (1) and (2) should be so written that the cons
tant terms and
are both positive. However, we can also
write the equations (1) and (2) in such a way that d^ and d^ are
both negative.
If(xi, yi^ Zi) be the coordinates of any point on the plane
bisecting the angle between the given planes, then the perpendi
cular distances of this point from both the planes should be equal
numerically. Since we consider perpendicular distance as positive
when measured in the direction from the origin to the plane, for
points on the plane bisecting the angle in which the origin lies,
the perpendicular distances will have the same sign, and for points
on the other bisector, opposite signs. Therefore if the point
(Xi* yi* ^i) lies on the bisector of the angle in which the origin lieSy
we have
and
Since this relation is satisfied by every point (xi, y^y
this bisector, the equation of this bisector plane is
axX-\-b^y+c^z-\-dx aiX-\-biy->t-c^-\-dz
on
V(a.*+V+Ci«)
...(3);
Similarly if (jcj,
Zj) lies on the bisector of the other angle
between the two planes, we have
ajXx+byyi+CxZx+dx
Viai‘+V+«i*) ~
02X,-f-Vi+CgZi-l-^/a
VW+V+c.*) ■
The equation of this bisector plane will be
giX-fdi>>+CiZ-|-</t
aiX-\-b^-\‘C,fl-{-di
VW+V+^2®)
...(4)
Hence the equations(3) and (4) are the required equations of
the planes bisecting the angles between the given planes. The
plane (3) bisects the angle in which the origin lies and the plane
(4) bisects the angle in which the origin does not lie. But we
should not forget to write the equations (l).and (2) in such a way
that di and d^ are of the same sign.
To distinguish between the two bisecting planes as regards the
bisector^ of the acute or obtnse angle between the given planes.
. If we are required to find which of the two bisecting planes
given by (3) and (4) represents the plane bisecting the acute or
obtuse angle between the given planes (1) and (2), we find the
75
The Planes
value of cos
where 9 is the acute angle between the bisecting
plane and any of the given planes. From this value of cos 0, we
find the value of tan B by using the formula tan d=-^^(sec* ^—1)
(or otherwise). In case the value of tan $ < 1 then B
and
hence this bisecting plane bisects the actue angle between the
given planes. Again in case the value of tan 6 > I then ^
and hence this bisecting plane bisects the obtuse angle between
the given planes.
An important remark. To find whether origin lies in the acute
or obtuse angle between the given planes (1) and (2).
The costant terms dx and d^ in the equations (1) and (2)
should be of the same sign,
(i) If the origin lies in the acute angle between the given
planes(1) and (2), then the angle 9(say) between the normals to
these planes is obtuse and therefore the value of cos $ is negative
i:e, aia2+biba+CiCa=negative is the condition for the origin to lie
in the acute angle between the planes,
(ii) If the origin lies in the obtuse angle between the given
planes(1)and (2), then the angle d (say) between the normals to
these planes is acute and therefore the value of cos $ is positive
i.e„ aiag+bib|-l-CiCj=positive is the condition for the origin to lie
in the obtuse angle between the planes.
Thus to distinguish between the two bisecting planes, we
should first find whether the origin lies in the acute or obtuse
angle by the above method. Then we should find the plane bisec-,
ting the angle which contains the origin and the plane bisecting the
angle which does not contain the origin. In this way we can dis
tinguish between the two bisecting planes with respect to the posi. tion of origin and the acute and obtuse angles.
Solved Examples(3) D
Ex. 1. Find the equations of the bisectors of the angles between
. the planes lx~y -Iz-6=0 and 3x-|-ly-6z-12=0 and distinguish
them.
Sol. Writing the given equations in such a way that the
constant terms are both positive, the equations of the given planes
...d)
are
—2x.+J'+2z+6=0,
...(2)
and
-3x-2;;+6z-M2=0. .
The equation of the - bisector plane of the angle between the
planes (1) and (2) which contains the origin is given by
76
Analytical Geometry 3-D
-2.y+j;+2r+6
-3x-2v+6z+n
V(9+4+36)
7(~2x+;v+2z4-6)=3(-3A:~2;/+6z-f 12)
5a:~13;;4-4z-6=0.
V(4+l+4)
or
or
or
or
The equation of the other bisector plane is
-2x+y-^2z+6_ “ 3jc 2;;+6z+12
V(4+l-j-4)
V(9+4-1-36)
7 (^2;c-r j^+2z+6)= -3 (-3A:-2;;-f6z-|-I2)
23x'-y-32z-7S=:0
...(4)
Now let 6 be the acute angle between the plane (4)and the
bisecting plane (3). Then
15
●-2.S+1.(
-13)4-2.4
^ _______
_
,
tan
-M-
v'(4-t- iT4jV(25-t- 169TT6) Tv2i0
V(sec2 d-l>=V[(42/5) - l J-V(37/5)>1, so' that d>45”.
Hence the plane 5x- 13>*-|-4z-6=0 bisects the obtuse angle
between the given planes (1) and (2) so that the other plane
23x—y- 32z—78—0 bisects the acute angle. ,
We also note from the given planes (l; and i2), that
2) (^3)-l-1'(-2)-f.2’(6)
-6—2.-1-12--16=-positive.
Hence the origin lies in the obtuse angle between the given
planes.
[See remark to § 17].
This confirms thai the plane (3) is the bisector of the obtuse
angle. Hence the equation (4) is the bisector of the acute angle.
Ex. 2. Show that the origin lies in the acute angle between the
planes a:-|- 2>- H- 2z - y - 0 and 4jc -- 3y+12z*' *r^3:^0. Find the planes
bisecting the angles between them and point out the one which bisects
the acute angle.
Sol. In order that the constant terms
are positive, the
equations of the given planes iiiay be written as
~.v-2;—2z-b9--0
...(1) .
and
4x-.3>;-H2z+I3==.0.
Vve have
(-l).4+(-2).(-3)-l.(-2).(J2) .
—4-1-6 — 24=—22e=:negative.
Hence the origin lies in the acute angle between the planes
(1) and (2; [See remark ic § 17].
fhe equation of the plane bisecting the angle between the
given planes (1) and (2) which contains the origin is
77
The Plane
- X ~ly-'lz+9 4x~3:f+12z+13
y'(16+9+144)
V(‘+4+4)
O'*
13(-a:-2v-2z4-9)-3 (4x-3:k+12z+13)
2Wl7y+62z-78=0.
or
...(3)
We have pioved above that origin lies in the acute angle bet
ween the planes and so the equation (3) is the equation of the
bisector plane which bisects the acute angle between the given
planes.
The equation of the other bisector plane (/.e., the plane bisec
ting the obtuse angle) is
-;c-2)' 2Z-1-9
V(l+4+4)
or
x+35y
4x—3y+\2z+\3
“V(i6+V+144>
Or- 156=0.
...(4)
.. the equations (3) and (4) give the planes bisecting the
angles between the given planes and the equation (3) is the bisec
tor of the acute angle.
Ex. 3 Find the bisector of the acute angle between the planes
2x ~y-\-2z+3=0 and 3x - 2y-h6z+^-=-0.
Sol. Proced as above in Ex. 1 or Ex 2. The required bisector
plane is
23jc-13)>-i-32z-f45=0.
Ex. 4. Find the equation of the plane that bisects the angle
between the planes 3x—6jj^-l-2z+5=0 and 4x—\2y+3z—3 which
contains the origin. Is this the plane that bisects the obtuse angle ?
Sol. Proceed as above in Ex. 1 or Ex. 2.
The required
bisecting plane is 61x ~ I62y-j-47z-r44=0, and this, is the bisector
of the acute angle,
§ 18. Combined equation of a Pair of planes.'
Tofind the condition that the general homogeneous equation
ofsecond degree in x, y and z namely
ax^-\-by^-]rCz^-\-'2'fyz-\'2gzx-)-2hxy^0.
may reprsent a pair of planes and tofind the angle between them
Also tofind the condition of perpendicularity of these planes.
The general homogeneous equation of second degree is
...(1)
ax^4-by^-\-cz^+2fyz+2gzx-\-2hxy=0
Let the equations of the two planes represented by (1) be
/ix-fmij-f «iz=0 and l^x+m2y'{ntZ=0.
78
Analytical Geometry 3-D
These equations will not contain the constant terms for other
wise their product will not be homogeneous. Thus we have
ax*-\-hy^-{-cz^+2fyz-\-2gzx+2hxy
=(/iX-bmijK+Wiz) OiX’^-m^yArni^).
Comparing the coefficients of like terms on either side, we
have
/i/j=fl,
«iWa=c, Wi«2+W8ni=2/,
«i/a+«2/i=*2g', lim^+lzm^=2}i
...(2)
The required condition is obtained by eliminating /j, mj,
and /a, ifij, «a from the relations (2). This is conveniently done by
considering the following product of two zero-valued determi
nants :
/i
/i
0
mi
m.
0
n2
0
X
h
h
0
mg
mi
0
Ma
«t
0
0.
[Remember]
Multiplying the two determinants by row-by-rdw multiplica
tion rule, we have
2/i/a
/imo-h/ami
/iWa+Za^i
Wi/i-f-ma/i
2mima
niU+thh
miW2+W|«i
=0.
2«iWa
On putting the values of /i/g, lim2+kmi etc. from (2), we have
2a 2h 2g
a h g
2h 2b
or
2f
—0 or
h
b f
2g 2f 2c
8 f c
abc+2fgh -aP- hg^ ch^=^^.
..(3)
This is the required condition that the equation (1) represents
a pair of planes passing through the origin.
To find the angle between the planes. Let 9 be the angle bet
ween the two planes represented by the equation (1).
Then B is the angle between the planes lix-\-miy-\-nxZt=*(i and
/|JC+may+/i82=0 and so is given by
Va+Wima+«ina
where /i/|+mima-f ni«a=a+^+c
and
S (mi/ig—ma/i,)2=j:[(Wi«»+Wani)*^4mima/7ina]
The Plane
79
=S(4/2-4irc)
=4(/a-ftc)+4(^~ca)+4 {h^-ab),
so that [27(/w,W2 -/Mani)«]»/**=2V(/*+^*+*®-*c-ca-flZ»).
tan 0=2A/(P+g^+h«-bc-ca-ab)
...(4)
a+b+c
2^/(P-{-g^+h^-bc-ca-ab)
‘
or
dc=itan-i
<i4*b+c
Condition of perpendicularity. The two planes given by (I)
will be perpendicular if 0=^it. i.e. tan P=tan |te=oo. Hence the
relation (4) gives
a-}-b+c=0.
Thus two planes given by (1) will be perpendicular if
the coefficient of x^+tbe coefficient of y*+the coefficient of z*=0.
Solved Examples 3(£)
Ex. 1. Prove that the equation x^-\-Ay^ -z^4r4xy—0 represents
a pair of planes andfind the angle between them.
Sol. Comparing the given equation with the homogeneous
equation of second degree in x, y. z [See equation (1). § 18], we
have
a=l,6=4,c=-l,/=0,^=0, A=2:
abc-\-2fgh—aP—bg^—ch^
«1.4.(-l)+2.0.0.2-0-0-(-l)(2)2=-4+4=0.
Hence the given equation represents a pair of planes.
If d is the angle between the planes, then
tand=:^V(f^4-g^Arh^—be—ca—ab)
[See result (4), § 18]
fl+6+c
2-x/(0+0+4+4+l-4)
, putting for a, 6, c etc.
1+4-1
=iV5.
d=tan-i(iV5).
Alternative method. The given equation may be written as
(x2+4xy+4y2)-z2=o, or (x+2y)2—z*=0
or
(x+2y+z)(x-\-2y—z)=0.
x+2y+z=0,x+2y-z=0.
These being linear equations in x, y, z represent two planes.
If d is the angle between these planes, then
l.l+2;2+l.(-l)
4 2 ^
cos 0
V(l+4+l)V(J+4+l)“6“ 3
tan d=|v'5.
Ex. 2. Prove that the equation
2x2-6y2—12z»+18yz+2zx+xy=0
represents a pair ofplanes andfind the angle between them.
80
Analytical Geometry 3-D
Sol. Comparing the given equation with the equation (1) of
§ 18, we get
a=2,6= -6, C--12,/=i18-9,^g-4-2=1,
abcA-1fgh —a.p
ch^
-2(-6).f ^12)f 2.9.1
2.81 +6.1+ I2.J
=^1444 9-162+6 1-3-162—162=0.
Hence the given equation represents a pair of planes.
If ^ be Oie angle between the planes, then
tanl9=
h^—bc^g-^ab) [See eqn.(4)of§ 18]
fl+6+c
2y^(81 +1+^^72+24+12) 2Vri.U85)l
2-6-12
- 16
\/(l85)
“"
16
■
185 441
sec2 g= 1 +tan^ 0=1+ -''
256'256
21
16
sec 0=- or cos 0=7rr or d=cos
16
21
■■ i?
Alternative method. Arranging the given equation as a quad
ratic in X, we have
2x3+.v (2z+v)-(6v*+12z*- 18;^z)=0.
(2z-ry)±^[(2zi-y)^+4.2.(6y^-t-12z^'-18yz)]
X
2.2
or
4x=. 2z v+-\/[4z®+4zv +v*+484‘2+9622_1443;.?]
or
or
=-2z ± V(49;'2- 140vz+ lOOz^)
=-2z-j;+ V(7;»-10z)2
=-2z-;;±(7;;—lOz).
4x= “2z—;;+7;» ~10z and 4.y^=5—2z+v—7>’+10z
4x—6;V+12z=0 and 4,v+8>'—8z=0
2.Y—3j>+6z=0 and a-+2v—2z=0.
These being linear equations in .y, y and z represent the two
planes. If 6 is the angle between these planes then using
cos 0=
V (^1*+■ ■t:' j V
"b
2.r+(-^3)42>+6.(—2)
-16
cos0=
V(4+9+36) Vn + 4+4)“"21 ’
giving the obtuse angle between the planes.
If g is the acute angle between the planes, then cos <1=16/21.
g=cos-i (16/21).
8]
The Plane
Ex. 3. Sltow that the equation —^ -}
^
y—z z—x x-y
represents a pair ofplanes.
Sol. Multiplying the given equation by
(y-2)(z-x)
we have
a(z-x)<x~-;0+* iy-z)(x-j')H-.c (y-^z)(z -x)=0
or a (zx—yz—x^+xy)-hb(xy.-y^-zx+yz)
(yz -xy-z^+zx)-0
or
«) yz (c+a - b)zx
-(a+b-c)xy=0.
Comparing the equation (1) with the general homogeneous
equation of second degree in x, y,z l.e., the equation
^x^+ByHCz^+2Fyz+2Gzx-{-2ffxy=^0,[Refer eqn.(1) of § 18]
we have A~a, B~b, C=c, F=—^(^+c-a),
(c+a-b),
(a-hb-c),
[Note that we have used capital letters A, B etc. because
small letters b, c etc. are Used in the question.]
The equation (1) will represent.a pair of planes if
A
H
G
H
B
F
G
F
C
[Refer § 18]
f=0.
Putting for Ay By C etc. we have
A
H
G
H
G
B
F
F
C
a
(fl+6-c) -J(c+n-6)
b
--1 (b-\-c- a)
\{bArc-a)
0
c
0
-i(a+b-c)
ic+a~b)
0
-\{a+b-c)
b
kib+c-a) ,
c
-i(c+fl-fe) -I(b+c~a)
adding the second and third rows to the first row
=0.
Hence the given equation represents a pair of planes.
82
Analytical Geometry 3-/)
Ex. 4. If the equation
<ft (x, y^ z)=ax^+by^+cz^-\-2fyz+2gzx+2hxy=0
represents a pair ofplanes^ then show that the products of the distan
ces of the two planes,from (a,
y) is
y)
^f[r.a^-^ATp-~2Sbc\ ‘
Sol. Let the equation
^(x, y, z)^ax^A-by^-\-cz^+ 2fyz-\-2gzX’\-2lvcy^(i
represent two planes given by
liX-\-miy-\-niZ—Q
...(2) and l^x-^-m^’^-n^z^Q,
so that we have
4^ (x, y, 2)s(/ijf-fw,>'+n,z){kx+m^y+n^z),
where 4(x, y, z) is given by (1).
...(1)
...(3)
...(4)
Comparing the coefficients of like terms on either side of(4),
we have
l\li=a, m^m^^b, n\n%=c
w,«a+ma«i=2/,.«i/2+«a/i=2g, /im2+/ami=2A 1
...(5)
Letpiand pa be the perpendicular distances of the point
(^» A y) from the planes (2) and (3) respectively.
Then we have
Pi=
lx9.+mS
Multiplying, we get
(/ta4PiPi=
V(li^+
Wiy
» Pi—
V(V+Wg2+«a^)
(/ga+mg/g-f n^y)
V(V+
«g2)
...(6)
Now substituting a,f, y for x, y, z respectively in (4), we get
(/ia+w,^+wiy)(/ga-|-mg^+«2y)=^ (a, fi, y).
-(7)
Also
+{nhW-¥m^n^)A-{nx^l^-\-n^lx^)
—a^-\rb^+c^A-S {(/i/«2-|-/2Wi)2 -2/j/amjm3}
=a8-j-6»+c«+^ (Ah^-2ab),
using the relations (5)
=a^+b^-\-c^-^(4h^-2ab)H¥^-2br)-\-i4g^-2ca)
=(fl"+^'+c2)+4(/3+gH/«")-2 (6c+ca+fl6)
=Sa^-^4sp-^22:bc.
V(V+WiH«i**)\/(/a2+w2H»9*)=\/[^«*+42:/2-2r^c]. ...(8)
Substituting the values from the relations (7) and (8) in (6),
we get
PiPi=
(g> yg, y)
Vl2:a^+W^-2Sbc]
\
The Plane
83
§ 19. Projection on a.plane.
Recall the definitions of the projection of a point and the
projection of the segment of a line on a plane (see § 6, § 7 of
chapter 2).
i
Similarly the projection of an area A on sl given plane is
defined. Let ^4 be an area enclosed by the curve PQR...» Let
P\Q\R\ ... be the feet of the perpendiculars drawn from
P,Q,R,... to the given plane. Then the projection qfthe area>4
enclosed by the curve PQR ... on the given plane is the area A'
enclosed by the curve P'Q'R' .. Iftfis the angle between the
plane of the area A and the plane of projection, then A'=A cos 9.
Now we shall discuss two theorems on the projections.
Theorem 1. Let the projections ofan area A on the co-ordinate
planes yz^ zx and xy be A^, A,and At respectivelyy then
A^A^+Af^+A^. y
^
Proof. Let the.direction cosines of the normal to the plane
of urea A'bcl, m, n. Alsa the normal to the yz-p]une is *-axis
whose d.c.’s are 1,0, 0. If a be the angle /between the plane of
area A and the >>z-plane, then a is the angle ^^etween the normals
to these planes and so
cos a=/.l+w.0+n.0=/.
- Now the projection A„ of the area ^4 on the ;^z-plane is given
by
A,f=A cos ol=AI.
Similarly we have
Ay=Am, At=An.
Squaring and adding, we have
Theorem 2. The projection ofa given plane area A on a given
plane ^ is equal to the
of the projections of An* Ay and At on the
given plane I, where An, Ay and At are the projections of the area A
on the co-ordinate planes viz. yz, zx and xy-planes respectively.
Proof. Let /, m, n be the d.c.’s of the normal to the plane A,
and let /', m', n’ be the d.c.*s of the normal to the plane
Now if
9 is the angle between these two planes, then
...(1)
cos 6=W-{-mm'-}-nn\
Now let the projection of the area A on the plane | be A';
then we have
..(2)
A'==A cos 9 or A'—A (//'-f-iwm'+wn').
Also by definition and in view of theorem 1, we have
...(3)
Ax^Al, Ay^^Am, Ag-^An.
84
Analytical Geometry 3-D
From (2), we have
^'=^(^0 r+(Am) m'+(An) n'
==AJ'+A,m'+A^'
[using the relations (3)]
=(the projection of the area Ajg on the plane
+(thc projection of the area A,on the plane
+(the projection of the area As on the plane i).
Proved.
§ 20. Area of a triangle.
Tofind the area of a triangle ABC the co-ordinates of whose
vertices are A (jCi, y^, ,B
y^, z^) and C ix^\ Vs, Zg).
Let /, m, n be the d.c.’s of the normal to the plane of the
triangle ABC and let A denote the area of this triangle.
Let Ax, B^ arid C, be the projections of the three vertices A,
B and C respectively on the ^z-plane. Clearly the co-ordinates of
these points are given by A„(0, y„z,), BJfi, y„ z,\ C„(0,>>3, zj).
Let A» denote the area of the triangle AxB„Cx i.e. A» is the
area of projection of the area A on the j;z-plane, so that we have
A*=A./.
...(1)
Also by the co-ordinate geometry of two dimensions,
I
yi
yz
1
Z2
(2)
1
Similarly if A» and A* are the areas of the projections of the
area A on zjc and jrj;-planes, then
Av=A»w
(3)
and
...(4)
1
I
Xi
Zy
y^
where
A»=i ●
Xz
Xz
Zg ■
1 » Ac==*i
xz
n
1
1
^8
yz
1
Squaring (1),(3) and (4) and adding, we get
A«?+ A,®+A*^= A"*(/®+wH«®)=A".l
or
A«=A,HA,HA/.
This gives area A of the triangle'^^C. *
...(3)
Solved Examples 3(F)
Ex. 1. Find the area ofthe triangle whose vertices arc ^(1, 2, 3),
5(2, -1, l)fl«dC(l,2, -4).
(Aga 1979)
Sol. Let A*» Av» As be the areas of the projections of the
The Plane
85
area A of triangle ABC on the yz, ix and xv*p1anes respectively.
We have
●
1 I
3^1
A»—^
Pi
Za
1 =\
Za
1
. i
Av—
=i
2
3
1
—I
1
1
1
2-4
3
1
2
I
1
21
2*
1
7
-2
^3
Za
1
^8
Za
1
1
-4
1
yx
1
1
2
1
3^2
1 =\ 2-1
1 =0.
Vi
1
1
(numerically).
and As=i
●^2
1
2
the required area A=V a»ha,ha»’^]
7(^ + T+”)=iV(490) square
units.
Ex. 2. A plane makes intercepts OA=Ot OB=b and OC=c
respectively on the co-ordinate axes. Show that the area of the tri
angle ABC is ^y/{b^c^-\-c^a^-\-a^b^).
Sol. The points A, B and C lie on the axes of x, y and z
respectively, so that their co-ordinates are A(a, 0, 0) .5(0, 6, 0),
C(0, 0, c).
Let A denote the area of the triangle ABC. The projection
of triangle ABC on the ;;z-plane is the triangle OBC and if A»
denotes its area then
Ax^hOB.OC==lbc.
The projection of the triangle ABC on the zx-plane
triangle OCA and its area Ay is given by
Ay=^.OC.OA=^a,
Also the projection of the triangle ABC on the xy-plane
triangle OAB and its area A* is given by
A,-=‘hOA.OB==lab.
the area A of the triangle ABC is given by
A*H- A*H A/+ A*®=i
or
...(1)
is the
is the
...(3)
Ex. 3 Find the area of the triangle included between the plane
3jc—4j;+z = 12 and the co-ordinate planes:
Analytical Geometry 3-2)
86
Sol. The equation of the given plane is
...(1)
3x—4j^+z=12 or x/4-^/3+^/12=1
The plane (1) meets the co-ordinate axes in the points
A(4,0, 0), J(0, -3,0) C(0, 0, 12).
Using the notations of Ex. 2 above, we have
A,=iOAOC=i(-3).12=18 (numerically),
A,=J.OC.O^=il2.4
=24,
As=iOi4.0B=i.4.(-3) =6(numerically).
area A of the triangle i45C=V(A»“+ A»®+A«*) =1 V(1S*+24H6*)=3V(26)square units.
Ex. 4. From a point P{x\ y\ z') a plane is drawn at right.
angles to OP to meet the co-ordinate axes at A,Band G. Prove that
the area of the triangle ABC is r»/(2xyz'), where r is the measure
(Kanpur 1983, 84 ; M.U. 1990)
ofOP.
Sol. The d.r.’s of the line joining O (0,0,0) and P{x\ y',z')
arejc'-0,/-0,z'-0/.e., x',y, z'.
..
/. the equation of the plane through P{x\ y\ z') and per
pendicular to OP is given by
jc'(x—*')+/
(z—z')=0,
or
XX*-\-yy ■\‘Zz'
or
XX*4-yy*
^r^ [V r=OP=V(*'*+3''®+*'*)]
X
^+—
= 1.
r^fx* r^ly* r*Iz*
The plane (1) meets the co-ordinate axes in the points
A{r^jx*, 0. 0), B{0,rVy\ 0) and C(0, 0, r^lz*).
Let A
file area of the triangle ABC, Also let A» he the
area of projection OBC on the j^z-plane of the area of triangle
ABC. We have
r* r
ty,^\OB,OC^\
...(2)
yz
or
No^.r.*s of the normal to the plane of ^ABC i.e., d.r.’s of
thei^e OP are x*, y*, z', so that the d.c.’s of this normal are
[V r=V(*'*+P'*+a''*)]
x*lr,y*lr,z*lr.
Also d.c.*s of the normal to the- plane of triangle OBC i.e.,
d.c.*s of x-axis are 1,0, 0.
If a be the angle between the planes of tringles
and
OBC, we have
cos a=1 ,{x*lr)+0,(y*lr) -f 0.(zV)=xV.
' Now
A»=A cos a.
87
The Plane
●●
2y'z'=A.f.
or
r®
Exercises
1. . A plane makes intercepts 9, 9/2» -9/2, upon the coordi
nate axes. Find the length of the perpendicular from the origin
(Ans. 3)
on it.
2. Show that the four points(0,-1, 0),(2, 1, —1),(1, 1, 1)
and (3, 3, 0) are coplanar and hence show that the equation of
the plane passing through these points is 4x—3^+2z=3.
(Agtal978)
3. Show that the four points (0, 4, 3), (—1, — 5,-3),
(-2, -2, 1) and (1,1, —1)are coplanar.
4, Find the equation of the plane through the origin and
parallel to the plane 3x-l-9j>-'}z4-5=0.
(Ans. 3x-f9;^—7z=0)
5. Show that the planes 3X-1-4;;—5z=9 and 2x-\-6y+6z=l
are at right angles.
Hint Two planes are at right angles if their normals are at
right angles. The d.r.’s of the normals to the two given planes
are 3, 4 —5 and 2, 6, 6 respectively and we see that these lines
are perpendicular.
6. Find the equation of the plane which contains the line of
intersection of the planes x-\-y-\-z—6=s0 and 2x4-3;>4-42-1-5»0
and is perpendicular to the plane 4x-|-5j>—3z—8=0.
(Ans. x4-7y-l-13z-l-96=0j
4
The Straight Line
§ 1. General equations of a straight line.
We know that every equation of first degree in.v,.v and z
always represents a plane (see chapter 3).
Now let us take two equations of first degree together
i.e.
az;c+^+«?a^+</2=0.
...(1)
Any point which simultaneously satisfies both the equations
given by (1), will lie on the curve of intersection of the planes
given by (I). Since the two planes intersect in a straight line,
therefore the equations (I) represent the equations of a straight
line. The equations (1) are called the general equations of a,
straight line. Therefore any two equations offirst degree inx^y
and z taken together always represent a straight line.
§ 2. Symmetrical form of the equations of a straight line.
(A) Tofind the equations of a straight line which passes through
a given point (ati,
Zj) and whose direction cosines are I, m, n.
- Y
Let OX, OY, OZ be chosen
as rectangular co-ordinate axes.
Let A be the given point
>'i» ^i) on the line. Choose
a general point P with co-ordi
»
nates
y, z) on the line at a
1
A,'
distance r (say) from the given
M
X
point A. Draw perpendiculars
AM and PN on the jc-axis from
the points A and P respectively,
so that MN is the projection of the segment AP on the .v-axis
and is given by
MN.-AP-1 or x-Xi=^rl.
Similarly projecting the segment AP on the y-axis and z-axis.
9
The Straight Line
we get
V—3^1=rm and z—ri=r«.
Therefore the co-ordinates (x, y, z) of any point P on the line
satisfy the equations
^
~r
..:(!)
/
m
n
'
It should be noted here that ‘r’ is the actual distance of any
point P(x, z) on the line front the given point (xi, yi, ^i)*
x-xj y y,
Hence
...(2)
m
n
1
are the equations {symmetricalform) of the straight line.
Vector method. We have
^=position vector of P—position vector of A
«(xi-{-yj+2k) -<xil+yij+zik)
=(x-x,)i+(y -yi) j+(r 2i) k.
_
Since /, m, n are the direction cosines of the straight line >4r,
therefore a unit vector along the line AP is /!+/«].+wk.
Now the vector AP is collinear with the unit vector /i-j-mj-l-nk.
/. 5p=r (/i-f/«j+»k), where the scalar r is the distance AP.
(x-Xi)i-l-(y -yj) j-f(2-21) k=r/H-rmJ-l-r«k.
Equating the coefficients of.i, j, k on both sides, we get
x-Xi=r/, y ~-yi=rm,z-2i=rn.
. x--xi y-yi
n
m
are the required equations of the given straight line.
(B) Symrnetricalform in terms of airection ratios.
Let the direction ratios of the required line be a, c. Hence
c
a
b
are the direction
cosines of the line. Thus the equations (2) of the line become
or
X—Xi
y-yi
a
6
a
b
2-Zi
c
(say).
...(3)
The equations (3) are the required equations of the straight
line. It should be noted here that *r' is not the actual distance
of any point P(x, y, z) on the line from the given point(xi, yi, z,).
90
Analytical Geometry 3-D
From equations (2) and (3) we observe that the form of equa
tions of the straight line remains unaltered if we use direction
ratios instead of direction cosines.
Corollary. From equations (1) and (3), the general co-ordi
nates of a point on a line are given by
(jCi+Zr, yx-\-mry Zj+zir) or (jCi+or, yx+br, Zi+cr).
(c) The parametric form.
In the equations (1) and (3), r represents a real number which
changes a the position of the point P on the line changes, so
that r is a parameter. For convenience let this parameter r be
denoted by *t\ Hence parametric equations of the straight line
are given by
.
X=Xi-|-lt, y=yj+mt,Z=Zj+nt
...(4)
x=Xi-Fat, y=yi4-bt,z=zi+ct
...(5)
where /, m, n and a, bt c are direction cosines and direction ratios
of the line respectively.
Note. If we use the actual direction cosines then the co
or
ordinates of a point (on the line) distant r from the given point
(^i» Pi,
and (xi+/r, >>i+wr, Zi+nr). From this point of view
the equations (I) or (2) are also called distance form of the equa
tions of the straight line.
§ 3. Line through two points.
Tofind the equations of a straight line passing through two
points whose co-ordinates are (xi, yx, Zx) and (xa, y^y z^.
Let A (xi, ^1, zi) and B (xa, y^^ za) be the given points through
which the line passes. The direction ratios of the line passing
through the points and B are Xg-Xi, y^^-yx^ z^~Zx.
Hence the required equations of the straight line are given
by [from equations (3), § 2(B)]
x-Xi
xa-xi
y-yi _ z-zt
y2-yi Zj—Zj
...(1)
SOLVED EXAMPLES(A)
Ex 1. Find the point in which the line
4
meets the plane x—2j?+z==20.
Sol. The equations of the line are
x-2 y-^-l z-2
, ,
—
4 «^ =r (say).
The equation of the plane is x—2;;-|-'2i=20.
12
...(1)
...(2)
The Straight Line
91
The co-ordinates of any point Q bn the line (1) are
(24-3r,-H-4r,2+12r).
-(3)
Suppose the line (1) meets the plane (2) in this point Q, hence
we have
(2+3r)-2(-1-|-4r)+(2+120=20, or r=2.
Putting this value of r in the co-ordinates of Q given by (3)
the co-ordinates of the required point are given by
(2-f 3.2, -1-1-4.2, 2-f 12.2) or (8, 7, 26).
Ex. 2. Find the co-ordinates of the point where the line joining
the points {2, -3, 1) and {3, -4, -5)cuts the plane 2x+>'+z=-7.
Sol. The direction ratios of the line joining the points
(2, -3, 1) and (3, -4,. -5) are 3-2, -4-(-3), -5-1 fe.
1,-l»-6.
Hence the equations of the line joining the given points are
...(1)
x-2_y+S_ z~\ r. (say).
3 -1 -6
The co-ordinates of any point on this line are
...(2)
(r-f2, -r-3, - 6r-f 1).
It this point lies on the given plane 2x+y+z=l. we have
2(r+2)-|-(--r-3)-l-(-6r4-l)=7, orr=-l.
Putting this value of r in the co-ordinates of the point given
by (2), the co-ordinates of the "required point are given by
(1-2,7).
Ex. 3. Show that the distance of the point of intersection of
the line "
and the plane x-y-^z^Sfrom the point
(-1,-5,-10)15 13.
Sol. The equations of the given line are
l„z-2_ .
...{1)
■^4
12
The co-ordinates of any point on the line (1) are (3r-|-2,
4r-l, 12r-l-2). If this point lies on the plane x->'-fz=5, we
have 3r-l-2-(4r-l)-M2r-|-2--=5, or llr=0, or r«0.
Putting this value of r, the co-ordinates of the point of inter
section of the line (1) and the given plane are (2, -1, 2).
.*. The required distance=distance between the points
(2, -l,2)and(-l, -5, -10)
=V{(2+I)H(-1 + 5)2-K2-|-10)2}
=V'(9+i‘6 + 144)=V(169) = 13.
I
92
Analytical Geometry 3-D
Ex. 4. Find the points in which tne line
_y -12 z-l
-f” 5
Y~ cuts the surface 1 \ x^-~5y^+z^=0.
Sol. The equations of the given line are
^+1 ;' -12 z-'l
, ,
...(1)
The co-ordinates of any point on the line (I) arc
( - r-l,5r-H2, 2r-l-7).
If this point lies on the given surface
11 -5y~4- z2~0, we have
or
or
or
●
..(2)
!1(-r -1)2-5 (5/-M2)2+(2r+7)2=0
11 (r2+2r-f 1)—5 (25r2-f 120/*-!-144)-}-(4r2-{-28r-f49)=0
-110r2- 550r- 660-0, or r2f5r+6=:0
(r+2)(r-l 3)^0 or /
?.
P utting these values of r in (2), the required points of inter
section are (I, 2, 3) and (2, -3, 1).
,.
of
Straighl lines through the point
(a. b C) n'lnch are (i) parallel to z axis (i.e. perpendicular io rite
xvthnJ)
Perpendicular to the z-axis {i.e. parallel to the
Sol. Let the equations of any line through the point (a, h, c)
be x-~a y-b z — c
. I *= ni
...(1)
where /, /;?, n are the d.c.’s of the line,
(i) Thed.c.’softher-axis (/.e. ofthc line perpendicular to
the xj.-plane are 0. 0. l. Hence if the line (I) is parallel to z-axis
or perpendicular to the .vy-plane then !, m, n arc proportional to
u, 1. Hence the equations of the required line are
X-a
y-b z—c
0
fii) If the line (1) is perpendicular to the z-axis or parallel'
to the A>plane then we have
/.04-OT.0-i-/;.l=0 or w=0.
fherefore in this case the equations (1) of the required line
are given by
x—a
X
m
o~ '
Ex. 6. Find the distance of the point (1, —2, 3)from the plane
^
2 3“-6
x-y+z=-5
^ measured parallel to the line
93
The Straight Line
Sol. Note here that we are not required to find the perpen
dicular distance of the point (1, —2,3) from the given plane, but
we are required to evaluate the distance of the point(1, —2, 3)
from the given plane measured parallel to a line whose direction
cosines are proportional to 2, 3, —6. For this we proceed as
follows :
The equations of the line through the point (1, —2,3) and
parallel to the line whose direction cosines are proportional to
2, 3, ~6 are given by
The co-ordinates of any point on it are (2r-|-l, 3r—2, “6r-j-3).
If this point lies on the given plane x^y-\-z=5^ we have
2r+I -(3r-2)4-(-6r+3)=5, or 7r——1, or r=l/7.
/9 —111 15 \
●. The point of intersection is 1 y. -y ’ 7"j
The required distance=The distance between the. points
(1, -2. 3) and (9/7, -11/7,15/7)
7K'4)V(-4)V(>-^n
= 1/7 V{4+9+36}=1/7.7=1.
Ex. 6. Find the equations of the straight lines which bisect the
angles between the lines xlli=y/mi—zlnif xjl^=ylmi=zlni.
Sol. Proceeding as in Ex. 17 page 38 chapter 2, the direction
cosines of the bisectors are proportional to /idr/2, Wi±/«a»
Clearly the given lines pass through the point (0, 0, 0) and hence
their bisectors also pass through the point (0, 0, 0) and so the
required equations of the bisectors are
z
X
y
ly-i-h
«i±W2
Ex. 8. Find the equations of ■ the line
(Xi, yi, Zi) at right angles to the lines
through the point
z
»1
Wh
'8
Wg
«a
Sol. Let the equations of the required line through the
point (x„ 7i, Zj) be
X- Xi
/
■
y -yi__^-Zi
m
" n
’
...(1)
94
Analytical Geometry 3-D
where the d.r.’s /, m, n of this line are to be determined. Since
line (1) is perpendicular to the given lines, hence we have
...(3)
UiA-mmi+nni—Q
...(2) and ll2+mmi+nn^=0.
Solving the equations (9) and (3), we h?(ve
/
m
n
— n^li~Iim2—l^i
Putting these proportionate values of /, m, m in (1), the equa
tions of the required line are given by
X-Xi_ _ ,y-yi _ Z-2i
mjWa — maWi «i/a—
*
Ex. 9. Find the equation of the plane through the point
(a, y) and if) perpendicular to the straight line
{x-~xfifl={y-yx)lm={z-z;)ln,
in) parallel to the lines xlli=yfmi—zlni and xfl2~ylm2=z}n2.
Sol. Let the equation of any plane through the point
...(1)
(«, p, y) be A (x-c/.)-{-B(y~-p)+C(z-y)=0.
(i) The equations of the given line are
...(2)
(jc-Xi)//=(y-y,)/m=(2~zi)/n.
If the plane (1) is perpendicular to the line (2), then the
normal to the plane (1) is parallel to the line (2); and hence A,
Bf C, the d.r.’s of the normal to the plane (1), are proportional
to /, m, n. So the equation of the required plane is given by
/(x-a)-l-m (y-j9)+« (z~y)=0.
(ii) The equations of the two lines are given as
xlli=ylmi=zlni and xik=y/m2-=zln2.
The plane (1) will be parallel to these lines, if its normal
whose d.r.’s are A^ D, C is perpendicular to both the given lines.
Hence we have Ali-^Bmi+Cni—O and Ali+Bm^+Cnz^O.
Solving these equations, we have
A
B
B
/WiWj-W2«i~" Wi/g—
Putting these proportionate values of A, B, C in (1), the
equation of the required plane is given by
(miWa-WaWi)(x-a)4-(/ii/a-W2/i)(y-P)-h(km2- knh)(z-y)=0.
Some Examples on the foot of perpendicular from a point to a
plane.
Ex. 10. Find the co-ordinates of the foot of the perpendicular
drawn from the origin to the plane 3x+4y-6z-f-l =0. Find also
the co-ordinates of the point on the line which is at the same distance
from thefoot of the perpendicular as the origin is.
95
The Straight Line
Sol.
The equa^tion of the plaae is 3x+4_)/—6r-t*l=0. ...(1)
The direction ratios of the normal to ●he plane (1) are
.3,4, —6. Hence the line normal to the plane (1) has dr.’s
3, 4,-6, so that the equations of the line through (0, 0, 0) and
perpendicular to the plane (1) are
...(2)
jc/3=j/4=z/-6=r (say).
The co-ordinates of any point P on (2) are
...(3)
(3r, 4r, -6r).
If this point lies on the plane (1), then
3(3r)+4(4r)-6(-6r)-M«0, orr=-l/61.
Putting the value of r in (3), the co-ordinates of the foot of
the perpendicular P are (—3/61, -4/61, 6/61).
Now let Q be the point on the line
which is at the same distance from the
foot of the perpendicular as the origin.
Let (x,,
zi) be the co-ordinates of the
point Q. Clearly P is the middle point
of OQ. Hence, we have
4 zi+0 6
Xi-t-0
3_ ^
■'61’
2 “61
2
"61’
2
or
/
E7
O
xi= -6/61,
-8/61, Zi=12/61.
.●. The co-ordinates of Q are (—6/61, --8/61, 12/61).
Ex. 11. Show that if the axes are rectangular, the equations
to the perpendicular from the point (a, p, y) to the plane
ax-Vby-\-cz-{-d=Q are (x-a.)la=(y ~^)lb={z—y)lc.
Deduce the perpendicular distance of the point (a, P, y)from the
plane. Find also the co-ordinates of the foot of the perpendicular.
Sol.
The equation of the plane is
...(1)
flX+hv-|-cz+f/=0.
The direction ratios of the normal to the plane (1) are a, b, c.
If the line is perpendicular to the plane (1), then it is parallel to
the normal to the plane, Hence the d.r.’s of the line perpendi
cular to the plane (1) are a, b, c. Also it passes through the
point (a, p, y), therefore the required equations of the perpendi
cular are
x—v.
a
y—p z—y
=r (say).
b ~~ c
...(2)
The co-ordinates of any point on the line (2) are
ihr+ff.,br-\-p,cr+y).
...(3)
96
Analytical Geometry 3-D
If this point lies on the plane (1), we have
a (ar+a)+b {br-\-fi)+c (cr-\-y)-\-d=0
or
r (a^-\-b^i-c^)z= —(aa+b^+cy+d)
or
...(4)
r=—(acf.+afi-\-q>-d)l(a^+b^-^c^).
Putting this value of r from (4) in (3), we get the co-ordinates
of the foot of the perpendicular.
Now the perpendicular distance of the point (a, /?, y) from
the plane (l)=the distance between the points (a, y) and
(ar+«, br+pf cr-\-y)
=V[(a^+«- a)H(br+P-fi)^+(cr+y~y)^]
=r’\/(a^-^b^-jrc^)
(axA-b/i+cy+d)
a»-hbp-\-cy+d*
“ (a^+b^-\-c^)
Ex 12. Find the equations of the line through (I, —1,2) per
pendicular to the plane 3.v+ 5y—Az= 5 and deduce the length of the
perpendicularfrom {\^
1,2) upon the plane and also the co-ordi¬
nates of thefoot of the perpendicular.
Sol. Proceeding as in Ex. 11 above, we get
(i) the equations of the required line as
x—\ ■ y-^l 2 2 ^, ,
3
Y=Z^=^(say)
...(1)
(ii) The co-ordinates of any point on the line (1) are
(3r-l-l, 5r-l,-4rH-2).
...(2)
If this point lies on the pl^;ne 3x-|-5j> - 4z=5, we have
5(3r-H)4-5(5r-l)-4( -4rH-2)=5. or r=3/10.
Putting the value of r in (2), the co-ordinates of the foot of
the perpendicular are (19/10, 1 /2, 4/5).
(iii) The required distance
=V[9/10)H(3/2)H(6/5)2J=3V./2.
Ex. 13. Find the incentre of the tetrahedron formed by the
planes
x=0^y=a,z=Q and x-^y-{-z=a.
Sol. The three planes a:=0,;'=0. and z=0 meet in the
point (0, 0, 0). Hence the incentre of the tetrahedron lies on the
perpendicular from (0, 0, 0) to the plane .v-f-yrfz=a.
The d.r.’s of the normal to the plane x-^y-^ z=a are 1, 1, 1.
Hence the equations of the perpendicular from (0, 0, 0) to the
plane x-i-y^z=,a are
97
The Straight Line
...0)
f=f=l=r (say).
The co-ordinates of any point on (1) are (r, r, r).
If this point lies on the plane x+y+z=a^ we have
r+r+r-a, or r=ia.
Hence the foot of the perpendicular is [or the perpendicular
from (0,0,0) meets the plane x+y+z=a ini (ia, ia, ia),
Let the incentre of the tetrahedron be (Xi,yi,Zi). Now we
know that the incentre divides the join of(0, 0,0) and (-Jn, io, ia)
in the ratio 3: 1 [3 on vertex side and 1 on plane side]. Hence we
a
3.ifl-H.0
have
^1=
=yi='Zi
or
Xi==yi=zi==
^
3-4-1
The co-ordinates of required incentre are (Jo,
itt).
Ex.14. A variable plane makes intercepts on the co-ordinate
axes the sum of whose squares is constant and equal to kK Show
\that the locus of thefoot of the perpendicular from the origin to the
plane is ix-^+y^^+z-^)ix^^-y^-\-z^)=k^.
Sol. Let the equation of the variable plane be
...(1)
xja-{-ylb^zlc—\
where a, b, c are its intercepts on the co-ordinate axes, so that
a^+bHc^=k^The d.r.’s of the normal to the plane (1) are 1/n, l/h» 1/e*
Hence the equations of the line perpendicular to the plane (1)
● [/.e. parallel to the normal to the plane] and passing through the
x.-O y--0 z-0
=r (say).
origin are
(3)
1/fl ■“ 1/6 " I/c
(4)
Any point on line (3) is {rfa^ rlb^ rfc).
If this point lies on the plane (1), we have
1
r/a2-l-r/62-l-r/c2=l or
Let (x^ j;, z) be the co-ordinates pf the foot of the perpendi
cular, then putting the value of r in (4), we get
ar^
1
or X—
c-i
Similarly
and.
Now x®-4-y*r+-^=
a-®+i-»4 C-*
●
1
...(5)
and
=(fir84 6-2+c"®)W,'uMng (2).
--
...(6)
98
Analytical Geometry 3-D
Multiplying (5) and (6), we have
(;C-2+j;-2+2-2)
This is the equation of the required locus.
Ex. 15. Find the equations of the line through the points
(a, b, c) and {a\b\ c') and prove that it passes through the origin if
aa’-\rbb''\-cc'=rr\ where r and r’ are the distances of the point
from the origin.
Sol. The equations of the line through the points (cr, fe,c)
and (a', b*, c') are
x—a y—a z—c
a'-a"‘b'-b"'c'-c
...(1)
The line (1) passes through the origin, hence we have
0-fl 0—6 0—c
a* a b'—b c*—c
or
a*~a V—b c'—c
—a
b
c
6‘
c
or
.a
or^_
6+
' 1=_^+1
c
a
b .c
From these relations, we.immediately get
.
o'6=0, 6c'—6'c=0, ca'—c'a=0.
...(2)
By Lagrange’s indentity, we have
(a*+6»4.c8)
-(flfl'+66'-|-cc')*
=(fl6'-a'6)a+(^>c'-6'c)2+(ca'
=.0+0+0
or
(fl»+6a+ca)(a'H*'Hc'2)*=(flfl'+66'+cc')2
...(3)
Now r=the distance of(a, 6, c)from the origin
==V[(«-0)H(6-0)H(c-0)8]=V'(uH*Hc*).
Similarlyr'=»V(«'*+i>'HOPutting these values in (3), we have
(flw'+66'+cc')*«=3(rr')*. or afl'+66'+cc'=rr'.
This is the required result.
Ex. 16. Find the image of the point (1, 3, 4) in the plane
2x—y+z+3-0.
Sol. The equation of the plane is
2x-j;+2+3=0.
...(1)
Let (1, 3,4) be the point P. Draw a perpendicular PN from
the point P to the plane.(1). , Take a point Q on this perpendi
cular on the other side of the plane(I). Then 0 is called the
image ofP if JVr the foot of the perpendicular^ is the middle point
of PQ. Let the co-ordinates of Q be (Xi,
Zi).
/
<6
The Straight Line
99
The d.r.’s of the normal to the plane (1) are 2, —1, 1. Hence
the equations of the line through (1, 3, 4) and perpendicular to
the plane (1) are
x—\
z—4
r (say).
2 ~-l “ 1
...(2)
Any point on (2) is (2r+l, — r+3, r+4). If this point is the
foot of the perpendicular i.e. the point N, then it will lie on the
plane (1) and we have
2(2r+l)-(_r+3)+(r+4)+3=0,
or
6r+6=»0, or r= —1.
Putting this value of r, the co-ordinates of are (— 1, 4, 3).
Now as explained above, N is the middle point of Pg. Hence
we have
-12 ’
2 ’ ~
or
Xi
3, yi—5, Zi~2»
The image of P(1, 3, 4) is the point Q(—3,5, 2).
§4. To transform the general form of the eqnations of a straight
line to symmetrical form.
Let the general form of the equations of the straight line be
given by the equations
...(1)
Now we are required to write down the symmetrical form of
the straight line given by equations (1). For this we must know
(i)the direction cosines or direction rajtios of the line and.(ii)the
co-ordinates of a point on the line. To find these two we proceed
as follows:
Step 1. To find direction cosines or the directiah ratios of the
line given by equations (i). Let /, m, « be the direction cosines or
direction ratios of the line. Since the line is comnion to both.the
planes and therefore it is perpendicular to the normals of botb the
planes. The direction ratios of the normals to the planes given
by equations(1) are Oi,
c, and Og, hg, c# respectively. Hence
we have
/ni+mhi-}-nci=0 and /aa+mha-l-«C8=0.
Solving these equations for /, w, n, we have
I
m
n ●
bxc%—b^x~Ciai-c^i~~axb2~af)f
6»-
100
Analytical Geometry 3-D
the direction ratios of the line are
...(2)
Step 2. Tofind the co-ordinates ofa point on the line given by
equations (1). The co-ordinates of a point cn a line can be chosen
in many ways. One of these ways is that we choosey the point as
the one where the line cuts the jry-plane {i.e. z=0 plane), provi
ded the line is not parallel to the plane
i.e. provided
biC2—bjfiiy
aib^—a^X’
Putting z=0 in both the equations given by (1), we get
ayX-\-bxy-\-di=0, OgX-f
</a=0.
. Solving these equations for x, y^ we get
1.
X
y
byd^—b^i
dxO%—d^x
^1^2 — ^2^1
Hence the co-ordinates of a point on the line (1), where it
cuts the z=0 plane are
bxd2~b^x difl^—d^Ox
...(3)
( a-Jy^-^ajbx
Hence the equations of the line in symmetrical form are
dxU2—d^x\
2—0
—tijji /
qipi—tj^bx
bxC^—bjfix
^1^2 ^2^1
Note. If fljfta—Oa6i=0, then instead of taking 2=0 we should
take the point where the line cuts x=0 plane or y=0 plane.
■4
SOLVED EXAMPLES (B)
Ex. 1.
Find in symmetrical form tfie equations of the line
3x-\-2y~z-T4=0=4x-\-y—2z+3andfind its direction cosines.
Sol. The equations of the given line in general form are
-(1)
3;c-l-2y- 2-4=0, 4x-l-y - ?2-{-3=0.
Let //m, /t be the d.c.’s of the line. Since the line is common
to both the plaiies, it is perpendicular to the normals to both the
planes.
Hence we haVe
3/-f 2^-n=(), 4/-f-m- 2«=().
I
n
m
Solving these, we get
-4-f l“-4+6”3-8
1
/
m
n
or
-3 ~ 2 "“-5 -^(9+4-b25) :^VW
the d.c.*s of the line are -3/V(38), 2/V(38), -5/V(38).
Now to find the co-ordinates of a point on the line given by
101
The Straight Line
(1), let us find the point where it meets the plane 2^0. Putting
2=0 in the equations given by (I); we have
3jc+2;>—4=0,4x+j^+3=0.Solving these, we get
1
, X
y
, or x= -2, y—5.
6+4“-16-9“3-8
/. The line meets the plane 2=0 in the point(—2,5. 0).
Therefore the equations of the given line in symmetrical form
are
jc+2 y—5 2--0
=T““2"”-5 ‘
Ex. 2. Find the equations to the line through the point (Ij 2, 3)
parallel to the line x—y-\-2z—5~0; 3x+^-f;2—6=0.
.
[Jodhpur 1967]
Sol. The equations of the given line in.general form are
...(1)
X -y+2z—5—0,3x-\-y+z
Let /, m, n be the d.c.’s of this line. Then we have
/-m+2n=0,3/+m+rt=0.
Solving these, we have
/
m . n
n
m
- /
-1 -1""2X^1""rUKLS’
"" 4*
Since the required line is parallel to the line (1), the d.c.*s of
the required line are proportional to /, m,» i.e. —3, 5, 4.
Hence the equations of required line are given by
(jc-1)/-3=(;.-2)/5=(2~3)/4. .
Ex. 3. Find the equations of the line through the point
2i) and parallel to the line
flxX+hlJ'+Ci2+</i=0, ffaV4 ^8.V+C22-1-<4=0.
Sol. Let /, w, n be the d.c.’s of the given line, then procee
ding as in § 4, we get
/
w
n
(Xiy
hipi—
Cjfl^—CtjCii
—hjfi
..■(1)
The required line passes through (Xi, yi^ Zi) and its d.c.’s afe
proportional to/, m,R given by (1). Hence the equations of the
required line are given by
Ex. 4. Find in symmetricaiform the equations of the line
x=ay’{‘b^z=cy-\-d.
102
Anatyticat Geometry 3-D
. Sol. The equations of the given line may be written as
X—qy+0.z-h=0,0.x+cy-z-\-d=^0,
...(■1)
Let /, w, n be the d.c.*s of the line. Then we have
!./-fl.m+0.rt=0,0./+c.w—l./i=0.
.*. l=an^ cm^iiy so that —=~=-^.
«
1
c
...(2)
Now let us find the point where the line (1) meets the plane
Patting
in the equations (I), we get x=h and z=rf.
The line meets the plane
in the point (^,0, d).
Therefore the equations of the given line in symmetrical form are
{x-b)/a=(y-0)fl={z~d)lc.
Alternative method. The given equations of the line may be
written as
x^b—ayyZ—d=cy
x-^b _ y z—d _ y
or
1 » c
a
x—b __ y _z~d.
a
I
c
These are the required equations of the line in symmetrical
form.
Ex. 5. Show that the lines
x=fl>+h, z=cy-\-d and x=o>-l-6',
are perpendicular if a(^ArCc'-\-\=0.
Sol. Proceeding as in Ex. 4 above, if the d.c.*s of the line
x^ay-\-b, z=cy+d are /j, imj, /i„ we have
lifa'=mjl=^nilc\
...(2)'
Similarly if /j, m,, «a are the d.c.’s of the line x=a'y-\-b'\
then we have
i.e.
§ S.
lila'^mj\=n2lc,\
The given lines will be perpenicular, if
lili-{-mjn7i+nin^^0
aa*+\.\-\:Cc*^0t or oa'+cc'+l=0;
...(2)
The plane and the straight line :
To find the co-ordinate of the point of intersection of a given
line and a given plane and to deduce the conditions that:
(0 the line may be parallel to the plane^
{ii) the line may be perpendicular to .the plane, and
{Hi) the line may be lying in the plane
103
The Straight Line
Let the equations of the given line in symmetrical form be.
x-xi y-yj z-zi
/
m
n =r (say)
...(1)
and the equation of the given plane be
ax+by+cz+d=0.
...(2)
The co-ordinates of any point i» on the line (i) are
If this point P lies on the plane (2), we have
a(Xi-\-lr)+b{yi'^mr)+c (z^-f nr)+rf=0
or
r(ai'\-b,n-{cn)=—(axi+byi-\-czi-^d)
r= (axi+byi-\-czi-{-d)l(al-hbm-\rcn}.
or
...(3)
Thus the co-ordinates of the point of intersection P of the
line (1)and the plane (2) are (xi-\-lr, Ai+mr, z,-l-/ir) where r is
givn by (3).
(i) The conditions of paralelism; If the line (1) is parallel
to the plane (2), then this line must be perpendicular to the nor
mal to the plane (2) and hence we have ● ajrbm-\rcn=0.
Again the point
Zi) should not lie on the plane i.e, we
must have axx-\-byx-^cz\-\-d^Qt for otherwise the line (1)^ will not
be simply parallel to the plane (2) but it will lie in the plane (2).
Hence the required conditions that the line (1) is parallel to
the plane (2) are
al-i-bm+cn=0, axj-l-byi-i-czi-fd^tO.
(ii) The condition of perpendicularity. If the line (1) is
perpendicular to the plane (2), then the line (1) must be parallel
to the normal of the plane (2) and hence the required condition
of perpendicularity is given by
b
c
a
m
1
(iii) The conditions that the line may lie in the plane.. If the
line (1) lies in the plane (2), then for all values of r the point
P(X\-\-lr^
Zi+«r) will lie on the plane (2) i.e,
fl(JCi+fr)+h(yiH-wr)-f-c(zi-|-nr)d=0
or
6yi+czi+d)=0
should be true for all values of r, and hence we must have
Hence the required conditions that the line (1) lies in the
plane (2) are
104
Analytical Geometry 3-Z)
al+bm+cn :i=0, axj4-byi+czi4 d=0
These conditions can also be deduced as follows :
If the line (1) lies in the plane (2), then the line is perpen
dicular to the normal to the plane (2), for which we have
a/+Z>/«+c/i —0.
Again the point (Xi.yi, z,) must lie on the plane (2)[since the
line (1) is passing through the point (ati, yj, Zj,)], hence we have
SOLVED EXAMPLES tC)
Ex. 1. Find the equation of the plane through the line
P=ax+by-{-cz-\ d=0, Q=a'x-\-by-\-c'z-\-d'==0
and parallel to the line xll^ylm=zjn.
Sol. The equation of any plane through the line P=0,2=0
i.Q. through the line of intersection of the planes P=0 and 2««0
is
P=.A2=0
...(1)
or (flX+6y+cz+f/)-f A
or
(fl+Aa') x+(6+AZ>')y-{r(c+Ac')z+</+Arf'=0
...(2)
The d.c.’s of the normal to the plane (2) are proportional
to </+Aa',6+A^', c+Ac'. The plane (2)[or (1)] will be parallel to
the line x//=y/m=z//i if the normal is perpendicular to the line
x//=y/ffi=r/«, hence we have
(a-fAnO /+(6+A6'J m-|-(c+At') /i=0
or
A(aT+h'm-|-c'/i)= - {al+bm-\-cn)
or
A «—(fl/+ bni^cn)f{a'l-\- b'm -j-c'n).
Putting this value of A in (1), the required equation of the
plane is given by
P-{(al+bm+c«)/(rt74 b'm+c'n)} Q=0
or
P {a'l-\-b'm-\-c'n)~Q (al^bm-}-cn).
Ex. 2. Find the equation of the plane through the line
3x-4y+5z«i0, 2x+2y-3z=4
and parallel to the line x^2y=^Zz.
Sol. The equations of the given line are
3x-4y+5z=10,2x+2y—3z==4..
■●●(I)
The equation of any plane through the line (1) is
(3x-4y+5z-10)+A (2x+2y-3z-4)=0
or
(3+2A)jc+(~4+2A) y-|-(5-3A) z-10-4A=0.
(2)
The plane (1) will be parallel to the line
X
y
if
6
3
C3+2A).6 + (-4+2A;.3+(5-3A).2=0
x—2y—2z
i.e.
— cs JL.
The Straight Line
105
or
A(12+6-6)-l-l8 -12+10=.0,orA=-|. ,
Putting this value ofAin(2), the required equation of the
plane is given by
(3 -4);c+(-4 -%) y-l-(5-|-4) z-lO-:-V =0,
or
x-20;;+27z=14.
Ex. 3. Find the direction cosines of the line whose equations are
and
and show that it makes an angle of 30°
with the plane y—z+2^0.
Sol. The equations of the line are
...(1)
x+y-^3, x-\-yi-z=^0.
Let /, m, n be the d.c.’s of the line (1), then we have
/-l-wi-l-0.«=0, /+m+n=0.
Solving, we get
1
n
m
/
/=1/V2, m=-l/V2, w=0.
...(2)
The equation of the plane is;^—z+2=0.
The d.r.’s of the normal to the plane (2) are 0, 1, —1, i.e.
directioncosinesare0, l/\/2,—l/\/2Now suppose d is the acute angle between the line (1) and the
normal to the plane (2).
Then using the formula cos y=/i/2+wiiWo+niWa, we have
1
1
+0.
cos 6=
V2
V0=60°.
Now the angle between the line (1) and the plane (2) is the
complement of the angle 0. Su the required angle between the
line (1) and the plane (2) is 9O"-0 i.e.. 90“-60® i.e., 30°.
Ex. 4. Find the equation of the plane through the points
(2, -1,0),(3, -4, 5)
and parallel to the line 3x~2y=z.
Sof, The equation of any plane through the point (2, —I,0)
...(1)
is n (x—2)4-6 0»-l-l)+c (z—0)=-0.
If the plane (1) passes through the point (3, —4, 5), we get
(3-2)46( - 44-1)+c(5)=0; or a-364*5c=0. ...(2)
z
X
y
The equations of the line are 3x=2y«=z, or ■^="^
...(3)
The plane (1) will be parallel to the line (3), if
...(4)
2a+3646c=0.
106
Analytical Geometry 3-Z)
Solving (2) and (4), we get
Putting these proportionate values of a, b, c in (1), the requi
red equation of the plane is given by
-33(;«-2)+4(y+l)+9(z-0)=0, or 33a:+4;;-9z-70=0.
Ex. 5. Find the equation of the plane through (2, 1, 4) perpen
dicular to the line of intersection of the planes
3x-|-4;>-|-7^-f-4=p and x-;;-|-22:H-3=0
Sol. Let /, /w, n be the d.c.’s of the.line of intersection,of the
two planes 3x4-4>>+7z+4=0, jr—;>-|-2z+3m0.
Then we have
3/+4m-|-7n=0, /-m-}-2n=0.
/
n
m
/ m
n
Solving,
8+7 "7-6 “-3-4
Thus d.r.’s of the normal to the required plane are 15, 1, -7.
Also the required plane is to pass through the point (2, 1,4).
Hence its equation is ,
15(je-2)+l (:>;-l)-7(z-4)=0
or
15a;+j^~7z—3=0.
Ex. 6. Prove that the lines 3x+2^+z-5=0«=aX+>>—2z—3
and
2x—y—z=0=lx-\-\Qy-%z—l5 are perpendicular.
Sol. Let /i, #Mi, ni be the d.c.’s of the first line. Then
3/i+2wi+«i=0,/i+Wi-2«i=0. Solving, we get
k
mi
or A=^_2s.
-4-l~l+6~3-2
-5 7 - 1 ●
Again let /a, Wa» »2 Ije the d.c.’s of the secon I line, then
2/a-ma-«a=0, 7/2+10m8-8«a=Q.
/
or
2 “1 “3 *
8H^""-7+16 “20^
Hence the d.c.’s of the two given lines are proportional to
-5, 7, 1 and 2, 1,3. We have ●
-5.2+7.1+1.3=0
the given lines are perpendicular.
§ 6. Tofind the equation of the plane through a given line
whose equations are g^ven in (i) general form and («) symmetrical
form.
(i) Let the equations of the line in general form be given by
PsaiX+Z>iv+CiZ+</i=0, 0=fl2A:+^ay+CaZ+</a=O:. ...(1)
Then P+Afi=0 /.e.,
aiX’\rbiy+CiZ-\-di-\-\(aaX+ftaP+CaZ+</a)=0
is an equation of first degree in x, y and z and hence represents a
107
Tke Straight Line
plane. Also P+\Q=0 is satisfied by all those points which satisfy
(1). Hence the equation of the required plane is given by
P+AQt=0.
(ii) Let the equations of the line in symmetrical form be
x~xi y-yi z-zi
n
...(2)
m
/
The required plane passes through the line (2) and hence it
passes through the point (xi, y^ Zi) which is a point on the
line (2).
The equation of any plane through the point (Xj, y^t fi) is
given by
...(3)
a (x--Xi)+6
(z-zi)"*®Again, the required plane passes through the line (2)» hence
the normal to the plane (3) is perpendicular to the line (2).
fl/+6w+c«=0.
Therefore, the required equation of the plane is
a (x-^xi)+b(y-yO+c(z-Zi)=0, where al+bm+cn=0.
§ 7. Tofind the equation of the plane through a given line and
parallel to another line.
Let the equations of the given line be
...(1)
(x -Xi)//i=Cf-yi)Mh=(z-zi)/«iProceeding as in § 6 (ii), the equation of any plane through
the line (1) is
...(2)
a(x-x,)+b (y-yi)+c(z-zi)=0,
...(3)
where
ali+hmi+cni—O.
Again let the plane (2) be parallel to another given line whose
equations are given by
...(4)
(x-Xa)//2=(y-;'2)/wa=(z-Zg)//!..
Since the plane(2) is parallel to the line (4), the normal to
the plane(2) will be perpendicular to the line (4) and so we have
...(5)
fl/a+bma+c/iai=0.
Eliminating a, 6, c between the equations (2),(3), (5), the
required equation of the plane is given by
x-Xx y-yi z-zi
0
iWl
A
or
ma
«a
(X“Xi)(wi??a“Wani)+(y~yi) Wi-nilf^
+(z-Zi)(/im|~/aWx)=0.
108
Analytical Geometry 3-D
SOLVED EXAMPLES(D)
Ex. 1. Find the equations of the planes through the line
(x—2)/2=(^- 3V3=(z—4)/5, are parallel to the co-ordinate axes.
Sol. The equations of the given line are
(x-2)/2=(;;-3)/3=(z-4)/5
...(1)
The equation of any plane through the line (1)[see § 6] is
. .
(x "2)+6
-3)H-c (z - 4)=0
..(2)
where
2fl+3A-f5c=0.
...(3)
We are required to find the equations of the planes parallel to
x-axis, ^-axis and z-axis respectively,
(i) The d.c.’s of x-axis are 1, 0, 0. If the plane (2) is parallel
to the X-axis, then the normal to the plane (2) is perpendicular to
the x-axis i.e., we have
ar.H-6.0-l-c.0=0.
...(4)
Solving (3) and (4), we have
a
b
c
or
0 ""5-0 ~0-3
0 “‘5 “-3’
Putting these proportionate values of a, b, c in (2). the requi
red equation of the plane is given by
0.(x-2)-i-5 (.v-.3)-3.(z-4)=0,
or
Sy—3z—3=0.
(ii) The d.c.’s of the j'-axis are 0, I, 0, If the plane (2) is
parallel to the ;/-axis, we have
fl.0-|-6.1-i-c.OcsO.
...(5)
Solving (3) and (5). we get
a
b
c
b
c
or £
0-5 “ 0 “2-0
Putting these proportionate values in (2), the required equa
tion of the plane is given by
-5 (x-2)-l-0.(;;-3;-|-2(z-4)=0, or 5x-2z-2=0.
(iii) The d.c.’s of the z-axis are 0,0, 1. If the plane (2) is
parallel to the z-axis, we have
fl.0H-6.0-bc.1=0.
...(6)
c
Solving (3) and (6), we get -j=-^2=“o ●
Putting these proportionate values of a, 6, c in (2), the requi
red equation of the plane is given by
3 {x-2)-2(;^-3)=-0, or 3x-2j;=0.
Ex. 2. Prove that the equation of the plane through the line
(x^i)/3«(j;-t-6)/4=(z-|-l)/2flWi//;nrfl//c/ to (;c-2)/2=(:y-l)/-3
The Straight Line
109
=(^4.4)/5 is 25a:—lljp-lVz-109=0 and show that the point
(2, —A) ties on it.
Sol. The equations of the given line are
...(1)
(x-l)/3=(;;+6)/4=(z+l)/2.
The equation of any plane through the line (1) is
...(2)
a{x-\)-\-b (y+6)-\’C(z+l)=0
...(3)
where
3a+4ft+2c=0.
The plane (2) is to be parallel to the line
...(4)
(a:-2)/2=(j;-1)/(-3)=(z+4)/5.
Hence the normal to the plane (2) is perpendicular to the line
(4), so that we have
...(5)
2a-3b-f-5c=0.
Solving (3) and (5), we get
c
C'
a
b
b
a
26“-ll “-17
20+6 ■“4 -15 “9-8
Putting these proportionate values of a, b, c in the equation
(2), the required equation of the plane is
26.(a:-1)-11 (j^+6)-17(z+I)=0
or
26a: -lly~17z-119=0.
Substituting the point (2,1, —4) in the equation (6) of the
plane, we get 26x2 11 X1 -17 X -4—119=0, or 0=0 i.e. the
point (2, 1, -~4) satisfies the equation (6) of the plane.
Remark. In the above Ex. 2, the point (2, 1, —4) lies on the
line (4) and also on the plane (6). Hence the line (4) will wholly
lie on the plane (6).
Therefore both the lines (I) and (4) are coplanar and the
equation (6) gives the plane containing both of them.
Ex. 3. Find the equation of the plane which contains the two
. parallel lines
, x-3 y+4 z—1
x+l y-2 z
t ●
^
2 “
3
Sol.
The equations of the two parallel lines are
(a:+1)/3=(>^-2)/2=(z-0)/1
and
(x-3)/3=(;;+4)/2=(z-l)/l.
The equation of any plane through the line (1) is
a (x+l)-\-b [y- 2)-{-cz=0,
where
3fl+2b+c=0.
...(1)
...(2)
...(3)
...(4)
The line (2) will also lie on the plane (3) if the point (3,^ —4,
1) lying on the line (2) also lies on the plane (3), and for this we
110
Analytical Geometry 3-D
have
«(3+l)+^> (-4-2)-f-c.l=0 or 4fl-66-|-c=0.
...(5)
c
Solving (4) and (5), we get y=-y -26*
Putting these proportionate values of a, b, c in (3), the requi
red equation of the plane is
8(x+l)-|-l.(j;- 2)-262=0, or Sx+y-26z+6=0.
Ex. 4. Find the equation of the plane which contains the line
3)=I(^-5) and which is perpendicular to the plane
2x-l-7v-3z=l.
Sol. The equations of the given line are
(x-0)/1=(j;-3)/2«(z-5)/3.
-(1)
The equation'of any plane through the line (1) is
a (x).f6(:y-3)+c(z-5)=0
...(2)
where
1 a+2.6+3.c=0.
...(3)
Also the plane (2) will be perpendicular to the plane
2x4-7y—3z=l
if
2.fl+7.6-3.c=0.
...(4)
Solving (3) and (4), we get
a
b
c
a
b
c.
or
-9“
3
”
1●
-6-12”6+3“7-4
Putting these proportionate values of a, b, c in (2), the requi
red equation of the plane is
—9x-f-3(;'-3)4-(z—5)=0, or 9x-3y-z+14=0.
Ex. 5. Show that the equation to the plane containing the line
^+1 y—'i z-l-2
-3 - 2 “T and the point ,0, 7, -7)is x-f-;;-|-z=0. Hence
1 “
show that the line i-—^-—=^-^
also lies in the same plane,
●Soi. The equations of the given line are
(x-f-l)/-3=Cv 3)/2=(z^.2)/l.
.. .(I)
The equation of any plane through the line (1) is
a (^+l)+fe (j'-3)+c (z+2)c=0
.. .(2)
where
-3a+2^+l.c=0.
.. .(3)
The plane (2) is also to pass through the point (0, 7, —7).
a+4i»-5c=0.
.. .(4)
Solving (3) and (4), we get
a
b
c
a ^ c
—10^4 1—15 —12—2 or 1 “ 1 “1 ●
Putting these proportionate values of a, b, j in (2), the required equation of the plane is
111
The Straight Line
...(5)
l.(jc+l)+l.(j;-3)+l.(z+2)=0 or
The equations of the second line are given to be
1 “ 2 “T”*
...(6)
The line (6) passes through the point (0, 7, —7) and this
point also lies on the plane (5). Now the line (6) will lie in the
plane (5) if the normal to the plane (5) [whose d.r.’s are 1, 1, 1]
is perpendicular to the line (6), the condition for which is
‘«ia9+6A4-CiC8*=0, i-e. l.l+1.2+l.(-3)=0,
which holds good. Hence the line (6) also lies in the plane (5).
This proves the required statement.
. '
Ex. 6. Find the equation to the plane through the point
(«',
y') and the line (x~a)fl=>(y—fi)lm={z-y)ln.
(Gorakhpur 1981 ; Ranchi 68)
Sol: The equation of any plane through the given line is
...(1)
a(x—(i)-\-b(y~fi)-\-c(z-y)=0
where
a.l-\-b.m+c.n=0.
...(2)
The plane (1) will pass through the point (a', jS', y') if
...(3)
a(a'-a)4-P)+c(y'—y)=0.
The equation of the required plane will be obtained by elimi
nating a, b, c between the equations (1),(3) and (2). Hence elimina;ting the constants a, b, c between the above equations, the equ
ation of the required plane is given by
X—a■ y-p
z-y
a'—a
P'—fi
Y-y
=0
/
m
n
S (x-a) W-P) ~m(y'-y)}=0.
Ex. 7. Show that the plane through the point («, /?, y) and the
line x=py+^=rz+j is given by
X
py-)rq
rz-^s.
or
PP+q
ry+s
1
1
1
(Meerut 1983 S)
The equations of the given line are
x=py+q==rz+s,
x-0 _y-\-(qlp) z+(j/r)
or
1
_ 1/p
1/r ●
..(1)
The equation of any plane through the line (1) is
a{x -- 0)+b(y-{-qlp)+c(z+s/r)=0
...(2)
where
l.a+(l/p).i»+(l/r).c=0.
...(3)
Sol.
if i
112
Analytical Geometry 3-D
The plane (2) will also pass through the point (a, [5, y) if
...(4)
aa-\-b{p+q!p)-\-c{y-\-sfr)=0.
The equation of the required plane is obtained by eliminating
the constants c, 6, c between the equations (2),(4) and (3) and
hence it is given by
X
y-\q!p
z+sir
a
P-\-<llP
7+sjr
=0.
1
1/p
1/r
Multiplying second and third columns by pand r respectively,
we get the required equation as
^ ■ py-\-9
a
pp-^q
1
1
=0.
1
Ex. 8. Show that the equation of any plane through the Vmt
X—a
0
n
/
where
A-{-/*+v=0.
Sol. The equations of the line are
...(1)
(x-a.)ll^{y-P)/m=-iz-y)ln.
The equation of any plane through the line (1) is
...(2)
a{x-a)-\-b(y-P)-^c(z-y)=0
where
...(3)
al-\-bm-\-cn=0.
Now choosing the values a—.Xjl, b^jilm^ c.^vfn^
the equation (2) of the plane is given by
(x-a)(Xll)+iy-P)(ti!m)+{z-y)(v/«)=0
where
-r.
-wiH—
^ -n=0, . or A+p+v—0.
/ /+—
m
This proves the required result.
Ex. 9. Find the equation of the plane through the line
ax-\-by-\-cz~Q~:a'x-\-b'y-\-c'z
and
oLX-\-Py-!ryz=^^--a:x-{-Py-^y'.z. ‘ ^
Sol. The equations of the first line are
ax+by-{-cz=0=a'x-^b'y-i-c*z.
This line clearly ^passei through the origin. If /,’ »r, n be the
d.c.’s of this line then we have
al+bm+cn=0
a'l+b*ni-{-c'n=b.
113
The Straight Line
Solving these.
n
m.
/
— sa ■■
""77 *
bc*‘—b'c ca’ c'fl ab'—ab
Hence the symmetrical form of the first line is
z
X
y
bc'-b'c ca'—c'n 'ab'-a'h'
...(1)
The equations of the second line are
aJC+j8>/+y2=0=a'x4^>+/2.
This line also passes through the origin. Hence its symmetri¬
z
X
y
cal form is
...(2)
jSy'—j3'y yflc'—y*a aj5'—ot'/3
The equation of any plane through the line (1) is
...(3)
Ax+By+Cz=0
...(4)
where
^(dc'-fi'c)+J?(ca'-c'a)+C(a6'-a'!>)=0.
Also if the line(2) lies in the plane (3) then the following two
conditions must hold :
I. The line (2) passes through the point (0, 0,0) and so the
plane (3)should also pass through (0,0, 01 and clearly it is true.
IT. The normal to the plane(3) should be perpendicular to
the line (2), the condition for which is
...(5)
^(jSy'-)3'y)+5(ya'-/a)+C{aj3'-«;j8)=0.
The equation of the required plane is obtained by eliminating
Af
C between the equations (3),(4)and (5). Hence the equation
of the required plane is
z
X
y
bc'-b'c
ca'—c'a
ab'--a'b
I5/-I3V
y«'—y'a
ajS'-a'^
=0.
Ex. 10. Find the equation of the plane through the point
(2, —1, 1) and the line 4x—3y+5~0^y‘—2z—5.
Sol. The equations of the line are
4x—3;i>+5=0, y—2z—5ca»0.
The equation of any plane through the given line,[using
...(1)
P+A0*O]is 4x-3;;+5+A (y-2z-5)=0.
or
If the plane (1) passes through the point(2, —1, 1), we have
4(2)-3(-l)+5+ A {-l-2.1-5}=0
16—8A=0 or A-=2.
Putting the value of A in (1), the equation of the required
plane is
4x-3;>+5+2 iy-7z-5)=0 or Ax^-y-Az^S.
114
Analytical Geometry 3-D
Ex. II. Prove that the equation to the two planes Inclined at an
angle (K to xy-plane and containing the line y=0, z cos p=x sin is
tan^ ^^z^-2zx tan p=y^ tan* a. (M.U. 1990P)
Sol. The equation of any plane through the line
z cos
sin p is given by
(x sin jS-z cos p)+Xy=zQ.
. .(1)
The other plane is the .v;>-plane, whose equation is z»=>0 i.e.
0.Ar+0.;;+l.z=0.
...(2)
The d.r’s of the normal to the plane (1) are sin
X, —cos ]8,
and the d.r.*s of the normal to the plane (2) are 0, 0, 1. Also the
angle between them is a and hence we have
cos ae=3
or
Or
or
or
or
sin jS.O+A.O—cos j8.1
V(sin* iS+'A»H-cos» ^).V(0+0+l)
cos a <\/(sin* 15+cos® j8+A®)=-cos jS
cos a <^(1 +A^)=—cos p
(A®+|) cos* a=cos* p
A* cos* a=cos* p—cos* a
A=>±-^(cos* 13—cos* a)/cos a.
It gives two values of A and hence there are two planes con
taining the given line and inclined at an angle a to the x>'-plane.
Substituting these values ofAin(l) and multiplying both the
equations thus obtained the combined equation of the two required
planes is given by
{
X sin p-z cosj84
\/(cos* j3—cos* g) I
cos a
^\
{
X sin p-z cos P —
or
or
or
or
(x sin p—z cos p)*
■y/fcos* P - cos* a) 1
cos a
^ f
cos* j8—cos* g 2
cos* a
^
0
0
X* sia* 13+z* cos* P~2xz sin B cos /3+>>*=_j;« cos* p sec* a.
Dividing throughout by cos* p, the above equation becomes
X* tan* iS+z* —2xz tan jS+y* sec* p=y* sec* a
X* tan* l3+z*-2zx tan p-{-y* (1 + tan* p)=>y* (1+tan* a)
(x*+/*) tan* j8+z*- 2zx tan p=y* tan* a.
Ex. 12. The plane lx^my=^0 is rotated about its line of inter
section with the plane z=0, through an angle <x.
equation of the plane in its new position is
lx-{-my±z^/(l*-\-m*) tan a=>().
Prove that the
115-
The Straight Line
Sol. The equations of the given planes are
and
2=0.
...(2)
/x+m>'=0
...(0
The equation of any plane ihrough the line of intersection of
the planes(1) and (2) is
...(3)
lx+my+Xz=0.
Suppose the plane (1) when rotated through an angle a about
its line of intersection with the plane z=0 has the equation (3).
Thus the angle between the planes (1) and (3) is a.
/./4-m.m+0,A
cos a«=
V(*'”+w*)V(/®+m2+A2)
6r
cos a=3
y/(l^Arm^)
Or
cos* a
/*+m*
/*+w*+A*
or
or
(I—cos* a)=A* cos* a
A*=(/*+m*) tan* a, A=±v'(/>+m*)tan a.
Putting this value of A in (3), the required equation of the
plane in its new position is given by
/Jc+my±(/*+/w*).^tan a=0.
§ 7. Foot and length of perpendicular from a point to a line.
(A) Line in symmetrical form.
Tofind the equations and length of the perpendicular distance of
a point P(Xi, yx, Zi)from a given line
/
m
n
The equations of the given line in symmetrical form are
X—
/
z-y
m
n
r (say).
The co-ordinates of any point N on the line (1) are
(«+/r,/3+mr, y-l-nr).
...(1)
...(2)
Let this point N be the foot of the perpendicular from the
point P(^i, yi, Zi) to the line (1), so that the line PN is perpendi
cular to the line (I).
The direction ratios of the line PN are given by
a-l-/r-x„ j3+mr-y„ y+nr-z,.
.(3)
Since PN is perpendicular to the line (1), using the condition
we have
or
or
I(a+/r-Xi)H-m (j8+mr-y,)+« (y+nr~z,)=0
r (/*+m*-!-«*)=/(jf,—a)+m (yi-j8)H « (Zj—y)
rc=[/(xi-a)+m (y,_j8)+n Ui-y)]/(i*+m*+/i*).
...(4)
116
Analytical Geometry 3-^
The equations of the perpendicular from the point
/*(●<!* yu 2i) to the line (1) are given by
X’-Xi
ct+lr—Xi
where r is given by (4).
y-yi
Z—Zi
pi-mr~yi
y^nr—Zi
..(5)
Also substituting the value of r from (4) in (2), the co-ordi
nates of the foot N of the perpendicular are determined, and then
the perpendicular distance PN is easily calculated. See Ex. I
> which follows after part (B) of.this article.
(B) Line in general form :
To find the equations of the perpendiculdr line from the point
P(Xy, Pit Xi) to a line whose equations are given by
ax-\-by-\-cz-{-d=0=a’x-\-b'y-\-c'z+d\
The equations of the given line in general form are
qx+by-{-ez+d~Ot o'jc+b‘>4-c'2+<^'=0.
.(1)
The perpendicular from a point P to the given line (1) Is the
intersection of the two planes namely (/) the plane through the given
point P and also through the given line and {ii), the plane through
[Remember]
the point P perpendicular to the given line.
Let /. m, n be the d.c.*s of the given line (1), then we have
a/+6w-l-c«=0, and a7-fb'w+c'»=0.
Solving these, we have
m
n
/
tss _
be'— b'c ca'— c'a ab’ — <tb
...(2)
Now the equation of any plane through the line (1) is
given by
...(3)
{iix-\-by-\-cz-\^d)-^\ {a'x-\-b'y^c’z-\-d')=^0.
If the plane (3) also passes through the point Pix^y^Zx)
then
or
A=
+c2i4-</)/(a%+
Substituting this value of A in (3), the equation of the plane
through the point P and the line (1) is given by
flx-i-by-fex+rf ^ a'x+b‘y^c'z-\-d’ ^
...(4)
axi+byx+czx-^d^ a^Xy-^byy+c'Zx-^d'
Now we are to find the equation of the second plane which
passes through P and is perpendicular to the line (1).
Since the plane is perpendicular to the line (1), therefore the
d.r,*s of its normal are proportional to /, m, n given by (2).
'
117
The Straight Line
Therefore the equation of the plane perpendicular to the line (1)
and passing through P(x^yi, Zi) is
i(x-Xi)+miy-yi)i-n{z-Zi)=>0,
...(5>
Therefore the equations of the perpendicular line from the
point
z,) to the line (1)are given by ● the equations (4)
and (5).
SOLVED EXAMPLES(E)
Ex. 1. Find the equations of the perpendicularfrom the point
X y-2 z-3
(3,
11) to the
. Find also the co-ordinates
3
4
of the foot ofthe perpendicular. Hencefind the length of the per
pendicular.
Sol. The given point Is P (3, 1, li) and the equations
of the given line are
Af—0 y—2 z—3
2 “3
^
r(say).
...(])
The co-ordinates of any point N on the line (1) are
(2r, 3r+2. 4r-|-3);
(2)
Let this point N be the foot of the perpendicular from the
point P(3, -1, II) to the line (1).
Then the d.r.’s of the perpendicular PM are
2r-3.(3r-H2)-(-l),(4r+3)-ll.
or
2r-3, 3r+3,4r -8.
...(3)
The d.r.’s of the given line (1) are 2, 3, 4, Now PH is per
pendicular to the line (1). The condition of perpendicularity gives
(2r~ 3).2-l-(3r+3).3+(4r-8).4=0.
or
29r-29=0, or r=l.
Putting the value of r in (2), the foot N of the perpendicular
is the point (2,'5, 7).
Putting the value of r in (3), the d.r.’s of PN &re —1, 6, —4.
Hence the equations of the perpendicular PN from the point
P(3. — 1, 11) to the line (i) are
x~3 y+1 z-11
“ 6 “-4 *
The length ;Of the perpendicular Pi\f=the distance between
the points P(3^ —1, 11) and JV(2, 5, 7)
=V[(3-2)H(-I-5)>+(ll-7)«J
=\/(l+ 36+16)=\/(53).
118
Analytical Geometry 3-D
-Ex. 2. Find the equations of the perpendicular from origin to
the line
ax+by+cz+d=0=r a'x-\-by+c'z-{-d'=Q.
Sol. First read § 7(B)again and then solve thiio problem.
The equations of the given line are
...(1)
flx-f6v+cz+d=0, a'x-^b'y+c'z+d'—O.
We know that the perpendicular from a given point P to a
given line is the intersection of the two planes, namely (i) the
plane through the given point P(0,0,0) and also through the
given line and (ii) the plane through the point P perpendicular
to the given line. Thus we proceed as follows:
The equation of any plane through the line (1) is
axA'by-\-cz-\-d-\-X
..(2)
If the plane (2) passes through the point P(0,0,0), then
</-i-Ad'=0, or A=3-d/d'.
Putting this value of A in (2), the equation of the plane
through the origin and the given line is
{ax+by+cz+d)—(d/d'){a’x+by-\-c’z+d')=0
{ad'-a'd) x-\-{bd*-b^d) y+{cd'--c'd)z=0.
or
...(3)
Now let /, /n, n be the d.r.'s of the given line (1), then
fl/4*^>w+c/j= 0, and fl'/+6'm+c'«=0.
/
m
n
Solving,
bc'~b'c
c'a
ab’-a'b
...(4)
Thus the equation of(he plane through P^O, 0,0) and per.pendicular to the line (1) is
/(x—0)+m O'-0)+n (2—0)«0
{bc'—b'c) x-f(cfl'—c'a) y-\‘(ab'-a'b) z=0.
or
...(5)
Hence (3) and (5) together are the equations of the perpendi- .
cular from the origin to the given line (Ij.
Ex 3. Show, that the distance d of the point P(a, j8, y) from
the line {X'^Xi)jl^{y ■~yx):m—{z—z\)jn measured parallel to the
plane
ox-\-by r'
i.s given by
U’(X| ~oi)(bn—cm)f
{al~[-brir^cnf
Sol.
The equations of the given line are
(X—X, ill=>{y—yi)lm—{z—zi)ln=r, (say).
Any point Q on (I) is
(-Va-}-//', >iT'/ir, z,+nr).
. .(1)
119
The Straight Line
Now P is the point (a, j8» y) and hence d.r.’s of PQ are
x,+/r-a,
z,+nr~y.
It is required to find the distance PQ measured parallel to
the plane ax+6;;+cz+</=0. Now
is parallel to this plane
and hence PQ will be perpendicular to the normal to the plane
ax+by+cz+d*=^0.
Therefore we have
(/r+Xi-a)
+
^>+(zi+«r-y)c=0
or
r=-[a(Xi-a)+6 (yi-j3)+c(2,-y)]/(a/+f>w+c«). ...(2)
2
Now rf*=Pfi*«=(Xi+/r—a)*+(^i+mr-j3)*+(Zi+ nr-ry)
or
or
(/*+»!*»+"*)+2r {/(Xi-a)+m (>»i-/3)+n (Zj-y)}
-/3)8+(2i-y)2).
Putting the value of r from (2) in (3) and using Lagrange’s
identity, we get the required result.
Ex. 4. Find the distance of the point P{3, 8, 2) from, the line
~
measured paraiiel to the plane
3x+2;>—2z+17=0.
Sol. The equations of the given line are
(x-l)/2=(;/-3)/4=(z-2)/3=r.(say).
—(I)
Any point Q on the line (1) is
(2r+1,4r-|-3, 3r+2).
Now P is the point(3, 8, 2) and hence d.r.’s of PQ are
2r+l-3,4r+3-8,3r+2-2 i.e. 2r-2, 4r-5, 3r.
It is required to find the distance PQ measured parallel to
the plane
3x+2;y-2z+17=G.
...(2;
Now PQ is parallel to the place (2) and hence PQ will be
perpendicular to the normal to the plane (2). Hence we have
(2r-2l (3)+(4r-5)(2)+(2r)(-2)=0.
or
8r— 16=0, or r=2.
Putting the value of r, the point Q is (5, 11, 8).
.*. Required distance=Tbe distance between P(3, 8, 2) and
6(5, 11,8)
= V[(3-5)H(8-n)H(2-8)=»l-V(4 f9+36)=7.
Ex. 5. The equations to AB referred to rectangular axes are
x/2=^/—3=z/6. Through a point P(l,2, 5),
drawn per'
pendicular to AB and PQ is drawn parallel to the plane 3x-\ Ay f 5* -'*
120
Analytical Geometry 3*X>
to meet AB in Q. Find the equations to PN and PQ and the co
ordinates of N and Q.
●Sol. The equations of the line /IB are given as
...(1)
x/2=;;/—3==z/6=»r (say).
Any point on (1) is (2r, —ir, 6r).
Let this point be N. Also
P is (1. 2, 5).
The d.r.’s of PJVare 2r~l, —Sr—2, 6»’—5.
Now PN will be perpendicular to the line AB given by the
equations (1) if
2 (2r-l)+(-3) (_3r-2) + 6 {6r~5)=0, or r=»|S.
Putting the value of r, we have N
52
-78
and d.r.’s of PN are -7:- 1,
49
or
49
.49
^
3,-176,-89.
Thus the equations to PN are
z—5
a:-1
y~2
3
176 ~-8y ■
.♦ . A
...(2)
Again the co-ordinates of any point Q on the line (1) are.
(2r, —3r, 6r) and the point P is (I, 2, 5).
The d.r.’s of PQ are 2r-l, -3r-2, 6r-5.
Now PQ is drawn parallel to the plane
3x-\-4y^5z=^0.
The line PQ is perpendicular to the normal to the plane
(3J, whose d.r.’s are 3, 4, 5.
Therefore we have
3 l2r—1) + 4 (—3r—2)-{-5 (6r—5)--0, or r-- .3.a*
Putting this value of r, we have Q (3, —9/2, 9) and d.r.’s of
PQ are 3-1, (-9/2)-2, 9-5 ie. 4, -13, 8.
.
The equations of PQ are given by
■y—1 y-2
z—5
4
”-13 “ 8
’
The projection (or image) of a line on (or in) a given plane.
The method is explained by the following examples.
The line is in symmetrical form.
Ex, 6. fVhat do you understand by the projection of a line on
a given plane 1 Find the equal ions of the projection of the line
X— 1 y-j-1 z—3
,
,
V
'~2^^ ITJ” “
on the plane x-i- 2y+z= 6.
Ill
7'he Straight Line
Solution. Definition I. The projection of a line on a given plane
is the line of intersection of the two planes namely(0 the given plane
and (ii) the plane through the given line and perpendicular to the
given plane.
Definition 2. Let P be the point of intersection of the given
line with the given plane and let Q be the foot of the perpendicular
from any point on the line to the plane, then the line PQ is setfd to
be the projection of the given line on the given plane.
Now
give the solution of the given problem, using both
definitions one by one.
Using definition 1.
The equations of the given line are
2 -“-I ■
. ^
4
and the equation of the given plane is x+2y-\-z=6.
Ihe equation of any plane through the given line (1) is
A{x-i)-\-B(y-\-\)^-C (z-3)=0
lArB\-^C=0.
where
...(1)
(2)
(3)
(4)
The plane (3) will be perpendicular to the plane (2), if
...(5)
A~\"2B
A
BC
Solving (4) and (5), we get
*
or
Putting these proportionate values of A,B^ C in (3), we.have
9(jc-l)+2 O'-l-D +S (z-3)=0
. (6)
9x —2j;—5z+4«0.
The equations (2) and (6) together are the equations of the
line of projection.
Using definition 2.
The co-oidinaies ol any point on the line(l) are (2r-l-l,
_f_1^4r4.3). Lei this be the point of intersection P of.the
given line (1) with the given plane (2). Then P(2r+l, —r-^1,
4;.4-3) will lie on the plane (2).
2r+H-2(—r-l)-h4r+3«=»6, or r=l.
Putting luis value of r, the point P is (3, —2, 7).
Now the line (1) dearly passes through -he point (1, - 1, 3)
and hence this is a point on the given line (1). Now we are to
find the foot of the perpendicular Q from fl, — 1, 3) to the
plane (2).
The d.r.’s of the norma! to the plane (2) are 1, 2, 1 and
122
Analytical Geometry 3~D
hence these are d.r.’s of the line through (1, -l, 3) and perpendi
cular to the plane (2). and therefore the equations of this perpehdicular line are
1
2
1
■=r, (say).
Any point on it is (riH-1, 2ri-l, ri-|-3). Let this be the
point Q and so it will li^ oh the plane (2).
/’1+1+2 (2ri-l)-l-r,-|-3=6. or r,=|.
Putting this value of ri, the foot of the perpendicular Q is
(5/3, 1/3,11/3).
The equations of the projection/.e. the equations of the
line PQ joining the points P (3, -2; 7) and Q Q]
Y) art
x-3_ y-{-2 _z-7
x-3_y-^2_z-7
3-1 -2-i
4
rT“To'‘
...(7)
Remark. If the equations (2) and (6) are transformed to
symmetrical form, we shall get the equations (7) of the projection
PQ. Wt; note that if we want to find the equations of the projec
tion in general form then we use definition one, and if we want
symmetrical form then we use definition two.
Ex. 7.
If L is the line
— -~ , find the direction
cosines of the projection of L on the plane 2x+j>—3z=4 and the
equation of the plane through L parallel to the line
2x+5>'+3z=4^x -3' -3z=6.
Sol.
The equations of the line L are
x—l
0_z-f2
=r, (say).
2
...(1)
Any point on the line (1) is (2r-t-l, —r, r—2).
If it lies on the given plane 2x-f>>- 3z=4,
we have
2 (2r+l) + (~r)-3 (r~2)=4, or 0,r+4=0.
...(2)
This relation is impossible and we do not get any value of r
as the coefficient of r is zero. This shows that line (1) is parallel
to the plane (2). Hence method of definition 2 (see Ex. 8) cannot
be used here. Now we shall use n^ethod of definition 1.
The equation of any plane through the line (1) is
(X-1)-H5;»+C (z+2)=0,
where
2A-P-^C=0.
The plane (3) will be perpendicular to ilje plane (2) if
2A-{-B-3C^0.
...(3)
...(4)
..^(5)
12^
The Straight Line
Solving (4) and (5). we get
2
8
4
A BC
1 “4 ^2
Putting these proportionate values of A, B and C in (3), the
equation of the plane through the line £ (1) and perpendicular to
the plane (2) is given by
l.(x-l)+4.:y+2(z+2)=0
...(6)
or
x+4y+2z+3=0.
The equations(2) and (6) together are the equations of the
line of projection of the line L on the plane (2).
Let /, m, n be the d.c.’s of the line of projection given by (2)
and (6). Then we have
2/+m—3n=0 and /+4m +2nt=»0.
n
/
m
Solving these, we get
or
m
n
i
T" -l'" 1
1
V{(2)»+(-l)*+(l)*}
l—2ly/6,m’= — \ly/(t^n<=\i‘\/C>.
Hence d.c.’s of the line of projection are 2/V6,
1/^6.
Now to find the equation of the plane through the line L and
parallel to the line
2x-f-5>-+3z^A, x-ry-
...(7)
Let
mu wi be the d.c.’s of the line (7).
Then
2/,*h5wi,+3«i=0,/,-w,-5«i=0.
/i
rnx
Solving these, we get 22 ~13 7
Now the equation of any plane through the line L(1) is given
by (3) provided the condition (4) holds. If the plane (3) is parallel
to the line (7), then the normal to the plane (3) will be perpendi
cular to the line (7), the condition for which is
22/4-13B-l-7C=0.
Solving (4) and t8), we get
C
B
A
-7+ 13 22 - 14 -26+22
A
B
C
3
Substituting these proportionate values of A^ B and C in (3),
the equation ot the plane through the line L (1) parallel to the
line (7) »s given by
3{x—l)+4y—2(z+2)=0, or 3.X+4;;—2z<=>7.
.124
Analytical Geometry D-3
The line in general form.
Ex. 8. Find the projection of the line 3x—>-|-2z—I,
x-\-2y — z=2 on the plane 3x-t2y-tz~0.
Sol. The equations of the given line are
-f-2z =● !, X f 2;/-z-.-2.
...(I)
The equation of the given plane is 3x-|-2j;+z=d.
...(2)
The equation of any plane through the line (I) is
(3jc—^-l-2z—[).-\r^{x-i-2y—z—2)“0
or
(3+A) x+( - l+2A);;-i-(2-A) z-l-2A=0. ^
;..(3)
The plane (3) will be perpendicular to the plane (2), if
3 (3-HA)+2 (_I + 2A) + 1 .(2--A)=0, or A _ a8*
Putting this value ol A in (3), the equation of the plane
through the line (1) and perpendicular to the plane (2) is given by
(3-f) x + (--l-3»>; + (2-i-f) z-1+3-0
or
3jc--8;;+7z+4=-0.
...(4)
/. The projection ol the given line (1) on the given plane
(2) is given by the equation.s (2) and (4) together.
Note. The symmetrical form ol the projection given above
by equations (2) and (4) is
II
9
15’
§ 8. Coplanar Lines. To find the condition that the two lines
whose equations are given may be coplanar Le. should Intersect and
to obtain the equation of the plane containing them.
(Gorakhpur 1982)
Two lines are called coplanar if they intersect. Even if the
two lines are parallel, they are also coplanar because they will
intersect at inhnity and will satisfy the condition of coplanarity
given below.
The equations of two lines may be given in three ways :
(A) Both lines in symmetrical fotm, (B) one line in syrnmetrical form and the other in general form, and (C) both lines in
general form.
Now we shall discuss these cases one by one.
(A) Both lines are given in symmetrical form ;
(Avadh 1981, $2; Allahabad 80; Kanpur 76, 81)
Let the equations of the two given lines be
(X—Xj)//i=ty->i)//Mi=(z - Zi)/«i,
...(1)
and
(x-x^)lk=>\y—yi)lm2 {z-Zi)/n.i.
...(2)
The equation of any plane through the line (1) is
125
The Straight Line
...(3)
A (x—Xi)-\-B(y~yi)-hC(z—Zi)=0,
...(4)
Alii-Bmi-rCni==0.
For the lines (1) and (2) to be coplanar, the plane (3) should
be such that the line (2) also lies in this plane. Then the normal
to the plane (3) will be perpendicular to the line (2) so that we
...(5)
have
i4/2+jfftWa+C»2 — 0.
where
Again the point (xs,j>2» Za) through which the line (2) passes
will also lie on the plane(3) so that we have
...(6)
A{xz~ Xi)T B(yi — y\)
Zi)~0.
Hence the required condition that the lines (1) and (2) are
coplanar (i.e. are intersecting) is obtained by eliminating A, B^ C
from the relations (6),(4) and (5), and is given by
X2—X1
Xi-yi
Za-Zi
...(A)
m,
/2
hti
"2
The equation Of the plane containing the lines (1) and (2) is
obtained by eliminating A, B,C between (3), (4) and (5) and is
given by
z-z,
1 X-Xi
y-y%
=0.
/2
nt2
na
The point of intersection of the lines (1) and (2).
The co-ordinates of any point F on the line (I) are (Ari-f *Ci,
miri+yi, Wifi+ zi) and those of a point Q on the line (2) are
(kr^^Xi, mzri-^yi, fiarz+Zz).
If the lines (1) and (2) intersect, then for some values of ri
and fa, the points F and Q should coincide,./.c. we have
+
or
and
/Wiri+yi«/Wara+j-2, Wiri-fZi=nar2+Z2 _
...(7)
(xi—X2)+liri-kr2^P,
(8)
{yi-ya)+fnirx-mir2=^0
(9)
(Zi—Za)-h «iri-Wara—0.
Solving the relations (7) and (8) the values of ft and ra are
obtained and if these values also satisfy the relation (9). then the
lines (1) and (2) are coplanar otherwise not. In case the lines (1)
and (2)are coplanar, on substituting the value'of Ti (or ra) the
co-ordinates of the point of intersection P (or Q)are obtained.
126
Analytical Geometry 3-D
Also by eliminating ri and -r^ between the relations (7),(8)
and (9) the condition that the lines (1) and (2) are coplaoar is
given by
Xi-Xz
/i
/.3
3^1->'2
OTi
W2
^1—^8
«1
W2
Zi—Zz
or
h
mi
ni
mj
W2
«=»0.
(B) One line IS;given in symmetrical form and the other line
is given in general form.
(Agra 1973 ; Meerut 82S ; Allahabad 76, 81)
Let the equations of the two given lines be
...(1)
(x—xi)fl=(y^yj)jm=(z—zi)fn
and
...(2)
OiX-h biyA-CiZ+di «=0=cfax+h2>'4-Csz+dz.
The equation of any plane through the line (2) is
(fli^+h,j;+ciz+</,)+A {azX-\-hzy+CzZ+dz)=0
...(3)
or
(Oi+Xaz)*+(h,+Ada);;+(ci+ Aca) z+{di-\-Xdz)^0.
...(3')
If the line (1) is parallel to the plane (3'), then the normal
of the plane (3') will be perpendicular to the line (1) and so we
have
i(oi+Xaz)-\-ni (h]+Ah2)+n (Ci+ Ac2)=0
or
^ (Ozl-\-hzmTCzn)— —(ail-\-bim + Cin)
or
^=--{ail-{-btm-\-Cin)l[azl+bzmA-Czn).
...(4)
Putting this value of A in (3), the equation of the plane
through the line (2) and parallel to the line (1) is given by
*.
^jX-\-biy+ciz+dt OzxA-bzvA-czZ+dz
Oii+bim-\-Cin
aai+ ham-f C2«
...(5)
The line (1) passes through the point (^i, ^i, z,) which must
lie on the plane (5) if the lines(1) and (2) are coplanar /.e., inter
secting. Thus for the lines (1) and (2) to be coplanar the point
(Xi, yi, z,) should satisfy the equation (5). Hence the condition
that the lines (1) and (2) are coplanar is given by
^1-^1+hi ri -h c,Zi -f
02X14- bzyi+CzZi+dz
Ozl+bzm-^c%n
...(6)
If the condition (6) is satisfied the lines (I) and (2) are inter
secting (or coplanar) and the plane containing both the lines IS
given by the equation (5). .
127
The Straight Line
Alternative Method: The co-ordinates of any point P on the
line (I) are (/r-l-ifi, mr+y„ nr+2i). If the lines (1) and (2) are
intersecting then the point P should also lie on the line (2). Hence
we have
(rnr-\ryi)~hCi (nr4-ri)+</i=0,
Oi (/r+xi)+ha (wr4-yi)-|-C2(nr-\-Zi)+di^ 0:■
From these, we get
r (ail+bim+Cin)=-(aiXi+biyi-^CiZi+dO
and
r (ail+bzni-^czn)=—{a^Xx +bzyi+C2Zi+da)
Eliminating r, the required condition that the lines (1) and
(2) are coplanar is given by
ajXx + biyt -hciz-f -t-^i fl8^i+62Vi+C22i+</a
02/4-b2»* 4*^20
flri/+6jW^f,n
(C) Both the lines are given in general form :
Let the equations of the two lines be
flix
+Ciz+</,●=» 0=>OaX+b^y^ C2Z+
...(1)
and
asX+bsy+CsZ+ds^0=aiX+biy-\-C4Z+dt.
● (2)
If the lines (1) ond (2) are coplanar, these will intersect in a
point say the point (a, j8, y).
The co-ordinates of this point must satisfy the equations of
the four planes representing the two lines, and so we have
flitt+^+Ciy 4- = 0,
U20t-f 62^+Cay </a=0.
Caa -H bsP -t-Cay -f rfa=0,
and
rf 64)3 4* C4y+</41= 0.
Eliminating a,
y between the above four relations, the
required condition is given by
=0
ax
bx
Cl
dx
and
02
fls
bz
Ca
di
C3
ds
C4
d.
The expansion of a fourth order determinant is
usually difficult and therefore it is convenient to solve numerical
examples by fi rst reducing the equations of the lines to symmetri
cal forms and then proceeding as in case (A).
SOLVED EXAMPLES (F)
Ex. 1. Show I hat the lines H*4-3^)«=J (3'4-5)=*—J {z—7) and
J(x4-l)=J(y4-l)<=—(z4-I)fl/’e coplanar. Find the equation of
the plane containing them.
(Rohilkband 1982; Madras 76)
Remark.
128
Analytical Geometry 3-D
Sol.
The equations of the given lines are
...(1)
(x+3)l2=>(y+5)f3=>(z-l)f-3= r,(say),
...(2)
and
(x+1)/4=(j;4-1)/5 =(z+1)/-1-fi (say).
The co-ordinates of any point P on the line (t) are (2ri—3,
3ri —5, — 3ri4-7) and those of any point Q on the line (2) are
(4r2-l. Sra-l,-ra-1).
If the lines (1) and (2) are coplanar then they intersect and
hence for some values of ti and fa the points F and Q coincide.
Thus we have
2ri-3=4ra-l. 3r,-5=-5ra- 1, _3ri+7«-ra-l
or
Ti —2ra=>l, 3r,—5fa==4, 3ri —fa=8.
Solving the first two equations, we get rit=a3, ra^l.
These values of rj and ra also satisfy the third equation, and
hence the lines (1) and (2) are coplanar (i.e. intersect).
Putting the value of ri (or the value of ra) the co-ordinates of
the common point of intersection i.e., F (or Q)are (3, 4, —2).
Now the equation of the plane containing the lines (1) and
(2)[Le. the plane containing the line (1) and parallel to the line
(2)) is
x+3 y+^ z-1
2
3
-3
4
5
-1
=0
[See § 8(A))
or
(JC4-3.)(3(-1)-(-3) 5}-(>;4-5){2(-1)-(-3)-4)
-i-U~7){2.5-3.4H0
or
(x-l-3).12-(;;-l-5).10-Kz-7).(-2)=:0 . or 6x-5;»-z==0.
Ex.2. Frove that the lines J (x—1)=|(;^-2)=J Yz—3) and
I(x—2)=J (;>—3)=^(z—4) are coplanar ; find their point of
intersection.
(Berahmpur 198IS ; Lucknow 81 ; Meerut 75)
Alsofind the equation of the plane in which they lie.
Sol. The equations of the given lines ate
...(1)
(x-I)/2=(:F-2)/3-(z~ 3)/4=-r,(say),
and
...(2)
(X-2)/3=(;»-3)/4-(z-4)/5=-ra (say).
The co-ordinates of any point P on the line(l) are (2ri-|-l,
3ri-|-2. 4ri-J-3) and those of any point Q on the line (2) are
(3r2+-2, 4ra+3, 5ra+4).
If the lines (1) and (2) are coplanar then they iutcrsect and
hence for some values of r, and rg the points F.and Q coincide.
Thus we have
129
The Straight line
or 2ri--3r2=l
...O)
2ri+ l=3ra'|-2,
or 3rj--4ra=l
...(4)
3r,+2=4ra+3,
and
or 4ri —5r2=l.
...(5)
4ri+3=»5ra+4,
l,ra=-l.
Solving (3) and (5), we get rt
These values of tx and ra also satisfy the equation (4), and
hence the given lines(1) and (2) are coplanar. Putting the values
of fi (or offa), the point of intersection P (or 0)is(—1,— 1, — 1).
Now the equation of the plane in which the lines (l)und (2)
lie is given by
●
jc—1
or
y—1
z—3
2
3
4
3
4
*5
:=0
[See § 8(A)]
jc—2;y-fz=aO.
Ex. 3. Show that the lines
(x+l)/(-3)=(y^3)/2«(z+2)/l
and
jc/l=(j;-7)y(-3)«(z+7)/2
intersect. Find the co-ordinates of the point of intersection and the
elation toihe plane containing them,
[Agra 1979, Gorakhpur 82, Kanpur 81, Robilkhand 77]
Sol. The equations of the given lines are
(x+l)/(-3)=(y-^3)/2=(z+2)/l«r,(say)
and
(x-0)/l=^(y-7)/(-3)=(z+7)/2:±.ra(say)
...(1)
.(2)
The co-ordinates of any point P on.the line (1)are(—Sri—l,
2ri+3, ri-2) and those of any point 0 on the line (2) are (ra,
—3rt-4-7, 2ra—7).
If the lines(1) and (2)intersect {i.e. are coplanar), then for
some values of r^ and r% the points P and 0 coincide. Thus, we
have
—3ri--l=ra, or Sri+fa— —1
■ (3)
2ri+3«=>—3ra+7 or 2ri+3ra=*4
..(4)
and
fj—2=2ra—7, or ri—2ra= —5
...(5)
Solving(3) and (4), we get ri= — 1, ra=2.
These values of r^ and r, also satisfy the equation (5), and
hence the given lines(1)and (2) intersect. Putting the value of ri
(or of fa), the point of intersection P(or 0)is (2, 1, —3).
Now the equation of the plane in which the lines (1) and (2)
lie is given by
130
Analytical Geometry 3-D
x+l
-3
y-3
2
z+2
I
=0
2
1
-^3
or (Jc-H)(4H-3)-(>»-3)(-6-l)-|-(z+2)(9-2)«0
or
Ex. 4. In each of the following cases show that the two given
lines are coplanar :
(1) i(*-5)=i(y-7)=
(z+3); ^ (x-8)=(>^-4)«|(z-5).
Alsofind their point of Intersection and the equation of the plane
in which they lie,
(«)
(y-2)^i fz+3);
|
(x-2)=i(y-6)^i(z-3}.
Alsofind their point of intersection and the equation of the
[Indore 1979, Madras 75, 78]
plane in which they lie.
(Hi) i(x-l)=|(:v-l)=i (^-1); i(^-5)=i (;^-7)=(z-9).
Also find the equation of the plane containing them.
(Ranchi 1976]
[Gorakhpur 1978]
Alsofind their point ofintersection.
Sol. Proceed exactly as in Ex. 1 (2 or 3) above. The
answers are
(i) (1, 3. 2). 17x-47j;~242-|-172=0.
(ii) (2, 6, 3). x-2;^+2+7=0.
(iv)(2.5.7).
(iii) X—2>^-l-z=0.
Ex.5. Prove that the lines
(x-a)la'^(y-b)lb'=(z-c)/c'
and
(x--aya=>(y^byb=^iz-c')lc
intersect andfind the co-ordinates of the point ofintersection and the
[Agra 1982]
equation of the plane In which they He.
Sol. Any point on the first line is P(aVi-|-a, 6>i-|-6, cVi+c)
and any point on the second line is Q (art-\-a\ br2+b\ crs+c").
The given lines will intersect if for some values of and ra the
points P and Q coincide. Thus, we have
6Vi-|-he=>6r8+hV f. Theseequationsare clearly satisc'ri^c=crt-i-c'
field by ri«=ra«=il.
Hence the given lines intersect. Putting the value of ri (or
of r$), the point of intersection P(or Q)is (a-\-a\ h-f-h', c-j-c').
and
131
The Straight Line
Now the equation of the plane in which the given lines lie is
given by
z
X
z—c
y
x-^a
y-b
a'
c'
6'
a
=0,or
b
c
=0.
b' d
ar
c
b
a
Ex.6. Show that the lines
x—a-{-d_y—a_z—a+d
a
«-8
a+S
x—b+c_y—b_
z—b—c
and
P—y
P
P+y
ere coplanar andfind the equationj^ the plane in which they lie.
Sol. The given-Unes will be copisuSTir
h-fl
(ft+c)-(c+<f)
{b-c)-{a-d)
a—S
a
a+8
c=0. lSee§.8(A),
condition (A)]
P-\ry
P-y
, P
Adding the third column to the first column^ the determinant
on the left hand side
b—a
2 ib-a)
=2
2a
a
2p
b—a
P
b—a
a
a
a-j-8
P+y
b+c^a—d
<x+8
P+y
P
P
aO,the first two columns being identical.
Hence the condition of coplanarity of the two lines is satisfied.
Hence the given lines are coplanar.
Now the equation of the plane in which the given lines He. is
given by
z^a—d
x—a+d
V—fl
a-8
a
a+8
p-y
P
p+y
eaO.
132
Analytical Geometry 3-2)
Adding the third colomn to the first column, we get
*+z—2fl
z—a—d
y-a
2a
a
=0.
2jS
P
P+Y
Subtracting two times the second column from the first
colomn, we get
z^a—d
x-^-z-ly
y-q
0
. a
a-f 8
=0
0
P
P-^Y
or x+z—ly^O as the required equation.'
iSxrT. Prove that tHflines-. x^y^z
x_y
a p y' ax bp cy'1 m n
will lie in one plane if
(//a)(h-c)-Kw/i3)(c-a)-K«/y)(a-h)=0.
LGerhwal 1978; Kanpur 81, 83; Meerut 89]
Sol. We clearly see that the three given lines pass through
the origin O and therefore they will be coplanar if they are
perpendicular to a line through the origin O. .
£ be the d. c.*s of this line through the origin O:
Hence if this line is perpendicular to the. given lines, we have
==0»
...(1)
...(2)
and
^/4*ijw£u=0.
...(3)
Solving(l) and (2), we have
I
S
cpy—tpy axy—cay bxP—aap
hr
.c
(h—c)Py. (c-a) ay (o~h) xp'
Patting these proportionate values of 17, C in the equation
(3), we have.
Ipy(h—c)-fmay (c—u)+/taj3(a-^h)=0.
Dividing throughout by ajSy, we get
(//a)(6-c)-|-(w/i5)(c-fl)-|-(ii/y)
as the required condition.
133
The Straight Line
Ex. 8.
Show that the lines
JL -JL
^
are coplanar if a<=b or b<=e or c=a.
Sol. We clearly see that the three given lines pass through
the origin O and therefore they will be coplanar if they are per
pendicular to a line through the origin O. Let /, w,n be the d.c.’s
of this line through the origin O. Hence if this line is perpendi
cular to the given lines, we have
and
laa-^ntbp+ncY=0,
-d)
loLla+mP/b-\:nyfc— 0,
...(2)
/a-|-iw/5“I"ny«=»0.
..(3)
Eliminating Icc^ m)3 and ny between the equations (1),(2) and
(3). we get the condition for the coplanarity of the given- lines as
a
b
c
\Ja
l/b
lie
I
1
1
=ss0.
Multiplying 1st, 2nd and 3rd columns by a, b, c respectively.
we have
!
i
I
1
I
or
a*
b^
c2
1
1
1
a
b
c
I
=0.
or
1
1
a
b
c
b»
c*
! a>
eO.
Subtracting 1st column from 2nd and 3rd columns, we get
I
0
0 ● I
d-o
c—a
a
b-a
c—a joOi or
b^—a^ c*—a*
a*
c*—a*
(b-a)(c-a)
1
b+a
or
1
1
=0,
c+a
ib—a)(c—a)(c—b)=0, as the required condition.
Hence the given lines are coplanar if
b=a or cna or b*=»c*
134
Analytical Geometry 3-D
Ex. 9. Prove that the lines x=ay-\-b=cz-{^d and x=>oty+P
=»yz+8 are coplanar If(a/3—8a)(y—c)—(c8—Jy)(a—a)=»0.
(Bardv?ao irSO; Kaopor 82)
Sol.
Tbe equations of the given lines are
x—0 y+8/fl
X>=»ay-^b=ez+d or
1
“
z-frf/r
1/fl “ 1/c
'X—0 y+PI<t , z+S/y
Xi3«y-(-j8=y2+8 or
1 = i/d ” 1/y ’
and
...(1)
...(2)
The lines (1) and (2) will be coplanar if
0-0
/3/a-8/fl
Sfy-dlc
or
1
l/a
“0,
1
1/a
0
pja—bla
1/y
]●
[See § 8 (A), condition (A)]
Sly—d/c
0
1/fl-1/a
1/c-l/y
1
1/y
l/«.
subtracting the third row from the second row
I/C
=0.
or
0,
(a/S—8a) (y-c) (cB—dy) (ol-to)
^
sssU,
flacy
, cyoa
(fl/8—8a) (y—c)—(jc8—dy) (a—fl)=0.
or
This is the required condition.
Ex. 16. Prove that the lines
3jc—5«=»4^—9«=33z fl/M/.X—1
4«=»3z
meet Ui a point and the equation of the plane in which they lie is
Sx—8y-f-3z-l-13«=>0.
(Kashmir 1975)
SoL
or
and
or
The equations of the given lines are
X—5/3 y—9/4
z
3x—5*=4>.—9«3z or
j/3 “ 1/4 “l/i
x-5/3 ;;-9/4
z
4 “
3 " “4
X—l»2y—4e»3z or
X-1
1
y~2
1/2
z
1/3
X—1
y—2 z
3 “"2
...(2)
The Straight Line
135
Now proceed as in Ex.'1 above.
Ex. 11. Prove that the lines
9)^—
.
6x+4y—5z=4, x-5y+2z=12 are coplanar. Find also their point
of intersection and the equation of the plane in which they lie,
(Agra 1980)
Sol.
and
The equations of the given lines are
.0)
...(2)
6x+4j>—5z=4, x—5;'+2r=12.
The equation of any plane through the line (2)is
(6x+4;K-5z-4)+A (x-5j>+2z-l2)=0
..(3)
or
(6+A)x+(4-5A)>^+(-5+2A)z-(4+12A)=0.
Now if the plane (3) is parallel to the line (1)* then we have
(6+A)(2)+(4-5A)(-l)+(-5+2A)(1)«=0, or A«=*~i.
Putting this value of A in the equation (3), the equation of
the plane through the line (2) and parallel to the line (1) is
given by
(6-|)x+(4+t)>;+(-5-f)z-(4-12/3)=0
...(4)
or
17x+l7;>-17z«»0, or x+y-z=0.
Clearly the line (1) passes through the point (9, —4, 5). This
point (9» -^4,4) satisfies the equation (4) of the plane. This shows
that the lirie (I)lies in the plane (4). Hence both the given lines
.(i) aud.(2) are coplanar and lie in the plane given by (4).
' To find the point of intersection.
The co-ordinates of any point on the line (1) are
...(5)
(2r-f9,-r-4, r+5).
The lines (1) and (2)intersect if this point also lies on the
line (2) fe., if it satisfies both the equations of the line (2).
Heiice we have
6(2r+9)d-4 <-r-4)-5(r+5)=4. or r=*-3
and
(2r+9)-5(-r-4)-l-2(r+5)=l2, or r=:-3.
Since both the equations give the same value of r, the two
given lines intersect. Putting the value of r in (5). the point of
intersection is (3, — 1, 2).
Ex. 12. Show that the lines i(x—1)=J
2)«i(2-3)and
4x—3y-\-\=0=5x-3z}-2are coplanar. Alsofind their point of
(Burdwan 1978; Gorakhpnr 79)
intersection.
Sol. The equations of the given lines are
‘'■V'.
.V
●.'f'
■
136
Analytical Geometry 3-D
;c~l
y-2 r-3ss
r (say).
4
...(1)
and
4jv-3;;+1=0, 5jc-32+2=.0.
...(2)
The co-ordinates of any point on the line (I) are
(2r+l,3r+2. 4/ +3).
■
.(3)
The lines (1) and (2) will intersect i.e.^ will be copianar if
this point also lies on the line (2). This point satisfies both the
equations of the line (2), if we have
4(2r+l)~3(3r-h2)+I-0, or f=^= —1
and
5(2r+l)—3(4f+3)-l-2“0, or r -1.
Since both the equations give the same value of r, the two
given lines intersect. Putting this value of r in (3), the point of
intersection is(— 1, — 1, — i),
Ex. 13. Show that the lines \
A)
1(^-1)
3x—2>’+z+:?=i0>=2x-f-33'.-l-42—4 are copianar. Also find
their point of intersection and the equation of the plane in which they
lie.
Sol. The equations of the given lines are
ac+4
2-1
— =.r (say)
3
5
.(1)
and
3x-2j+2-f-5=0, 2jc+3;>+42-4«0.
(2)
The equation of any plane through the line(2) is
(3x-2;;4-2-|-5)+A(2x-|-3;;-f42-4)=0
or
(y+2^) x+(—2-f-3A)y-f (»-f-4A) 2+(5—4A)=0.
..:(3)
Now if the plane (3) is parallel to the line (1), then we have
(3+2A)(3)-K-2+3Aj(5;+(i+^A)(-2)=»0, or A=-a%.
Putting this value of A in the equation (3), the equation of
the plane through the line (2) and parallel to the line (1) is
given by
ISt:
0
or
45x—17;>+25zi-53~0.
...(4)
Clearly the line .(l) passes through the point(-4. -6, 1) and
It satisfies the equation (4) as
45 (-4)r-17(-6)+25(l)+53 =.0.
Hence the lines (1) and (2) intersect and they lie in the
plane (4).
T^find the point of intersection.
The co-ordinates of any point on the line (1) are
- (3r-4, 5r- 6,-2r+I).
●●(5)
137
The Straight Line
The lines (1) and (2) intersect, if this point also lies on the
line (2) /.a., if it satisfies both the equations of the line (2).
Hence we have
3(3r-4)-2(5r-6)+(~2r+l)+5=0, or r=2
and
2(Jr-4)+3(5r-6)+-4(_2r+I)-4=0. or r=2.
Since both the equations give the same value of r, the two
given lines intersect. Putting the value of r in (5), the point of
intersection is (2, 4, —3;.
Ex. 14. Show that the lines
16=0=4x4-3;?—2z+3
and
x-3y+4z+6^0=x—y-{-z+l
(Gorakhpur 1975)
are coplanar.
Sol. If the given lines are to be coplanar, then we must have
7
-4
7
16
4
3 -2
1
-3
I
-1
Applying
left hand side
3
4
[See § 8(C)]
6
I
I
R^-AR^ R^—R^y the determinant on the
0
3
0
9
0
7 -6 -I
I
0-2
3
5
1-1
3
1
0
1
9
7 —6 —1
2
3
3
0
.expanding the deter
minant along the first
5
column
0.
7 -6 -22
-2
3
» by Ca—3Ci
11
=-3{-66-(-66)}=-3(-664-66)=0.
Hence the given lines arc coplanar.
Ex. 15. Show that the lines x4*;?+z —3=0=2x4-3y44z—5
/
i
138
Analyttcai Geometry 3-D
and4x—y-\-5z—7—0=>2x—5y—z—3.are coplanar, and find the
(Agra 1975)
plane in which they lie.
Sol. We shall first transform the given equations to. sym
metrical forms. The equation's of the given lines are
...(1)
X-|-.y
—3=0=2a:+3y+4z—5
and
...(2)
4x->+5z—7=0=2x—5.V-Z--3.
l^t /, m, n be the d.c.’s of the line (1); then we have
/-|-/«4-n«=0,
and
.?/+3w-l-4n=0. .
Solving, we get
/ m
/
m
n
or
l“-2 T
4-32-4 3-2
.*. the d.c.’s of the line (1) are proportional to 1, —2, 1.
Now let.us find the point where the line (1) meets the plane
z==0. The co-ordinates of this point are given by
X -3= b, 2x+3x-5=0.
x=4,y>=-l.
Solving,
Hence the line (1) passes through the point (4, —1,0). Thus
the symmetrical form of the line (1) is
X—4 v+l 2—0
/ .
...(3)
r -2
Again let /j, /Wi, be the d.c.’s of the line (2); then we have
4/i—mi-l-5ni=0, and- 2/i—5mi—ni=0.
. or U
13 7 -9*
20+2
1 +25 10+4
. the d.c.’s of the line (2) are proportional to 13, 7, —9.
Now let the line (2) meet tue plane z=0. The co-ordinates of
this point are given by
4;c-;;-7=0, 2x-5y-3=0.
jcc= 16/9, >^=.1/9.
Solving,
Hence the line (2) passes through'the point (16/9', 1/9, 0).
Thus the symmetrical form of the line (2) is
1
16
* 9 ^ 9 z-0
7 ;;^=--r2(say).
...(4)
The co-ordinates of any point P on the line (3) are
(ri+4, —2ri—1,Ti)
and that of any point Q on the line (4) are
(13r*+16/9, 7r*+1/9, - 2f2).
If the lines(3) and (4)[i.e* (1) and (2)] are coplanar, then for
some values of ri and ra the points P and Q coincide. Thus we
have
Solving,
h
mi
The Straight Line
...(5)
...(6)
...(7)
ri4.4=13r2+(l6/9).
—2ri— l«=7r2+(l/9),
and
r,= -9f2.
Solving (5) and (7), we get r» -10/11, ra= 10/99.
The values of fi aad rg also satisfy the equation,(6) and
hence the given lines are coplanar. Putting the value of ri (or of
fg) the point of intersectloh P (or Q)is(— 9/11, 8/11, —7/11).
Now the equation of the plane in which the lines(3) and (4)
[f.e.(l) and (2)1 lie is given by
z
x—4
y+l
1
-2
I
13
7
-9
0 (See § 8 (A)]
+2{I.7-(-2)(13)H0
or
(.^;-4).ll+(;;+l).22+z.(33)=0 or ^+2j;-l-3z'=»2.
Ex. 16, A, A'; B, B'\ C, C are points on the axes. Show that
the lines of intersection of the planes A'BCy ABC; B'CA, BC A ,
CAB,CA'B'are coplanar.
Sol, Let the co-ordinates of the points and A’ be (u, 0,0)
and (fl'. 0, 0); B and B; be (0, b, 0) and (0, b\ 0); C and C be
(0, 0, c) and (0. 0, c').
The equations (in the intercepts foriti) of the
A*BC
and i4S'C'are respectively given by
^+S+a b c
I
and
a o
c
Therefore, the equation of any plane through the line of inter
section of these two planes is given by
...(1)
for sonie value of the constant A.
Choosing A=l, the equation (I) of the plane [containing the
line of intersection of the planes A*BC and AB'CJ becomes
. or
z=2.
...(2)
The symmetry in the equation (2) shows that the lines of
140
Analytical Geometry 3-/)
intersection of the other two pairs of planes also lie in the plane
given by (2).
Hence the lines of intersection of the given three pairs of
planes are coplanar.
Ex. 17. Find the equation of the plane through the,line
xll=ylm==zln
ond perpendicular to the plane containing the lines
xlm==yln =*zjl and xjn ~yjl—z(m.
(Lucknow 1982, Ranchi 79, Punjab 77, Agra 75, M.U. 89(S)]
Sol. The equation of any plane through the line
xfl=:ylm=zln is
Ax-\-By+Cz=0,
...(1)
where
Al+Bm-\-Cn=0.
...(2)
Also the equation of the plane through the lines
xlm=yfn^zll and xln^yjl^zlm
(note that-both of these lines are passing through the origin) is
^ ' y
z !
or
m
n
n
/
/ =0, [See § 8(A)]
m
\mn~F) x-\-{ln’-m^) y-\-{ml- n^) z=^0.
...(3)
According to the question, the planes(1) and (3) should be
perpendicular the condition for which is
...(4)
A{mn-^F)-^B(ln-m*)^C{mi-n^)=>{),
Solving the relations (2) and (4), we get
B
m (m/—n>)—/I \ln—rri^)
(mn—l*)- I
C
I(//i-w‘*)—m(w«—
or
or
or
A
B
2
C
Pn—Im*- m*n+mP
A
B
(m-n)
m±nl(«- /;
,
{P—m*)n+tm Xi—m)
__A
^
B
(m-n){mt-\^nT+mn)^in^i)(nm+im tnl)
141
The Straight Line
C
(i—m)(in+mh+im)
A _ B _ C
m—n n—l l—m
or
Putting these proportionate values of B,C in (1), the equa
tion of the required plane is given by
im—n)+.(nrri)
w)z*=»0.
Ex. 18. Find the foot and hence the length of the perpendicular
from the point(5, 7, 3) /o the line
(x-15)/3=(j;-29)/8=fz-.5)/(-5).
Find the equations of the perpendicular. Alsofind the equation of the
plane in which the perpendicular and the given straight line lie.
Sol. Let the given point (5, 7, 3) be P.
The equations of the given line are
...(1)
(X-l5)/3=(:»;-29)/8=(z-5)/(-5)=r (say).
Let N be the foot of the perpendicular from the point P to
the line (1). The co-ordinates of N may be taken as
(3r-H5, 8r+29, -5r+5).
...(2)
jt. the direction ratios of the perpendicular PN are
3r+15-5. 8f-|-29-7, -5r+5-3,
i.e. are 3r+l0, 8r+22, -5r+2.
...(3)
Since the line (1) and the line Pi^are perpendicular to each
other, therefore 3(3r+10)+8 (8rH-22)-5(-5r+2)«0
or
98r+196=0 or r=-2.
Putting this value of r in (2) and (3), the foot of the perpen
dicular N is (9, 13, 15) and the direction ratios of the perpendicu
lar PN are 4, 6, 12 of 2, 3, 6.
A the equations of the perpendicular PN are
...(4)
(^-5)/2=(>-7)/3=(z-3)/6.
Length of the perpendicular PN
=the distance between P(5, 7,.3) and iV(9, 13,15)
«=V{(9-5)«-|-(13-7)2+(15-3)*}=:14.
Lastly the equation of the plane containing the given line(1)
and the perpendicular (4) is given by
jc-15
>^-29
z-5
3
8
-5
2
3
6
0 [See § 8(A)]
142
or
or
Analytical Geometry 3-jD
(X-15)(48+15) -(j-29)(I8+10)+f7-5)(9-16)c=0
9x—4v—z—14=0.
§ 9. To determine the equations of a straight line intersecting two
given lines.
.
Case I. If the equations of the two lines are given in symmetri¬
calforms.
Let the equations of the given lines be
...(1)
(X-ai)/A=(y-Pt)lmi=^(z-^yi)lni=»ri (say),
and .
..(2)
(jc-»2)!k=(y(z~ya)/«2=*ra (say).
The co-ordinates of any point P on (I) are
(/iTi+ai, miZi+^i, Hiri+yi)
and those of any point 6 on (2) are
(/ara+aa, iWaZa+^a* W2^2+ya)*
Now we are to find the equations of a line which intersects
the lines(I) and (2). Suppose the required line intersects the lines
(l)-and (2)in the points P and Q respectively.
Then the required line is one which joins the points P and Q.
The values of ri and fa will be determined by some additional
given conditions.
Case II. If the equations of the two given line are given in
generalform.
Let the equations of the given lines be
f/jc=>0*=iVi
and
Wat=0=Va.
Now the equations of the required line intersecting both the
given lines are «i+iUiVi=0 and tta+PaVa=0,
where the values of and /ta are determined by some additional
given conditions.
SOLVED EXAMPLES(G)
Ex 1. A line with direction cosines proportional to (2, 7, -—5)
is drawn to intersect the lines
(x-5)/3=(:f-7)/(-1)=(z+2)/1
and
.(x+3)/(-3)=(y-3)/2=(z-6)/4.
Find the co-ordinates of the points of intersection and the length
intercepted on it.
[Kanpur 1980, Lucknow 79, Bundelkhand 78]
Sol. The equations of the given lines are
(x-5)/3=(y-7)/(-l)=(z+2)/l =r, (say),
...(1)
and
(x+3)/(-3)=(j;-3)/2=(z-6)/4=/:8 (say).
...(2)
Any point P on(1) is (3ri+5, ~ri+7,ri-2),
and any point Q oh (2)is (—3ra—3, 2fa+3, 4ra+6).
The Straight Urn
143
The direction ratios of PQ are
...(3)
(3ri+3ra+8, —ri—2ra4*4,ri—4fa —8).
Suppose the line with d.r.’s (2,7, —5)meets the lines(1)and
(2). in the points P and Q respectively.
Then the d.r.’s 2, 7, — 5 will be proportional to the d r.’s
given by (3).
3fi+3ra+8 — Ti—2i*aH~4 ^rt--4ra—8
-5
2
7
..(4)
From the first two of(4), we get
7(3rx+3ra+8)=2(-rx-2fa+4)
or
...(5)
23r,-f-25ra+48=0.
And from the 1st and 3rd of(4), we get
2(r,^4ra-8)=-5 (3ri+3ra+8)
or
...(6)
17fi-|r7fa 24=0,
Solving (5) and (6), we have
fi —>a= — !●
Now putting these values of r^. and rg the co*ordinates of the
points of intersection are
P(2, 8, —3) and g(0, 1, 2).
The required length intercepted by the lines (1). and (2) on
the line with d.r.*s (2, 7, — 5)
“PG=V[(2-0y»+(8-l)*+(-3-2)»]
= V(4+49+25) = V(78).
Note. The equations of the iine PQ in the above Ex. 1 are
given by
(JC~2)/2=(;;-8)/7=(z+3)/(—5).
Ex. 2. Find the equations to the planes through the point
{\, 0,—1) and the lines
4x—y—\2=0t==3y—4z—l andy-2Z’{-2=s^0=x—5
and show that the equations to the line through the given point which
intersects the two given lines.can be written as x=y+.\=^z+2.
Sol. The equations of the given lines are
...(1)
4x—;/-13-0, 3;;—4z—1=0
and
...(2)
;;-2z4-2=0, x-5=0.
The equations of any planes through the given lines (1) and
(2) are respectively given by
4x-:>;-13+/*, (3;^-4z- 1)=0.
and
(j^-2z+2)+/*a (JC-5)=0.
If these planes pass through the point (1, 0, — 1), we get
4-0-13+/ii(0-l-4-l)=0 giving /ii=3, and 0+2+2+/*a(l -5)=0
giving/£a=l.
Putting these values of pi and /xa, the required equations of
the planes are
144
Analytical Geometry 3-D
4x-:V~134-3(3);-4z-1)=0,
iy--2z+2)+\.{x-5)r=>0
x+2y—3z—4^0,and x+y~-2z—3»=>0.
...(3)
The equations(3) are the general equations of a line through
the given point (1. 0,—1) and intersecting the given lines (1)
and (2).
Transforming the equations (3) to the symmetrical form.
we get
and
or
1
I
1
or x=»;>-i-l*=iz-|-2.
,£x. 3. Find the equations to the straight line draim from the
origin to intersect the lines
2x4- +3z—4«?»0=*——5z.—6
and
3x—y-f2z—1=0=X-H2y—Z--2.
Sol. The equations of any line intersecting the given line are
(2x4-5y+3z-4)+/Ai(x-j;~5z-6)=0,
[See §9(11)1
and
(3x-;;4-2z~l)4-/£a (x+2>'-z-2)=.0.
If this line passes through (0, 0, 0), then we have
—4—6/t»i*=0 or /iit=r—2/3; and — l—2/tta=>0 or /xa=> —1/2.
Putting these values of and
the equations of the required
line are
(2xH-5j;+3z-4)-(2/3)(x-y-5z-6)=0
and
(3x—y+2z—1)—|(x+2y-z—2)=0
or
4x-i-17;>-|-l9z-0 and 5x-4;>+5z =0.
Ex.4. Find the equations to the line drawn parallel to Jx*=«y»z,
so as to meet the lines 5x—6«=»4j>+3=»z and
2x—4=3y-h5=z.
Sol. The general equations of the line 5x—6B4;;-f3—z
may be written as
5x-6-(4>>4-3)=D and 5x-6--z=0
or.
5x—4;;—9—0=5x—z—6.
...(1)
Similarly the general equations of the line 2x-4*=»3;^-|-5e=z
may be written as
2x—3)>—9=0=2x—z-4.
...(2)
The equations of any line intersecting the given lines [i.e● »
the lines (1) and (2)) are
(5x—4;>—9)-|-/4i (5x~z—6)«=i0
and
(2x—3;>—9)-i-/*a (2x—z--4)=0
or
5 (l+/tt|) X—4)>—/i4iz—(94-6/ti)*=a0
and
,..(3)
2(1-1-fta) x-3j;^/*a2--(9+-4/1*2)“0
}
145
The Straight Line
X
y
If the line (3) is parallel to the line -^=-f
z
1’
then this
latter line is perpendicular to the normals of each of the two
planes given by (3), so that we have
4-5(I+p.)+'(-4)+l.(-f«r)=0
Pi=-1«/19
and
4-2(l+(ia)+l.{-3)+!-{-f*8)='>giving|«s=--5/7.
Putting these values of(»i and pj in (3) and simplifying, e
required equations of the line are
15jLrr76;)>4-16a--75=0 and 4x-21:f+5z-43=0.
Ex. 5. A line with directiorT^ihes proportional ro2, 1,2
meets each of the lines given by the emotions
Jf+«=2y=2z.
Find the co-ordinates of each of ihe points ofinter^cHon.^^
Sol. The equations dTflier-gtven-UiifiS are
,
=z/.l=r,(say),
...(2)
‘ (jc+fl)/2=y/l =z/l.=r55(say). .
Any point P on (1) is (fi, ri—a, ri),
and any point Q on (2) is (2rg- fli/a, Ta).
The d.r.’s of7*6 are rj—2ra+o, r%—rx-a, riIf the d.r.’s of PQ are proportional to 2, 1, 2, then
r, -- 2ra-f fl r^—rz—a ri—r.
2
.-(3)
2
1
From the first two of(3), we get r,-3fl=0 or ri=3n and
from the last two of(3), we get ri —rg--2a=0'or r^=a.
Putting these values of
and rg, the co-ordinates of the
points of intersection are P {3a, la, 3a) and Q {a, a, a).
Ex. 6. Find the equations to the line intersecting the lines
ji:_ ]
I, 2x-|-2-*-=2;'=z+l
(Agra 1977)
and parallel to the line I (a:—!)«=(>» lj=i(z~2),
Sol. Here the required line being parallel to the given line
i {x -l)=(y -l)=i(z -2), will have its d.r.’s 2, 1, 3.
Proceeding as in Ex. 1 above, the equations of the required
line P6 are I (jc -1)—>7=:^ (z i).
Ex. 7. Find the equations of the line intersecting the lines
x—a—y=z—a,x-\‘a=y=^(z+a)and parallel to the line
k [x--a)={y-a)^i{z-7a).
Sol. The equations of the given lines are
(x-a)ll^yll={z-a)l]=ri{s&y).
...(1)
and
146
Analytical Geometry 3-/)
and
...(2) .
(:v+fl)/l =;p/l=(z+a)/2=ra (say).
Any point on (1) is P(rj+o, ri, rj+a) and any point on (2) is
0(fs-n, rg. 2r8-<i).
The d.r.’s of PQ are rx—r^+Ta^
Ti—2r,-|-2a.
X—a y—a z—7xi
If the line PQ is parallel to the giveiTltne
= -j—=-y-»
then the d.r.’s of PQ are proportional to 2, 1, 3.
—mir=i^2a'lT-r8 rx-2rg+2a
-3~
2
-.(3)
From the first two of(3), we get ri-rg=2fl
...(4)
and from the last two of(3), we get.2ri—rg=2a .
Solving the equations (4), we get ri=0, rg= — 2a.
Putting these values of r, and rg, the co-q^inates of the points
ofintcrs^gnjije P(5r6rs)^«V(-3a» -2a,^—5a).
the equations of PQ are (x-A)/2=y/l=^i-A)/3
Ex. 8. Find the equations ofthe straight line through the origin
and cutting each ofthe lines
{x-Xiyii-=(y-yi)lmi=‘iz-2i)fny
and
{X -X2)//g=(j'-y2)M=(^-Sol. The equation of a plane through the first line namely
{x-xi)lli^(y-yi)lmi^{z-zi)lniis
...(1)
A (*-Xi)^-B iy-yi)4-C {z-zj)=0,
...(2)
where
.Bwj+Cwi==0.
If the plane-(1) ptoses tlirpugh (0, 6, 0)i then
...(3)
Axi^^jf^Czi=0^
Eliminating^ B,C from (1),(3) and (^,the equation of the
plane through the or^h and^tHrough the fimt liUe is
Z-Zt
y-Pi
Xi
Pi
Zt
h
m
r*i
^0^
I
Adding the secondirpw to the firstrow; we get.
X
z
y
Xl
y^
zi
0
...(4)
h
or
● ●●(40
147
The Straight Line
Similarly the plane through the origin and through the second
line is
or
(fky2-
X
y
z
^8
y%
z,
h
/«2
...(5)
-0.
Wj
x-\-{IzZ^- WjjXa) y+(Wt^a-/iV*) «=0
(S')
The planes (4)[or(4')]and 5[or (5') togethtt give the required
line.
Ex 9. Find the equations to the straight line drawn through
the origin which will intersect both the lines
x-\ y+3 z-5
1
4
Sol. The equation of any plane through the first line is
..(1)
A (x-l)+5<;;+3)+C?(g-^=»0.
where
...(2)
l.^+4.B+3.C=0.
If the plane (1) passes through (0; Oj 0)|we have
...(3)
-/<+3B-5G=d.
Solving (2) and (3). we get zI/29=.^(-2)«C/(-7).
Substituting these proportionate valUfi^of iiiVjPj: C in (1)» the
equation of the plane through the origin aihd^ thei first line is
29x~2y-7ari.0.
-(4)
Similarly the-equation of the plane thrbugli the-origin’lahd
the second line is
..(5)
9x-2y—3g*=0.
The planes(4) and (5)together give the roquirediine.
Ex.10. Find the equdtidns of M /(»e ihrdi^ tha pdint
(3.1» 2)para//e/ to the fflane Axr{-yA:^5ii^ s^aS tdJ<m(- tf^ Uhe
3C+3=);+1=2(x—2). Find also tH^poini c^in^^
Soli Let the point P be (3j Iv 2)^
The equation of the given
.(1)
The eq^tions ofihe givih
Wr)f2=^(yr\-iy^
Letthe Imq through P^ l\
the line (2)iin^e point <Qf.
fhe pOjM 12 may he tai^a^(2r^3^2^^^
the d.r.’s ofPQ are 2r-6^ 2ir-2, ri
...(2)
tire plSne(1) cut
...(3)
148
Analytical Geometry 3-D
But PQ is parallel to the plane (I) and hence PQ is perpendicularlo the normal of the plane (1) whose d.r.’s are 4, 1, 5 and
so we have
(2r-6)4+(2r--2).l +-r.5=0 or r=26/15.
Putting this value of r, the co-ordinates of the point of
intersection Q are (7/15, 37/15, 56/15).
Also from (3), the d.r.’s of PQ are
-38/15,22/16.26/15 or 19,-11,-13.
.*. the equations of the line
are
x-3 y~\ z-2.
19 ~-ir“-i3
Ex. 11. Find the equations of the line through (a, b, c) which
is parallel to the plane /x-fm>+«z«=0 and intersects the line
OiX-\-b^y-\ CyZ-\V-1-CbZ+</2.
Sol. The required line is the line of intersection of the two
planes givew below. '
(i) The plane passing through the point (a, b, c) and parallel
to the plane /x+m>>4-wzc=0. The equation of this plane is clearly
given by
...(1)
/(x—
(v—h)-fn (z—c)=0.
(ii) The plane through the point (q, P, c) and the given line
(2)
aiX-|-6i7+CiZ+</i=0=qaX+62:V+C2Z-|-</a. .
The equation of any plane through the line (2) is
...(3)
(fliX-f
CizWj4 ja (^2^+
^22-i-</2)=0.
If this plane (3) passes through (q, c), we get
(qiq-f6i6-j- Cic4 dx)+p {a^-\-bzb-)rCip-\‘d2)=^
or
(qiq+ bJ}-\- CiC+i/i)/(qa«+/^2ft+ CaC+4)*
.. s. Putting this value of in (3), the equation of the required
second plane is given by
oi^^biy-\-CiZ^dj _qaX+fe2j>4c^z-^%
qjq+Pi^-fCiC+f/i ^qjq+M-r C2<?+4^2
...(4)
Hence the equations of the required line are given by the
planes(I) and (4).
^
§ 10. To find the perpendicular distance of a point frdin a line and
the co-ordinates of the foot of the perpendicular.
(Punjab 1979; Meerut 84S)
Let P (xi, y^t Zi) be a given point and let
be a given line.
Let the equations of AB in symmetrical form be
●
x-g y—p-zy
/
m
n ’.
.
...0)
149
the Straight Line
where /,
n are the d.c.’s
PCX},¥r.^,)
of(1). The line (1) is passing
through the point A (a, y)
and has direction cosines
/, m, n. From P draw PN
perpendicular to AB. Now
it is required to find PN. From
.1^
the right angled A^APN, we
A
Al
e
have
PN^=AP^-ANK ...(2)
/
Now ^P=the distance between (a, y?, y) and P (Xj,.Vi, Zi)
.. (5)
and
^//=projection of AP on AB i.e. the projection of AP
on a line whose d.c.’s are m, n
...(4)
=(^i-«)/+(;yi-/?) w+(zi-y) n.
Putting the values from (3) and (4) in (2), we get
PN^^{{xr-cY+{y^-fif+{z^-yY)
-{(xi - «) l+iyi-fi) w+(zi- y) nY
={(^1 -
-{(xi—a) H CVi—y5) m+(zi--y)
[V
={m (zi -y)-n 0’i-/?)P+{n (Xi-«)-/(zj-y)}*
. . +{/(Pi-)9)-m(Xi-a)P
[By using Lagrange’s identity]
a
2
2
m
/
/
m
n
n
.+
+
j'l—«
zi~y
Zi-y
Xi—cc yi-fS
...(5)
Note, In the equations (1) of the line AB, I, m, n have been
taken as the actual direction cosines of the line, In case direction
ratios a, b, c of AB are given, we should either first find the direc
tions cosines of
or we should divide the R.H.S. of(5) by
(aHhHc*).
To find the co-ordinates of the foot of the perpendicular N.
Since
the foot of the perpendicular, is a point on the
line AB given by (1), its co-ordinates may be written as
-(6)
(/r-f a, mr+;9, wr-l-y).
The d.r.’s of PA^ are /r+a - x^_, mr.-\-fl—yi, nr+y-Zi.
Also PN is perpendicular to AB.
●V-
150
Analytical Geometry t-D
(/r+a-Xi)l-\-{mr.-\-P—y^.m-\-{nr+y-z^.n=a
V Ww»+«*)=/(:Vi-a)+/« {yx-^)-{-n (z,-y)
r=/(Xj-«)+m
-fi)+n (zi-y)
[V /2+wH«»=l]
Putti;ig I3iis value of r in (6) the co-ordinates of N are
obtained.
or
or
SOLVED EXAMPLES(H)
Ex. 1.. From the point P(l, 2, 3) PN is drawn perpendicular
(x-2)==.J (3'-3)-J (z-4). Findthe distance
PN,the equations to PN and co-ordinates of hi. (Rohilkhand 1981)
Sol, The equations of the given line AB(say) are
3)/4=(z—4)/5=r (say).
...(1)
The line (I) is passing through the point A (2, 3, 4). Since
JV, the foot of the perpendicular, is a point on the line (1)[ue.AB].
the co-ordinates of N may be written as
(3r+2, 4r-j-3, 5r+4)
...(2)
the d.r.»s of PN are 3r+2—1,4r+3-2,5r4-4-3 i.e.
are
3r+I, 4r;f 1, 5r+l.
...(3)
^e
d.r.’s
of
the
line
whose
equations
are
given
by
(1),
are
3, 4, 5.
Since PV is perpendicular to AB, we have
3.(3r+l)+4.(4r+l)-l-5.(5r-f-l)=0. or r=—6/25.
Putting the value of r in (2), we get
Ns{32l25, 51/25, 14/5).
Pi\r»the distance between the points P and N
um-1
A 25 )+(2r'^)+(r-^)
Pnttii^ the value of<-ia (3). the d.r.*s of PAT are 7/25, 1/25,
5/25 f.«. are7, 1,-5.
. the equations to PN i.e. of a line passing through
P(1, 2, 3) and having d.r.’s -7, 1, —5 are
^-1 y-2 2-3
, 7 - 1 --5 ■
Ex. 2. Hoy> forts the po/n/(4, 1, l) fnm the line tf intersection of x-\-y-\-z—A=^{ir==x—2y^z-4’l
Sol. Let ^e point(4, 1,1) be taken as P.
The equations of the.given line AB (say) are
(jc—4)-fy-i-2=;0,{x—At)—2y-z—Q,
Solving for x—4,y, z, we get
JC-4
y
y
z
jor JC-4
-l-f-2“l+l--2~l
...(1)
151
The Straight Line
Equations(1) are the symmetrical form of the given line AB.
The line (1) has d.r.’s 1, 2,-3 and passes through the point
^(4,0,0).
Draw PN perpendicular to the line AB given by (1), so that
Nt the foot of perpendicular, lies on (1). Hence the co-ordinates
...(2)
of iVmay be written as
2r, — 3r).
The d.r.*s of PN are
r-{-4-4, 2r-l, -3r-l i.e. r, 2r-l, -3r-l.
Since PN is perpendicular to ABy we have
r.l-K2r-l).2-bC--3»'-l)
r=-l/14.
' Putting the value of r in (2)*-we^g€t-'
^^
-1/7. 3/14)
the^sttee ofP GfT I,1) fro® lli® 6*^®” l™e (1)
=the distance between the points P and N
V(378) 3V(42)
14
14
Ex. 3. Find the equations of the perpendicular from (I, 3, 7)
(Papjaiil981)
on the line x~3—5ty y=2-{-5ty z——l+2t,
Sol. The equations of the given line, say AB, may be
written as
...(1)
(jc-3)/(-5)=(;;-2)/5=(z-|-7)/2=r.
Now proceed as in Ex. 1 above and get
N=i-53/54, 323/54, -212/54)
and the equations to PN are
. x-1 y-3 z-1
107 ”—161 ”670 '
Ex. 4. Find the locus of a point which moves so that its
distancefromihe line x=y= —z is twice its distance/from the plane
(AgHaSO)
x^y-{‘Z=\.
Sol. Let the given point be P (Xx,yi,zf^ whose locus is
required to be found.
The equations of the given line AB are
...(1)
x/i=:f/i=2/(—1).
The line (1) is clearly, passing through A (0,0,0).and has
d.r.’s 1, 1,—1. Hence d.c.’s /, ?n, n of (1) are 1/V<3» l/y3
—1/V3*
Pi
the perpendicular distance of P fnonx thetline
(1), then by § 10, we have
a
2
-1/V3 1/V3
1/V3 1/V3
1/V3 -1/V3
+
+
y,—0
Zi-0
Zi—0
Xi—0
Xi-0
Pi-0
Analytical Geometry 3-i)
152
or
=Hxi^+yi-\rZi^+yiZi+ZiXi-Xiyi)
Let Pi be the perpendicular distance of P (Xi, y^^
plane X ^-l-z=l. Then
Pz Xrt^+Zj^
V(l+HD
...(2)
from the
...(3)
According to the given problem,
Squaring,
or
or
Xi*+Vi?+Zi!+^J^^i^r-JCLVi=2(xi*+j>i*+ZiHL”2^'J'i
7^
-f2xiZi—2xi-2;>iZi^2>>i-iZi)
or
Xi®+;'i®+^i*“Wi+ 3zaXi—3xi)^i-4xf+4j?r^4zi+2“0. ■
the locus of P (Xx, y^, Zj) is given by
^j^y*+z^—5yz+3zx -3xy -4x+4y-4z+2=0.
Ex. 5. Find the locus of a point whose distance from x-axis is
(Agral981)
twice its distanceform the yz-plane,
Sol. Let the given point be IP (Xi, yi, z) whose locus is
required to be found;
The equations of x-axis are
(x-0)/1=(:h-0)/0=(z-0)/0,
because the x-axis passes through (0,0,0) and has d.c.*s 1, 0,0.
IfPi be the perpendicular distance of P (xji y^, Zi) from (1),
then
I
0 ;8
0
1 8
0
0 2
4/>!*=
y\
yx
Zl
Zl Afl
...(2)
or />!*=(-Zi)*-|-(yi)* or Pi®=:vi*+Zi*.
Let Pi be the perpendicular distance of P(xi, y^ Zi) from the
...(3)
j^z-plane. Then
pj=x-co-ordinate of P=Xi.
According to the given problem, we have
Pi=2p2 or. Pi2=4/?2»
or
>'i*+Zi^==4xi2.
[using (2) and (3)]
●V the locus of the point F(xi, yi, zO is 4x®Ex. 6. Find the length of the perpendicular drawn from origin
to the line x-f2;»4-3z+4=0=2x+3y+4z+5. Also find the equa
tions of this perpendicular and the co-ordinates of the foot-of the
perpendicular.
(Meerut 1984 S, 84 R)
Sol. The equations of the given line AB (say) are
x+2;>+3z+4=0=2x-|-3^+4z+5.
_ -The symmetrical form of the above given line is
+
1$^
the Straight Line
...(1)
(x-2)/i=(j'+3)/(-2)=(z-0)/l.
The line AB given by (1) is passing through A (2, -3,0) and
has d.r.’s 1,-2, 1. Here the point P is (0,0, 0).
Draw PN perpendicular to i45, so that AT, the foot of the
perpendicular lies on (i). Hence the co-ordinates of JV maybe
written as
(r-|-2, — 2r—3, r).
● ●(2)
The d.r.’s of PN are r+2 -0, -2r- 3 - 0^ r -0
-O)
i.e.
r+2, -2r -3, r
Since PN is perpendicular to AB, we have
(r+2). 1 + (- 2r - 3) (—2)+r. I=0 or r = - 4/3.
Putting the value of r in (2), we get
iVS(2/3, -1/3, -4/3).
Again putting the value of r in (3), the d.r.’s of PN are
2/3, -1/3,-4/3 i.e. 2,-1.-4.
the equations to PN Le. the equations of a line passing
through P(0, 0, 0) and having d.r.’s 2,-1, 4 are
jc-0 y-0 z-0 „
L.
The length of the perpendicular W
=:the distance between the points P and N
8
V(21)
3
*
Ex. 7. Find the equations of the two planes through the origin
which are parallel to the line (x “l)/2=Cv+3)/(—l)=(^+l)/(—2)
and distant Sjl from it.
(Meerut 1976, 89, 89 S)
Sol. Let the equation of any plane through the origin
be
..(1)
ax-\-by+cz=0.
The equations of the given, line are
X— 1 3^+3 2+1
...(2)
T" "" -1
-2 '
If the plane (I) is parallel to the line (2), then the normal to
(1) whose d.r.’s are a, .6, c will be perpendicular to the line (2) and
hence we have
...(3)
2.0—1./>—2.c=0 or 2c—h—2c=0.
According to the given problem, the plane (1) is at a distance
5/3 from the line (2).
Hence the distance of (I, —3, —1), a point on the line (2),
= 5/3
from the plane (1)
154
or
Analytical Geometry 3-Z)
fl.l-t-^.(~3)+c.(-l)
5_
“3*
On squaring. 9 {a-3b-cf=15
9 {a^+m+c^-6ab-2ac-\-(>bc)=15 (o*fZ>Hc*)
16a8-56*2-J-16c2+54a6+ 18flc-56&c=0
8fl2_2862+8c*+27flZ>+9nc-28^c=0.
...(4)
Putting the value of6=2 {a - c)from (3> in (4), we get
8fl3_28x4(a -c)2+8c8+27fl.2.(fl-c)+9flc-26c.2(a-c)=0
or
-50fl2+i25ac-50c2=0 or 2a2_5ac+2c*=0
or
(fl—2c)(2a—c)=0 or a=2c, Jc.
For a=2c, from (3). 6=2c
and for fl=|c,from (3), b=a—c.
■ Putting fl=2c, b=s2c in (I), the equation of one plane is
2c^+2c>;+cz=0 or 2x+2;^+z=0.
...(5)
Again putting a=|c, b= -c in (1), the equation of the second
plane is
icx-cy+cz=0 or x-2yi-2z=0.
.,.(6)
The equations (5) and (6) are the equations of the required
planes.
or
or
or
§ 11. Intersection of three planes.
(M.U. 1990)
Let the equations of three planes be given by
Ui=aiX+biy+ciZ+di=0,
...(1)
W2Sa2X+62y+C82+<4=0,
...(2)
and
«3=^8^+6jiy-j-CaZ+^a=0,
...(3)
where u^, u. and Wg denote respectively the left hand members in
the equations (1),(2) and (3). No two of these three planes are
parallel.
We know that two non-parallel planes intersect in a straight
line and hence we get three lines of intersection by taking two
planes at a time out of the three planes given by (1),(2) and (3).
There arise following three cases : ,
Case I. The three lines of intersection explained above may
coincide ie, the three given planes have a common line of inter
section.
Case II. The three lines of intersection explained above may
be parallel to each other and no two of them coincide. In this
case the three given planes.form a triangular prism.
Case III. The three lines of intersection explained above may
intersect in a common point. In this case the three planes inter
sect in a point.
\
155
The Straight Line
Before proceeding to prove the actual theorem, for conveni
ence, we make use of some notations given as follows:
Consider the matrix (or a rectangular array)
fli
bi
Cl
di
...(i)
C9
Qz
bz
C3
Let the determinant obtained by omitting the first column in
(i) be denoted by Ai
we put
Ai=
bi
Cl
di
bz
Cz
dz
bz
Similarly, the determinants obtained by omitting the second,
third and fourth columns will be denoted respectively by Aa» Aa
and A4- Thus we put
A*=
A4=
Cl
dt
Oi
bi
di
Cz
dz , Aa*=
«2
bz
dz
bz
dz
Cz
dz
Ox
bi
Cl
Oz
bz
Cz
«a
b3
Cz
The symmetrical form of the line of intersection of the planes
(1) and (2) is [See § 4]
diOz—dzOi
^ /bidz—‘b^i
Xdibz-Ozbi
z-0
(iibz—azby
b\Cz —b^i
CiQz — C^l
aibz-Oibi’ ...(4)
where
Oibz—Ozbi^O.
Now we shall discuss the three cases given above in detail as
follows :
Case 1. The three planes intersect in a common line.
The equation of any plane through the line of intersection of.
the planes (1) and (2) is given by
or
or
Wi+A«2=0
(aiX-{-biy+CiZ+di)-\-X {azX+bzyi-CzZ+dz)=0
(<»i+Aaa) x+ibfi-Xbz);'+(Ci4-Aca) 2+(rfi+A</9)=0....(5)
If the three given planes intersect in a common line, then for
156
Analytical Geometry 3-D
some value of X the plane (5) should represent the plane (3). Thus
comparing the coefiBcients in the equations (5) and (3), we have
fli+Agg
Ci-fAcg
Ca
=/t (say),
fli+Aa2-/i«8=0,
Ci+A^i—#*6a=0,
Cl+ACg —feCataO,
and
</i+At/*-ftd'a=0.
Now we are to eliminate two arbitrary constants A and y, and
this can be done from any three out of the four equations given
above. Hence eliminating A and /t from any three equations taken
at a time out of these four equations, we have the conditions as
fli
h
Cl
a-i
b%
Cj
fla
^
=0 i,e. A 1=0
bt
di :
a%
bz
dz
Oa
bz
da
ai
Cl
di
az
Cz
da
Oa
Ca
dj
bi
j'“0 i.e. A#*=^
=0 U 45—0
di
ba
Cz
dji
bz
Cz
da
=0 i.e. Ai=0,
Hence the three planes will have a common line of intersec* tion of A4=0, Aa=0, Aa=0 and Ai=0.
Alternative method. If the three given planes (1),(2) and (3)
intersect in a common line then the line (4)[the line of intersection
of the planes(1) and (2)] will lie in the plane (3), so that we have
[by § 5 (iii)]
^3(V9-Vi)+^a
CaCj+Ca (flif>a-i?a^i)=0
157
The Straight Vine
i.e„
and
i.e..
flfs
Ox
h
Cl
flg
b.
Cg
bz
c,^
=0 i.e., A*=0
)M
diO^—d^i
a^bg -- agbi
0\bg - aj)x
^+c,.0+</,=0
t7a {bxd%—b%d-j^-\‘h^ (d\a^~d.iax)‘\~d^
Qi
bt
di
Oi
bg
dg
Og
hg
dg
—0 i.e., A3=0*
[Note, While writing the symmetrical form of the equations
(4) of the line of intersection of the planes (1) and (2), we have
taken the point oh the line for which r~0. tf we take the point
for which jc=0 the condition Ai=0 is obtained instead of Aa==0.
Similarly by taking thie point for which y=0, the condition Aa^O
will be obtained.]
Hence the planes (1),(2) and (3) will intersect in a common
line If
A4=0, Aa=0» Aa=0
Case 11. the three planes form a triangular prism.
The three planes will form a triangular prism if the line of
intersection of any two planes is parallel to the third plane and
does not lie in it.
The line of intersection of the planes(1) and (2) is given by
(4). The line (4) will be parallel to the plane (3) if
^3 {biCg—biCi)+bg (citfa—C2«i)+^3 iaibg—agbi)=0
i.e„
bi
Cl
(ft
b.
C2
a,
b.
Cg
=0
A4“0-
The line (1) will not lie in the plane (3) if
bldg—bgd^
didg—<4^1
axbg—agbi
Oibg—Ogbi
j+c,.0+</s^0
i.e.,
i.e.,
ag (bldg—bodi)-{‘bg (diOg—dggi)-\-dg {pibg—
ax
bi
di
ag
b
i>g
dg
a%
bg
dg
7^0 i.e., A*?^0.
\\
158
Analytical Geometry Z-D
\
He\ice the three planes will form a triangular prism if
A4=0 and Aa^^O or Aa^^O or Ait^O.
Case ni. The three planes intersect in a point.
The three planes will intersect in a point if the. line (4) of
intersection of the planes (1) and (2), is neither parallel to nor lie
in the planers). Rather than the line (4)'must meet the plane (3)
in a point. \
.
.
Thus the condition that the three planes meet in a point is that .
A47»^0.
■
Alternative method. Solving the equations (1),(2) and (3) by
the method ofdeterminants.[This method is called Cramer’s Rule],
we have
X
-y
■
h
Cl
di
ax
Cl
dx
b%
c%
dz
«9
C2
dn
Cz
dz
z
az
Cs
bx
dl
dz
-I
ax
bx
a^
bg
d%
az
^8
^8
az
bz
az
bz
Cz
Cl
or
Ai As
As
A4
or
,_A8.
A4
...(6)
Hence the three planes will intersect in the point whose co> .
ordiimtes are given by (0) if
A 49^0.
l>ei^thb.thfee pkiies bej given by the equa¬
tions (1)^ (2^;a.hdv(^i; l^v^pirbceed asfolio#:
ev^lil^^ A^ If
tbhn^tliii5j threer {iUmes
int^^l^vin^a^ppiii^Ayiib^ cp-orHinates; are ^veh^l^ the r^ations
(6) above;
I^ Ai^Q^ #*n\eyal#te A#. A^ and^Ai(I) IffA^O (pTr Ai5^ or A8?^0)» tfe the three pknes
form i^tri|nj|^^'prism;^^
(ii)
then the three planes inter¬
sect iii a commdh:iiiic^
159
The Straight Line
Remark. If A4=0 and Aa=0 and at least one of the three
common minors
and ai^a—
of A4 and
As is not zero, then it can be proved algebrically that Aa=0 and
Ai=0. Consequently in this case the three planes will have a
common line of intersection.
SOLVED EXAMPLES (1)
Ex. 1. Examine the nature ofintersection of the planes
(i). SxJrIr-42+'2=0; Ax-ly-5z-2=0,
2x+8y-2z-l=0.
(//) x+2;;-|-3z-6=0, 3x+4y+5z-2=0,
5x+4y+324-18=0.
iiii) 2x+4;^+22-T^0, Sx+y -z-9=0, x-y-z-6*0.
Sol. (1) The given planes are
...(1)
5x+2;;-4z+2=0
...(2)
4x-2;;-5z~2=0
...(3)
2x+Sy—2z—l=0
The rectangular array of coefficients is
5
2
-4
2
4
-2
-5
2
2
8
-2
-1
We have.
5
A4=
4
2
-2 . -5
2
1
= 0
-4
8
-2
=
1
2
-4
-1
-2
-5
0
8
-2
on adding the 3rd column to the 1st column
—4 ,on adding the first row to the
second row
0
—9
2
0
8
*72:?fe0.
-2 1
Hence the ^y.en planes inters^t in a point.
Adding(1) apd (2), we get
9x^9z=0 <MT. x=z.
Putting x=z in (3), we-gct.
8y—1=6 of y=l/8;
160
Analytical Geometry 3-D
Putting x=z in (2), we get
—x—ly—2 or jc=—2;;—2.
Putting the value of y in (5), we get x— —9/4=z.
Hence the point of intersection is( -9/4, 1/8, —9/4).
(ii) The rectangular array of coefficients is
1
2
3
-6
!
3
4
5
I
4
2
3
3
3
4
5
5
1
4
3
5
We have, A4=
2
18
2
3
-2
-4
0
6
=24-24=0.
-12
0
...(5)
»
^3—
—5/?i
Since A4=0. therefore the three planes either intersect in a
line or form a triangular prism.
1
2
-6
Now Aa=
3
4
-2
5
1
4
2
18
-6
0
-2
16
0
-6
48
, by
= _96-(_96)=-96+96=--0.
Similarly we find that Aa=0 and Ai=0.
Hence the three planes intersect in a line,
(iii) The rectangular array of coefficients is
2
4
2
-7
5,
i
●1 .
-,9
- 1
: -6
i?3- 5/?i
161
The Straight Line
2
2
We have, A4=
5
1
1
-1
-I -1
0
6
1
6
, by Rt-SRa,
Ri-2Ra
4
1 -I -1
=24-24=0.
Since A*=*0» therefore the three planes either intersect in a
line or form a triangular prism.
2
6
5
2
4-7
Now Ab=
—9
5
6
21
1 -1 -6
1
0
0
5
1
by Ct+Ci, C,+6Ci
=l.(126-30)=96#0.
Hence the three planes form a triangular prism.
Ex. 2. Show that the planes
7x-Zy-7z=Q, 3x-14y-132=0, 8x-31;/-332=0
{Meerut 1977]
pass through one line andfind its equations.
Sol. The rectangular array of coefficients is
2
-3. -7
0
3
-14
13 0
8
-31
-33 0
We have, A4=»
2
-3
-7
3
_14
-13
8
-31
-33
2
-1
-1
3
-11
-4
8
~23
-9
,by Ca+C„
C.+3Ci
162
Analytical Geometry 3-jD
0
0
-5
-7
-4
-10
-14
-9
=
, by.C,+2C,l»
=-l(70_70)=0.
Since A4=0» therefore, the three planes either intersect in a
line or form a triangular prism.
2
Now Aa=*.
-3
0
3
14
0
8
-31
0
Similarly A2=0 and
=0.
Ai=0.
Hence the three planes intersect in a common line.
Clearly the three planes pass through (0, 0,0) and hence the
common line of intersection will pass through (0, 0, 0).’’The
equations of the common line are given by any of the two given
planes. Therefore the.equations of.the common line are given by
Ix—iy—lz^O^
and
3jc-14;;-13z=0.
the symmetrical form of the line is given by
X
z
z
X
y
y
39-98*~-21+26“-2S+9
_59" 5 “-19*
Ex. 3. Prove that the planes
x^2y+z -3^0,
form a triangular prism.
3-0, x-z—1=0
Sol. Proceeding as in Ex. 1 above, we get At—O and Ag9^0.
Hence the’given planes form a triangular prism.
Ex. 4. Prove that the planes
x=cy-{-bZt y=az-{-cx, z=bx-tay
pass through one line ifa*-^b^ ■\-c^+2abc=^ I, and show that the line
of intersection then has the equations
X
z
y.
V( i -a*) “ V( I - IP) “ V( 1 -- C") *
[Buttdclkhand 1978; Punjab 77; Rajasthan 75, 77]
Sol.
The equations of three given planes arc
■x-cy-bz=0. cx-y-\-az==0, bx-^ay—z—0.
163
The Straight Line
The rectangular array of coefficients is
-b
0
—c
1
c
-1
a
0
b
a
-1
0
We have, Ai—
1
—c
-h
c
-1
a
b
a
-1
=»1 (1
Also
A3=
Similarly
{^c-ab)-b («c+6)
1
—c
0
c
-1
0
b
a
0
and
0.
Ai=0*
Hence the given planes intersect in a line if A4=0
aH^“+c»+2aiw;=l.
if
Clearly the given planes pass through (0,0, 0) and hence the
common line of intersection will pass through (0, Oj 0). Let /, m,
n be the d.r/s of this line. It being perpendicular to the normal
of each plane, we have
/—cm—h»=0
...(1)
c/—mrt-flw=0
●..(2)
W+flm—n=0.
...(3)
m
/
n
Solving (I) and (2),
-ac~-b^~bc—a
— 1+c*
m
n
/
or
nc+h bc+a I—c**
...(4)
/
m
n
Solving (2) and (3),
\—a^^ab+c cai-b
...(5)
/
m
n
Solving (3) and (1),
ab+c°‘ I ~b*~bc-{-a
-(6)
Taking first two terms of each of (5) and (6) and then multi
plying, we get
m*
m*
or
1-6**
iab-^c) {l-a^)-{ab+c) (1-6*)
.. (7)
Similarly from (6) and (4),. we have
164
Analytical Geometry 3-D
m*
rt®
...(8)
Now from (7) and (8), we get
n8
./*
1-a*
n
m
/
●●
...(9)
"VCl-c”)
/. the d.r.’s of the common line of intersection of the given
planes are given by (9). Since the line passes through the.origin,
hence, its equations are given by
Ex. 5. Prove that the planes
x+ay+{b-\-c)z+d^O,
x-^by+(c-\-a)z+d=0,
x-\-cy-{-(a+b) z+d«=0.
[Allahabad 1978, Kanpur 82]
pass through one line.
Sol. The rectangular array (or matrix) is
1
a.
b+c
d
1
b
1
c
We have, A'4=
Also A3=
c+a
d
1
a+h
a
d
b+c
1
b
C+fl
1
1
c
a-\-b
fl+h+c
h+c
1
a-\-b+c
c+fl
1
a+h+c
a+h
1
1
b+c
1
1
I
1
adding 3rd
column to 2nd
=0.
c+a
l
a
d
1
a+b
a
1
1
b
d
1
b
1
1
c
d
I
c
1
=0.
1.65
The Straight Line
Similarly Aa=0 and Ai=0*
Since A4=0, Aa=0, Aa=0 and Ai=0, therefore the given
planes intersect in a line.
Ex. 6. Show that.the planes
ny—mz=X,lz-nx~fji and mx~ly=v
have a common line if/AH-/«/*+/iv=0, and the direction ratios of the
[Panjab 1980; Bebrampur 76 S]
line are /, w, n.
Show further that the distance of the linefrom the origin is
Sol. The equations of the given planes may be written as
..(1)
0.x+7i;^—/W2—A=0
...(2)
—,/ix+
=0
0.
(3)
mx-ly-\-0.2—V
The rectangular array (or matrix) of coefficients is
-A
—m
0
n
. —n
0
/
●m
-/
0
0
n
—n
0
I
m
-/
0
Wc have, A*=
-ft
—V
—m
=0—rt {0—lm)—m («/--0)=0.
the given planes will either meet in a line or form a
triangular prism.
A
n
0
Now Aa=
—n
0
- fi
m
-/
—V
=0—n
A («/—0)
= —«(/A+/Wft+nv).
If the planes are to meet in a line then As “'ust also be zero.
Then we have
n (/A-f mf*+7iv)=0
or
...(4)
/A+/Mfi+nv=0.
[We have assumed n^QJ]
If the condition (4) is satisfied, then we can see that A2=ll
166
Analytical Geometry 3-D
and also Ai=0. Hence (4) is the condition for the three given
planes to have a common lin6 of intersection.
Let a, by c be the d.r.’s of the line of intersection. It being
perpendiculer to the normal of each plane, we have
0 n-f-n6—mc=0
...(5)
...(6)
—na+0,b-\-lc=0 ma--lb+0.c=0.
...(7)
Solving(5) and (6),
a
b
c
nl'~mn~n^ or -j-a—j=«—,
/ m ,n*
showing that the d.r.’s of the line of intersection are /, my n.
To find the eqnations of the line of iotersectioo. The line of
Intersection meets the plane 2=0 in the point given by
[Putting 2=0 in(1) and (2)]
ny-\=Qy -/Iff—
or
x=-filnyy=\fn,
the line of intersection passes through the point
_
Xftty 0)and has d.r.’s /. »i, n and therefore its equations are
given by
x+ti/n y~Xfn 2-0
/
m
n
...(8)
The distance of the line (8)from the origin. Using § 10, the
required distance p is given by
n
m
1
0-A/« 0-0
t
/
m
n.
0—0
0+pjn
0-l-/i/n
0-A/n
1
—iX—mp
n
)■
1
[using (4;]
+
Ex. 7. Prove that the planes x^y sin ^-\-z sin y=z sin 6
●px sin and z=x sin
sin 0 wiil intersect in the line
_* _ y _
COS 0 cos^ cos ^
Sol.
[Allahabad 1982; Garhwal
The equations of the planes may be written as
x—y sin 0—z sin ^=0,
X sin
sin fi =0,
X sin
sin d—z=0.
78 Sj
...(1)
...(2)
...(3)
167
The Straight Line
Let us find the line of intersection of the planes (i) and (2).
Let l,m,n be the direction cosines of the line of intersection of
the planes (1) and (2). It being perpendicular to the normals of
both the planes, we have
l—m sin 0—n sin ^=0, / sin 0-m+« sin ^==0.
Solving, we get
n
m
/
—sin 0 sin ^—sin 0 —sin 0 sin 0 — sin 6 — I+sin* 0
or
I
TO
n
sin 0 sin ^+sin 0 ^sin 0 sin 0+sin
1 -sin* 0
...(4)
If +04-0=»j7c, then we have
sin 0=sin {^tc—(0-}-0)}=cos(0+0)
=cos 0 cos 0—sin 0 sin 0
or
sin d+sin 0 s-n 0=cos 0 cos 0.
...(5)
Similarly sin 0+sin 0 sin d=cos 0 cos 6.
...(6)
Using the relations (5) and (6),(4) becomes.
TO
n
/
cos 0 cos $ “cos 0 cos 0 cos* 0
TO
n
/
or
cos 6^ cos 0 cos 0 *
...(7)
Clearly the planes (1) and (2) pass through (0, 0, 0) and so
their line of intersection will pass through (0, 0,0) so that its
equations are given by
...(8)
x/cos 0=yloos 0=z/cos 0.
Now it remains to prove that the line (8) must lie on the
plane (3).
The point(0, 0, 0)through which the line (8) passes also lies
on the plane (3). Also the normal to the plane(3) whoss d.r.’s
are sin 0, sin 9, —1 must be perpendicular to (8), the condition
for which is
cos 9 sin 0+cos 0 sin 0+cos 0(—1)=0
sin(«+0)-cos 0=0, or sin (|ti-0)--cos 0=0
cos 0—cos 0=0 or 0=0 which is true.
Hence the line (8) also lies on the plane (3). Thus the equa
tions (8) are the equations ofthe required line.
or
or
Ex. 8. Show that the planes ax-Vhy-\-gz—Q^ Ax+6iy+/z=0,
gx-\‘fy-\-cz—Q have a common line ofintersection if
168
Analytical Geometry 3-X)
a
h
g
h
b
f «0
g
f
A=
c
and the direction ratios of the line satisfy the equations
^■■0A~0A’
da db
dc
Sol. The equations of the given planes are
ax-i-hy+gz=0
...(1),
hx-\-by-\-fz:=0
gx+fy-{-cz=0.
The rectangular array of coefficients is
a
h
g
0
or
A
6
/
0
g
/
c
0
We have
a
h
g
h
b
f
...(2)
...(3)
.
“ A (as given in the problem)
g
f
c
A*= A =abc-{‘2fgh—ap—bg*-ch\
...(4)
Also
a
h
.0 =0. Similarly Aa=0 and Ai=0*
A»^=
h
b
0
g
f
0
If the given planes intersect in a line then A4 must be zero
(as Asf Aa and Ai are already zero). Hence the given planes will
have a common line of intersection if A4=0 or A=0
or
...(5)
abc+2fgh ●ap—bg^—ch^~0.
Let /, m, n be the d.r.’s of the common line of intersection of
the given planes. It being perpendicular to the normals of the
planes, we have
fl/+/im+g/i=0, hl+bm-\-fn=0, g/+/m+c«=0.
Solving any two (say the first two) of these relations, we get
n
m
/
hf- bg ^gh—qf ~~ab~ li^ *
169
The Straight Line
m2
n*
...(6)
(hf-bgf~{gh-af)^ (ab-h^y
Differentiating (4) partially w.r.t. cr, 6, c respectively, we have
...(7)
d^lda=bc-f^ dMdb--=ac-g\ Zt^ldc=ab-hK
Now {hf—bgY=h^f^-¥b^g^ 2bfgh .
=/i2
b iqP+bg^+ch^^abc) using (5)
=A2f^Jrb^g^-abP~bY-bch^-\-abH ●
=h^f^-abf^-bch?-{‘abH
= -/2 {ab-h^pbc {ab-h^)
=(£7fe-/i2)(bc-P).
Similarly {gh aff={ab h^) {ca—g^).
Substituting these values in (6), we get
Squaring,
/«2
/2
\ab-h^){bc -P)^ {ab-h^){ca-^) "(a6~/*»)»
«
m2
/2
or
2
ab
h^
/>c-/2 ca-g
«2
»r
or
[using (7)]
^ ’9_A
0fl 06 0c
Ex. 9. /br what values of k do the planes
X -v+z-f 1=0, kx-{-3y-{-2z 3=0, 3x+A:j+z 2=0
(/) intersect in a point ; (/7) intersect in a line ; {Hi)form a
triangular prism ?
Sol. The rectangular array of coefficients is
I
I
-I
1
3
k
2
3
—2
k
1
3
Now we calculate the following determinants :
-1
I
0
—1
0
A4= 1
k
3
2
3
k
1
- =(A+3)
A'+3
5
k
1
3+^
adding 2nd column to 1st and 3rd
0
0
-1
1
.3
5
k-
k -}-1
1
-(^'+31 (/c-4;.
3
={k+3)ik+l-5)
170
A.39
Analytical Geometry i-D
1
—1
1
0
-1
0 , adding 2nd column
to 1st and 3rd
0
k
3-3
fc+3
3
3
k -1
3+A:
k k-2
1
0
=(^+3)(A:-2),
1
1
1
0
k
2
-3
k-1
.2 -5
3
1
-2
2
1 -3
adding(—1)times 2nd column to 1st and 3rd
{(/:-2)(-3)+10}=3A:-16.
and Ai—
—1
1
1
0
1
1
3
2 -3
0
2
-3
k
1 -2
k-2
I
.-2
adding 3rd column to 1st
5 {k-2).
(i) The given planes will intersect in a ^oint if A«t^0 and
so we must have
— 3 ^nd k^A. Thus the given planes will
intersect in a point for all real values of k other than —3 and 4.
(ii) If A:=-3, we have A4=0, Aa=0 but Av^O. Hence the
given planes will form a triangular prism if^= —3.
(iij) If Ar=--4, we have A«=0 but AsT^^O. Hence the given
planes will form a triangular prism if A:ca4.
We observe that for no value of k the given planes will have
a common line of intersection.
Ex. 10. The plane xla-[-yfh+zic=l meets the axes in B
and C. Prove that the planes through the axes and,the internal bise
ctors ofthe angles of the triangle ABC pass through the line
X.
z
y
a
"'by/{c^+a^)-c^/{a*-^ly^) *
Sol. Plane x/fl+3'/6+2/c=il meets the axes in A (a,0,0) ’
B(0,6.0),C(0,0,c).
The equation of any plane through x-axis is (i.e. y=0,2=0)
is
;;-fA2—0.
...(1)
The d.r.’s of
are —a, b*0
171
The Straight Line
—a
^
D.C.’s of AB are
.0
c
—a
,0,
VCfl'+c*)’ V(«Hc2>
D.C.’s of interior, bisector of LBAC are
at
\
1
\
*
^
2 V{a^+b^)’2 V(«*+c*)
D.C.’s of AC
Now if plane (1) passes through the internal bisector of
LBAC,then normal of(1) will be perpendicular to the internal
bisector of Z-5i4C.
c
b
-0
O+l.
●2V(«*+c*)
2v'(a>*4-^>«)
b\/ia^+c^)
or
cVia^+b^}
Put this value of A in (1), the equation of plane through x*axis
and internal bisector of LBAC is
^>V(g8+c^)
^ cyf{a^-\-b^)
y
^
or
...(2)
Similarly the equations of other two planes arc
y
X
z
...(3)
X
...(4)
Planes (2), (3), (4) pass through the lines
X
a
y
.
g
cy/{a^-^b*)
Exercises
1
2
Find the coordinates of the point where the line
(Ar-l)/2=(3'-2)/-3=(z+3)/4
meets the plane 2x+4;-—z+l=0.
Abs. (10/3, -3/2, 5/3).
Show that the distance of the point of intersection of the line
(x-3)/l=(;^-4)/2=(z-5)/2
and the plane x+y-\-2=\l from the point (3, 4, 5) is 3.
3. Find in symmetrical form the equations of the line
x+>’+2+1=0=4jv+v-2z.+2
172
Analytical Geometry 3-i)
and find its direction cosines.
1
X+-3 y+^ z-0
Ans.
2 ~ —1
-1
Direction cosines are —
2/\/6, —1/\/6*
4. Show that the equation of the plane which contains the two
parallel lines
x-3
x—4 y—3 z—2
, ..
_=
and ■■
1
4" “"5
1
IS
1U-V-3Z -35=0.
5. Show that the equation of the plane containing the line
and parallel to the line xfa-zlc=\, y^m^O is
a |--i
*
O
C +l=0.
6.
Find the equations of the perpendicular from the point
(1, 6, 3) to the line
X y~l z—2_
1 ~ 2
3 *
Find also the coordinates of the foot of the perpendicular.
Ans. Coordinates of the foot of the perpendicular are
(1,3,5).
Equations of the perpendicular are
x—i y—6 z-3
0
-3 "" 2 ●
7, Show that the following pairs of lines are coplanar :
(i) x—4^—^(y+\)=:z
and 4x—y+5z~l=0==2x-5y-z~-3.
Also find the equation of the plane containing them.,
(Kanpur 1980)
(iii)
(:«- 3)- --i(>-2)-(zH-l)
●
and jc+2>^+3z=0=2x-1-4;;+3z+3.
Also find the point of intersection.
Ans. (i) x+2y \-3z==2
(ii) (9,-6, 1).
8. Show that the lines
,v+2;;-5z+9=0=3;c-j^+2z-5
and 2x-{-3y—z—3—0=‘4x—5y-\-z-\~3 are coplanar.
9. Prove that the lines
x—3y-\-2z-^4=0—2x+y-^-4zi-l
The Straight Line
173
and 3x+2;;+5z—l=0=2j'+z
intersect at the point (3, 1, —2).
10. Find the equations of the line which can be drawn from the
point(2,-1, 3) to intersect the lines
i(x—l)«i(j'—2)=i(z—3)
and
Ans. J2X+4;;—9z+7=0= 1 lx11. Find the distance of(—2, 1, 5) from the line through (2, 3, 5)
whose direction cosines are proportional to 2, —3,6.
Ans. ^ *^(61).
12. Prove that the equations of the perpendicular from the point
(I, 6, 3)to the line
y-\ z-2 are x-1 y—^ z—3
^= -3 “ 2
3
2
and the coordinates of the foot of the perpendicular are
(Meerut 1977)
(1,3. 5),
13. Find the locus of a point which moves so that its distance
from the line x—y—z is twice its distance from the plane
x+;»+z=l.
Ans. .v®4->*4*z*+5xy+5;^z+5zx—4x—4;»—4z+2=0.
5
Shortest Distance
Definitions.
§ 1. Skew lines.
Two lines are called skew lines or non-intersecting lines if they
do not lie in the same plane. Skew lines never intersect and.are
not parallel.
Shortest distance. The straight line which is perpendicular to
each of the two skew lines is called the line of shortest distance.
The length of Jhe line of shortest distance intercepted between the
skew lines is called the length of the shortest distance. The
shortest distance is briefly written as S. D.
§ 2. Length the equations of the line of shortest distance.
To find the Shortest distance between two given lines and to
obtain the equations of the shortest distance.
(Allahabad 1978;Behrampur 81; Gauhati 78; Indore 78;
Kanpur 77, 78, 83; Punjab 76; Rajasthan 73, 77; Robilkhand 81)
Several methods, depending upon the forms of the equations
of the skew lines, are followed to find the shortest distance. They
are as follows ;
Method I. Projection method. The equations of the skew lines
being given in syrhmetricalforms.
Let the equations of the given lines
be
x-xi y-y^
mi
and
^-^8
Z—Zi
.. (1.)
S.l>
y-y2
ms
fit
...(2)
C
The line (1) is passing through the
point A {Xiy yi, z{) and has d.c.’s pro
portional to /i, mi, ffi. The line (2) is passing through the point
B{x2, y^, Zj) and has d.c.’s proportional to /j, m„ n^.
/i
Shortest Distance
175
Let PQht the line of shortest distance between the two lines
so that PQ is perpendicular to both the lines (1) and (2).
Let /, m, n be the d.c.’s of the line of shortest distance PQ.
Then we have
//i+wwi+/wi=0 and //a+mwa+nwa—0.
Solving, we get
/
m
n
/MiWa-
“Wi/g—
-/,Wi
1
Vmrrtint—mtni)^} ~
/Wj/ii)*}*
Now PC=the length of shortest distance between the given
lines (1) and (2)
=the projection of the segment AB on the line of
shortest distance PQ
(^f-.vi)+ w iyt-y^)+n (Za-Zi)
(min,-w2«i)
Wi/i){yt~>d
—
—
(Zj—Z\)
x^—Xi
yt—yi
Zz—Zi
h
nil
Wl
I2
/«j
/la
...(3)
This is the required length of the S.D. between the given
lines (1) and (2).
The equations of the shortest distance. Clearly the line PQ of
the shortest distance is coplanar with both the given lines (1) and
(2). Hence the line PQ of shortest distance is the line of inter
section of the two planes, namely,(i) the plane containing the
given line (I) and the line PQ of shortest distancj and (ii) the
plane containing the given lini (2) and the line PQ of the shortest
distance.
Now the equation of the plane containing the given line (1)
and the line PQ of the shortest distance whose d.c.’s arc /, /w, n is
X-rXi
y-yi
k
/Ml
/
m
Z-Zi
...(4)
n
Also the equation of the plane containing the given line (3)
and the line PQ of the shortest distance whose d.c.’s are /, m, n is
Analytical Geometry 3-Z)
176
x~x^
y-y^
2—Zo
It
»?2
«8
/
m
n
=0.
...(5)
The equations of the line PQ of the shortest distance are
given by the planes (4) and (5) taken together.
Note, If the shortest distance P0=O, then the two given
lines (1) and (2) will intersect i.e., they will be coplanar. From
(3), we observe that
F6=0 if
x^- xi
yi-y\
h
Wi
/8
W2
22~zx
=0
«9
which is also the condition for the two lines (1) and (2) to be
coplanar. [See chapter 4 § 8 (A)].
Hence we can give an another statement for the two lines to
be coplanar.
it
Two lines are coplanar if the shortest distance between them
vanishes.
Method IL General co-ordinates. The equations of the two
lines being given in symmetricalform :
Let the equations of the two lines be given by (1) and (2)
[See method I]. .
The general co-ordinates of the points on the two lines (1)
and (2) are given by
(hn+Xu Wirj-f
«iri-l-zi), say the point P.
and
(kfi+Xiy rntr^+yi, n^r^-j-Zi), say the point Q.
Let JP and Q be the points where the line of shortest distance
meets the given lines(1)and (2) respectively, so that the line PQ
is perpendicular to both the given lines (1) and (2).
Now find the d.r.’s of the line PQ and apply the conditions
that the line PQ is perpendicular to both the given lines(1) and
(2). Thus two equations in
and rg are obtained. Solve these,
equations to get r, and /g. Having found i\ and rg, the co-ordinates
of the points P and Q and also the d.r.’s of the line PQ are
known. Now we can at once find the length PQ of the shortest
distance and also the equations of PQ.
177
Shortest Distance
Method III. The equations of one iine being given in general
form and those of the other iine in symmetricalform.
..(6)
Let the equations of one line be Mi=0=Vi
and the equations of second line be
...(7)
(x~-Xi)lli=(y~yi)lmi=‘(z—Zi)lnx.
The equation of any plane through'the line (6) is
...(8)
Wi+Av,=0.
Find A so that the plane (8) is parallel to the line (7) and
substitute this value of A in (8). Then the length of shortest
distance between the given lines (6) and (7) is equal to the length
of the perpendicular from any point, say the point (Xj, yu Zi), on
the line (7) to the plane (8).
The equations of the shortest distance. The shortest dwtance
is the line of intersection of the two planes namely (a)the plane
containing the given line (6) and perpendicular to the plane (8)
and (b) the plane containing the given line (7) and perpendicular
to the plane (8). Hence the equations of the line of shortest
distance are given by the equations of the planes (a) and (b)taken
together.
Method IV. The equations of both lines being given in general
form.
Let the equations of the two given lines be
...(10)
Mj=0«=3V,
...(9) and tt2=0=Vj.
The equation of any plane through the line (9) is
tti + AiVi=0.
...(11)
The equation of any plane through the line (10)is
l/2"t'A2V2=0.
..(12)
Now Ai and A2 are determined with the conditions that the
planes(11) and(12) are paraliet.
The shortest distance is equal to the distance between the
parallel planes (11) and (12).
The equations of the shortest distance are given by the
equations of the two planes namely (a) the plane through the line
(9)and perpendicular to the plane (11)[or (12)] and (b) the plane
through the iine (10) and perpendicular to the plane(11)[or (12)].
Remark. For the sake of convenience we can . reduce the
'equations of one or both the straight lines to symmetrical formahd
then follow the methods 1, II or 111 as explained above.
178
Analytical Geometry 3-D
SOLVED EXAMPLES
Ex. 1. Find the shortest distance between the lines
(X-1)/2c=(;,~2)/3«=(2-3)/4;
(*-2)/3«(j;~4)/4={2-5)/5.
Show also that the equations of the shortest distance are
11x+2;F-7z+6=0=7x+>-5z+7.
(Agra 1974, 78;Berahmpnr 76S,81S; Madras 76;
Meerut 78, 86, 89; Vikram 78)
Sot. The equations of the given lines are
(*“1)/2=»(3'—2>/3ei(r—3)/4=rt .(say)
●..(1)
(x—2)/3«>(y—4)/4«=i(z—5)/5=ra(say).
...(2)
Method L (Projection method). Let /, m,n be the d.c.’s of
the line of S. D. Since it is perpendicular to both the given lines
(1)and (2), therefore we have
2/+3ni-{-4n=0; 3/+4m+5ncB0.
th
n
Solving, we get
15 -16 iz^lO 8—9
/ m
n
1
\/(F+m^+n*)
or
-1“ 2 -l"i(-l)>H-(2)«+(-i)»>=:76*
A The d.c.’s of S. D. are — 1/V6,2/^6, —1/^6.
Now i4 (1, 2,3) is a point on line (1) and j8(2, 4, 5) is a
point on the line (2). . Hence
the length of S.D.otbe projection of join of A and P on the line
whose-d.c.’s are-1/V6, 2/v'6, — 1/V6
=(-l/V6) (2-D+(2/i/6) (4-2)+(-1/v'6) (5-3)
=> l/V^*
The equations of S. D. (See § 2)
The equations of the plane through the line (1) and S. D. is
1 y-2 z-3
is
2
3
4
-I
2
-1
=0 or
llx+2y—7z^6»0
And the equation of the plane through the line (2) and S.D.
x-2 y-4 z-5 I
3
4
5
j«0 or 7x+y—52+7«=iu
...(4)
-1
2
-1
‘
s.a The equations (3) and (4) together are the
●
...(3)
equations of the
Shortest Distance
179
Method 2.
Any point on line (1) is
(3)
(2ri-fl. 3ri-i-2,4riH-3), say ?.
Any point on line (2) is
(3r»+2, 4ra+4. 5ra4- 5), say Q
The d.r/s of the line PQ ate
(3i*,+2)-(2r,+1),(4ra+4)~(3r,+2). (5r*+5)-(4ri+3)
or
...(5)
3r»—2tj+ I, 4i*j—3ri+2, 5ra~4f|+2.
Let the line PQ be the line of shortest distance, so that PQ
is perpendicular to both the given lines (1)and (2) andrtherefore,
we have
and
or
2(3ra-2ri+l)+3(4ra-3r,+2)+4(5r,-4ri+2)=a
3(3ra-2r,+l)+4 (4ra-3ri+2)+5 (5ra~4fi-f2)=0
38ra-29ri+I6=0 and 50ra-38r,+21=0.
Solving these et^uations, we get ri= 1/3, ra=—1/6.
Substituting the values of ri and fa in (3),^4rafld^5)rwciiSV?r
P(5/3, 3,13/3), Q (3/2, 10/3, 25/6)
and d.r.*s of PQ (the line of S.D.) are — }, -i i.e. — 1, 2, — 1.
The length of S.D.a the distance between the points/* and Q
1
V6
The equations of S.D. are either given by equations (3)and
(4) of method 1 above or we can write the equations of a line
passing through the point P and having d r.’s — 1, 2, — 1.
Ex. 2 Ftitd the shortest distance between the tines
X—3 y-8 2—3 x+3 y+7 2-6
3 “~1=~’'=3““2“=1“‘
Find also its equations and the points in which it meetsthegfven
lines.
(Avadh 1982; Garhwat 79; Indore 16, 79;
Kaopnr 74, 8(^ Lucknow 80,81; Madrad 78;
Meerut 72,80, 83, 89S; Rohilkband 78)
Sol. The equations of the given lines are
(jf—3)/3=*(j;-8)/-l*=(z-3)/l=ri (say)
. (1)
and
(:«+3)/(-3)=0»+7)/2=(z-6)/4=ra (say)
...(2)
Any point on line (1) is (3ri+3, -ri+8, r,+3), say P. ...(3)
Any point on line (2)is(-3fa-3, 2ra-7,4ra+6), say Q.. .(4)
The d r.’s of the line PQ are
(
3) (3ri-i-3),(2fa—7)—(—ri+Oh (4ra-i-6)—(f|+3)
or
—3rt—3ri—6,2ra+ri —15,4ra—ri+3.
(5)
m
Analytical Geometry 3-/)
Let the line PQ be the lines of S.D., so that PQ is perpendi
cular to both the given lines(1)and (2), and so we have
3(-3f,-3r,-6)-1 (2r,+f,-15)+1.(4ra-r,+3)=0
—«I^*-^t=.3fj-3r,-6)+2.(2rs+r,-15)+4(4r>-r,+3)=0
or
—Tfg—1lri=0 and 1 lr2-|-7ri=0.
Solving these equations, we get ri<=raoO.
Substituting the values of n and in (3),(4)and (5), we have
Ans.
^(3, 8. 3), e(-3,-7, 6)
Afid thed.r.’sof Pfi (theline ofS.D.) are -6,-15. 3 or
-2,-5. 1.
The length of S.D.<=>the distance between the points P and Q.
“V{(-3-3)>-i-(-7-8)*+(6-3)*}=3V30.
Now the line PQ of shortest distance is the line passing
through P(3, 8, 3) and having d.r.’s -2,-5, 1 and hence its
equations are giveiLby
^
^^
X—3
8 r—3
or x-3 y-B z-3
2
-2 “-5 = T5
1
Ex. 3. Find the length of the shortest distance between the lines
x-3 yS
y+\ z-M
1 “-2
1 * 7 “-6 "“-I ’
(Garbwal 1981; Kanpur 83; Meerut 83S,i86P; Robilkhand 82)
Show also that its equations are given by
(x-l)/2=(j^-2)/3=(z-3)/4.
(Meerut 1986P)
Sol. The equations of the given lines are
(x-3)/l«(jv_5)/(-2)=(z-7)/l=n (say);
...(I)
and
(x+l)/7=(:p+l)/(-6)=(r+l)/l4=ra(say).
...(2)
Any point on line (I) is(ri+3, -2ri+5. ri-f7), say P. ...(3)
Any point on line (2)is(7r8-l, -6ra-l, ra-l), say Q....(4)
liie d.r.*s of the line PQ are
(7ra-l)-(r»+3).(-6r*-l)-(-2ri-|-5),(ra-l)-(ri+7)
or
7ra-ri-4,-6r*+2ri-6, ra-r,-8.
...(5)
Let the line PQ be the.line of S.D., so that PQ is perpendi
cular to both the given lines (1)and (2), and so we have
l(7ra—fi—4)—2(—6ra+2rj—6)-{- l(ra—ri —8)es0
and 7(7r8-ri—4)-6(—6ra-t-2ri-6)-l-l.{ra-ri-8)c<=0
or
20ra—6ri«=0 and 86ra—20ri=0.
Solving these equations, we get ric=ra»0.
Substituting the values of r/ and fa in (3),(4) and (5), we have
P(3,5.7).C(-1.-1, -1)
aqd the d.r.*s of PQ (the line of S.D.) are —4, —6, —8 or 2,3,4,
181
Shortest Distance
The length of S.D.«Pe=v'{(-l-3)H(-l-5)*+(-l-7)*}
=V{(4)*+(6)*+(8)*}=2>/{29).
Now the line PQ of shortest distance is a line passing tbroogir
S, 7) and having d.r.’s 2» 3, 4 and hence its equations are
given by
*-3 y-5 z-7
2 “ 3 “4
or
z-7
3“+>=—■'■*
x-l y-2 2-3
2
3
4
Proved.
Ex. 4. Find the shortest distance between the iines
(x-l)/2=(j;-2)/3=(z~3)/4; (x-2)/3=(;;-3}/4«(z-4)/5.
Hence show that the lines are copianar.
(Meerut 198$)
Sol. We are solving this problem by projection method. The
equations of the given lines are
/t\ (X- l)/2«(y-2)i3i=(z_:r.3J/4_:_-:-.●●vi;
and
(x-2)/3=(;;~3)/4=(z~4)/5.
...(2)
Let /, m, n be the d.c.’s of the line of S D.
The line of S.D;
being perpendicular to both the lines (1) and (2), we have
2/+3m+4/1=0 and 3/+4/n+5/i=0.
Solving these equations, we have
/
m
n
15-16= 12-10 8-9
I
m
n
1
V^+w*+n*)
or
^"^^"'^'"V((-1)=‘+(2)*+(-1)*}‘=“V6*
...(3)
Clearly the line (1) passes through.the point A (1, 2, 3) and
the line (2) passes through B (2, 3, 4).
/. The length of S.D;=The projection of join of <4 and B on
the line of S.D. whose d.c.*s are
or
= /(2-l)+i«(3-2)+«(4-3)
-1
Since the length of S.D.=0, hence the given lines are coplanar/.e. intersecting.
Ex. 5. Find the points on the lines
x—6
,
A j
3^+9 ^—2
~--=,-{y^n)=xz-Aand
4
which are nearest to each other. Hence find the shortest distance
between the lines and also its equationsi
(Bundelkhand 1978)
182
Analytical Geometry 3-D
Sol. The equations of the given lines are
...(1)
<x-6)/3=(j;-7)/(-l)=(2-4)/l=ri (say)
...(2)
and
JC/(-l)=U'+9)/2=^{z-2)/4=r2 (say)
The points on the lines (1) and (2) which are nearest to each
other are the points where the line of S.D. meets the lines (1)
and (2).
...(3)
Any point P on (1) is (3r,+6, -ri+7, r,+4)
...(4)
and any point Q on (2) is (—3f2, 2rz-9, 4rg+2).
The d.r.*s of PQ are
...(5)
—3r2^3ri—6» 2rg+rj—16, 4f2—rj—2.
Let the required points be P and Q, so that PQ is the line of
S.D.- Hence PQ is perpendicular to both the given lines (1) and
(2) and so we have
3(^3r,-3ri-6)-l(2r2+r,~16)+l.(4r2 -r,-2)=0
and -3(-3r2-3ri-6)+2{2r2+ri- 16)+4.(4r8-ri-2)=0
or
-7r2-H.r,-4=0 and 29r2H-7ri-22=G.
Solving these equations we gei
I, ra‘='h —
Substituting these values in (3),(4) and (5), we have
(3. 8, 3). e(-3,-7, 6)
and d.r.’s oi PQ (the line of S.D.) are -6, -15, 3 or 2, 5,-1.
The length of S.D.=Pe= y/ii“3 -3)H(-7-^8)*+(6-3)*}
= V{.6)*+fl5)*-|-(3)*}=3V(30).
The line PQ of shortest distance is a line passing through
P(3, 8, 3) and having d.r.’s 2, 5. — I and hence its equations are
X-.3 y^S z-3
2
.5
—1
Ex.6. Show that the shortest distance between the lines
x-i^a^ly^—Uz and x-y-{-2a=6z -ta tS‘2a.
(Barahampur 1981)
Sol. The equations of the given lineis are
...(1)
(x+<3)/12=);/6=z/(- 1),
and
,..(2)
*/6‘=(j'+2fl)/6=(z-a)/I.
Let /, m,R be the d.c.*s of the line of S.D. The line of S.D.
being perpendicular to both the lines (1) and (2), we have
12/+6r»—R=0and 6/+6m+n=0.Solving these relations, we have
n
/
Rl
6+6 -6-12 72-36
1.
^
. Vl^^'+R»*+R*) _1
or
2
3:“6“.VK2/+t-:>)*+(6P)“7’
M
Shortest Distance
Clearly the line(1) passes through the point A(—a,0,0)and
the line(2) passes through J?(0,—2fl, fl).
/. The length of S.D.=The projection of join Ofi4and£.
on the line of S.D. whose d.c.’s are
/,
n.
«=-/(0+fl)+m(-2a-0)+ii(a-0)
=(2/7) a-(3/7)(-2ii)+(6/7) ae-2tl.
Proved.
Ex. 7. If the axes are rectangular,find the shortest distance .
between the lines y=az-{’b,
and y=>a'z+b\ z=a'x+^'.
Hence deduce the condition for the lines to be coplanar.
Sol. The equations of the given lines in symmetrical form
are given by
*+i8/?=3LZ?=,i..
l/«
a
I*
x+P'la_y-b'_z
i
a'
J/«'
and
...(1)
...(2)
Let /, m,n be the d.c.’s of the line of S.D. The line of S.D,
being perpendicular to both the lines (1) and (2), we have
/.(l/a)+m.fl+n.l=0 and /(l/a0+wa'+«.l=0.
Solving these relations, we get
n
m
I
a-fl’ (l/a')^(l/a)“U/a) a'-(l/«').a
n
m
/
or
a-a'
aV—fla
aa'(fl-fl')
1
.(3)
Clearly the line(1) passes through the point ><(—^/a, b,0)
and the line (2) passes through 5(—
h\0).
The length of S.D.=The projection of the join of A and B on
the line of S.D. whose d.c.’s /, m,n are
given by (3)
«/(-/37«'+j3/a)+w(b'-6)+n(0-0)
/+(4'-6)jn
”a«'
+
—a' 4-(a—«')“●+*
“~da.y
(b’-b)((t-a')
'
184
Analytical Geometry 3-1)
[Putting the values of /, m from (3)]
{a-a’)-¥{h-b')(a-a')
V(«V* (fl—
*
neglecting the negative sign.
This is the required S.D.
If the given lines are coplanar i,e. intersecting, then S.D =0.
Therefore the required condition is given by
(aj8'- «'j3)
-6')(a a')=--f0.
Ex. 8. Show that the mortest distance between any two oppo-^
site edges of the tetrahedron formed by the planes y+z=0,zf x=0,
x4->»=0, x+y+z—a is 2a/\/6 and that the three lines of shortest
distance intersect at the point
x-^-y-=^ z=^—a.
(Gorakhpur 1975; Lucknow 76,■ Meerut 87)
Sol. The equations of the edge of the tetrahedron determined
by the planes v+z—0, z+x=r-0 are
...(1)
The equations of the edge opposite to that given by (1) are
determined by the planes x-|-;?=0, x-r>^+z^-=a and hence are
given by
x/l=:y/(-l)=(z-fl)/0.
● (2)
Let/, m, « be the d.c.’s of the line ol S.D. between (1) and
(2). This line of S.D. being perpendicular to both the lines (1) .
and (2), we have
/.l+m.l+n.(~l)=0and /.l + /7:.(-l) + n.0=0.
„ , .
/ m n
V(/H
1
...(3)
Qearly the line (1) passes through the point .4^0, 0, 0) and
the line (2) passes through j9(0, o, a).
The length of S.D.= 1 he projection of the join of ^ and
iS on the Imc ofS.D. whose d.c.’s
_
soivmg. rr°ryu^)--Hi;‘+(yrv6-
/, w, n are given by (3).
(0 - 0)-i- /2 («-G)
c=r
The equations of S.D. (See § 2).
The equation of the plane through the iiuc 0; and S.D. is
X
2
y
1
●
1
i
i
2
^.a:
185
Shortest Distance
or
or
x(2+l)-;v(2+l)+z(l-l)=0
...(4)
x--y=0.
The equation of the plane through the line (2) and S.D. is
z—a
X
1
-1
0
1
1
2
=0
or
or
X (-2-0)-j^(2-0)+(z-a)(1+1)=0
...(5)
x+^'-z+a—0.
The equations (4) and (5) together are the equations of the
S.D. These equations are clearly satisfied by the point
x—y:^z= — a. .
By the symmetry of the co-ordinates of the point
JC'-;»=z=-a,
it follows that the other lines of S.D. between other two pairs of
opposite edges will pass through the point x^y^z^—a.
Hence the three'lines of shortest distance intersect at the point
x^^y=az^—a. ,
Ex. 9. Find the length and position of the S.D, between the
lines
x_y+l_z--2
,5x-2j-3z+fi=0=x-3j^-h2z-3.
4 3' 2 ’
Sol. The equations of (he given lines are
...(1)
x/4==(;;-!-I)/3=HZ~2)/2
and
...(2)
5x-3z-f6=0^X—3>’4-2z—3.
Here we shall use method 111 of 2.
The equation of any plane through the tine (?) is
(5x-2>-3z+6)+A (x-3;-+2z-3)=0
or
(3)
JC (5+A)4-j»(-2-3A)+z (-r-3+2A)4^(6-3A)=.0,
rjf the plane (3) is parallel to the line (1), then the. normal to
the plane (3) will be perpendicular to the li.ne (,1) and hence we
have.
. .
^
4(5+A)+3(-2-3A)+2(-3+2A)=0 or A=8.
Putting this value of A in (3), the equation of the plane
through the line (2) and parallel to the line (1) is given by
l3x-26;»+r3,z-18--0.
...(4)
● Clearly the line (i) is passing through the point
—1, 1).
186
Analytical Geometry 3*D
Length of S.D.=*The length of perpendicular from the
point
2) to the plane (4)
I3.0-26.(~l)+13.(2)-18
“ VUi3j*+(-26)»+(l3)»)
34 34V6 17V6
“I3V6“13x6
39 ■
The equations of S.D. The equation of the plane through the
line(1) and perpendicular to the plane (4) is given by
! X
y+.l
2—2
3
i 13 -26
X
y+I
or
13
4
3
2
=0,
13
2—2
2
=0
1
1
-2
*(3+4)-(;;+i)(4 -2)+(2~2)(-8-3)=0
7x-2y-ll2+20«=«0.
...(5)
Again the equation of any plane through the line(2) is
(5jc—2;>—32+6)+/*(x-3>>+22—3)=0
or
x.(5+/*)+:f(-2-3/*)+2(-3+2/*)+(6-3/t)=0. ...<6)
If(6) is perpendicular to the plane (4), then we have
13(5+/*)-26(_2-3/*)+13(-3+2/*)=0.
Dividing it by 13, we get
5+/*-2(-2-3/i)+(-3+2/t)=0, or /t=-2/3.
Putting the value of /* in (6), the equation of the plane
through the line (2) and perpendicular to the plane (4) is givefi by
I3;c-132+24=0.
...(7)
A The equation (5) and (7) together are the required equa
tion of the S.D.
Ex. 10. Find the length of the shortest dlstahee between the
z-^is and the line
x+j»+22—3«0=2x+3;>+42—4. (Gorakhpur 1981)
Sol. Here we shall use Method III of § 2.
The equations of 2>axis are
xi0=yj0=zl\.
.(1)
The equations of the other line are
*+J'+22—3=0,2x+3;>+4r—4=0.
...(2)
the equation of any plane through the line (2)is
(x+>+22-3)+A (2x+3>+42-4)=0
or
X (1+2A)+j>(1+3A)+2(2+4A)-3-4A=0,
..,(3)
or
or
m
Shortest Distance
If the plane (3) is parallel to the line (1)[i.e. z-axis] then the
normal to the plane (3) will be perpendicular to the line(I) and
hence we have
0.(1 -f 2A)+0.(H- 3A)+ l.(2+4A)=0. or A=—
Putting this value of A in (3), the equation of the plane
through the line (2) and parallel to z-axis [i.e,(1)] is given by
X 0+j^.(l--3/2)+z.0-3+2=:0, or y+2=^0.
-W
Clearly z-axis i.e., the line (1) passes through the point
(0, 0,0).
Length of S.D.-the length of perpendicular from the
point (0. 0,0) to the plane (4)
0+2
=1=2.
Ex. 11. Find the shortest distance between the z^axis'-and the
line
(Meernt 1984)
ax+by-^ cz i-d^^O =a'x+b'y-\-c'z-{-d\
"—
olso that it meets the z^axis at a point whose distance
from the origin is
(ab’-d'b){bc'-b'c)+(ca'-c'a)(ad'-a'd)
(6c'-6'c)H(ca'-c'fl)»
Sol.
(Agra 1979; Allahabad 78; Bundelkband 79;
Kanpur 75, 76, 79, 82; Lucknow 77,82; Rajasthan 78)
The equation.s of the z-axis are
X
b
y
2
(^^y)*
The equations of the other line are
aX’\-by-^cz-{-d=‘0^a*x-{-b'y’\‘C*z-\‘d\
The length of S.D. We shall find it by Method 111.
...(1)
...(2)
The equation of any plane through the line(2) is
(ax+b>+CZ+d)+A Xfl'x+'6>+ e'z+</')=0
...(3)
X (fl+A^/)+j» <i>-l-Ab')+z(c+Xc')+(d+m^0,
or
If the plane (3) is parallel to the line (1)[i.e. z-axis], then
0(a+-Ao')+0.(b-|-Ai>')-fl.lc-l-Ac')=0 or X^^-cIc'
Putting this value of A in (3), the equation of the plane
through the line (2) and parallel to z-axis is given by
X (a-‘ca'lc’)+y(b-cb’lc')+z(c--c)+id-cd^fc')^0
...(4)
or
X (ac''-ca')‘{-y (be’ -cbr)-\-(dc'--cd')=0.
Clearly z-axis i.e., the line (1) passes through the point
(0. 0,0).
188
Analytical Geometry 3-i)
/. Length of S.D.<=3the length of perpendicular from the point
(0, 0,0) to the plane (4)
^ y/{(ac'— ca')^+{bc'—b'c)^}
dc'-d'c
“V((flc'-- fl'c)*+{be'- b'e)^}
Proved.
The distance from origin of the point where the line of S.D.
meets the z-axis. It will be convenient to use method II.
The equations of line (2)in symmetrical form are
bd'-b'd
a'd-ad'
^ ab'—a'b
^
ab'—a'b
z—0
ca
ca'— c'a
bc'—b'c
ab'-a'b
..(5)
[See § 4, Chapter 4]
For convenience, let (5) be written as
X—Xi
y-y2 z—0
=ra (say).
1st
mg
«8
...(6)
Since 0,0, 1 are the actual d.c.’s of the z-axis [l.e, the line (1)]*
therefore ti is the actual distance of any point on the line (1)from
(0,0, 0). Let P (0,0, ti) be any point on the line (1).
Also any point on the line-(6) is (/arg-l-Xa,
say the point Q.
The d.r.’s of PQ are
/grg+Xa, niir^+yz,
njrg),
Let the line PQ be the line of S.D., so that PQ is perpendi
cular to both'the lines (1) and (2), and therefore, we have
and
or
or
or
0(4^3+^a)+0.(w2r2-ll.(«2r2~rj)=^0 or rx^n^fi
4(/2^s+^8)+W2 iWara+j'aHna (ffaf2-ri)«=0
4®ra+4?C2+wt2“r8+^W8j?a-|-0=0
[V ric=nara]
ra (42+^2^)=-(4-V8+Way2)
ra= —(4x2-It Wa>'2)/(4*+wia*)●
fx —n^rz
«2(4xa+W2j'2)/(4^4-
Substituting the corresponding values of /a, Wa, «a» Xa and j'a
from (5) and (6), we have
—{ab'—a'b)
{bd'-b'd)
Tifsa
●ab'-a'b
{be'—b'c)* {ca
,
y\ a'd-ad' )
{db'—d'b){bc'—b'c}^-ica'—c'a){ad'—a'd)
{be'—b'c)'^-i-(ca*—c'fl)*
Proved.
189
Shortest Distance
Ex. 12. Show that the shortest distance between the lines
x—Xi y-yi
z-zi x^Xj _y-yi_z—Zj
cos Cti'~‘C0S Pi“cos yi* cos ’^COS p2 cos yt
meets the first line In a point whose distance from (Xi, >i, Zi) is
[2{(Xi—Xi)(coseti-cosdcosct2)}]lsln*d
where 6 is the angle between the lines,
Sol. The epuations of the given lines are
Z-Zi
X'—Xi
y-yi ^
Ti (say)
...(1)
cos ai cos Pi “cos yi
Z~Zi
x-Xj^y—ya^
and
r» (say).
...(2)
cos a2™ cos ^^“cos ya
Any point P on (1) is
(fi cos ai+xi,ri cos Pi +yi, ri cos yi+Zi)
and any point Q on (2) is
(fa cos aa+Xa, r'i cos pa-\-ya, r% COS ya+Zs).
Let the shortest distance meet the line (I) at P and the line
(2)atQ.
Now, since cos ai, cos Pu cos yi are the actual d.c.*s of the
line<l). therefore ri is the actual distance of P from the point
(Xi, yi, Zi) on the line (1). It is required to find ri.
D.r.’s of PQ are ri cos ai+Xi—fa cos a'a—Xa, ri cos pi+yi
—fa COS pa—yi, ri cos yi+Zi—ra cos ya—ZaSince PQ is the line.of shortest distance, therefore, it is per
pendicular to both the lines(1) and (2), and hence, we have
cos ai (fi cos ai+Xi—ra cos aa—Xa)
+cos pi(fi cos Pi 4-yi-ra cos p» -y,)
+cos yi (fi cos yi+Zi — ra cos ya—Za)=»0
and cos aa (ri cos ai +Xi — ra cos aa—Xa)
4-cos pa (ri cos pi +yi —ra cos jSa—ya)
4-cos ya (ri cos yj4-Zi —ra cos ya—Za)*^©.
Now
...(3)
...(4)
cos* ai+cos® )8i4-cos* yi= 1, cos* aa4cos* j8a4'C0S* ya«=l
and cos ai cos aa 4-cos |3i cos jSa-i-cos yi cos ya^cos d,
}
...(5)
9 being given toj^be the angle between the lines (1) and (2).
From equation (3j, we have
fi (cos* ai4-cos* /3i-l-cos* yi)—fa (cos ai cos ai4'COS Pi cos Pt
-Hcos yi cos ya)4-(^i—^a) cos «i+(yi—ya)cos pi
4-(Zi—Za) cos yjc=:0
190
Analytical Geometry 3’D
or
(using (5)}. ...(6)
fi—rg cos 9+S {(X1—X2) cos ai}c=0
Similarly equation (4) may be written as
...(7)
— T2+ri cos <?+ r {(xi—X2) cos «2}*=0.
Multiplying (7)by cos 6 and subtracting from (6), we have
Ti (1 —cos* 5)+27{(xi—X2> cos ai}-ri^ {(JCi—X2) cos as) cos
or
Ti sin* d4-27{(Xi—X2)(cos ai—cos a2 cos d)}=Q
or
ri=f27{(Xi—X2)(cos tti—cos a2 cos 0)}]/sin* 6,
neglecting the negative sign because ri is the distance.
Ex. 13. Show that the equation of th^ plane containing the line
y/b'^zjct=\, x=0 andparaUei to the line
xla—zjc^ 1,7=0 is xfa—ylb~ z!c+1=0
and if2d is the shortest distance then show that d“*==fl“*4-6“*+c“*.
(Agra 1977, 81; Kanpor 82; Punjab 82; Ranchi 74;
Meernt 87P,90P
Sol. The equations of the given lines are
y/6+z/c=l, x=0
...(1)
x-a
and
— ——=1, v=0 or
a
c
*■
a
0
c’
...(2)
the equations of the second line being put in symmetrical form
The equation of any plane through the line (1) is
(>>/6+z/C“l)+Ax=0 or Ax+(l/6) y + {lfc) z-1 =0.
...(5)
- If the plane (3) is parallel to the line (2), then the normal to
the plane (3) whose dj.’s are A, 1/6, 1/c will be perpendicular to
the line (2), and so we have
u.A+0.(l/6)+c.(l/c)=0 or A=-(I/u).
Putting this value of A in (3), the equation of the plane con- .
taining the line (1) and parallel to the line (2) is
-4+4+J._|=0
or
a
b
c
+
a
b
c ^
...(4>
Clearly (a, 0, 0) is a point oh the line (2). Hence the length
2d of the S.D.athe perpendicular distance of (o, 0, 0) from the
plane (4).
2
a n/a)-0—hi
2d^
or
</*=- l/(fl-*+6-*+e-*) or ^-*=tr*+6-*+c-*.
Ex. 14, Show that the shortest distance between the diagonals
of a rectangular parallelopiped and the edges not meeting it are
6c/*\/(6*+c*), cfl/V'(e*+a*), o6/\/(o*+6*)
where a, b, c are the lengths of the edges.
(Punjab 1981; Garhwal 78S)
191
Shortest Distance
Sol. Consider a rectangular parailelopiped whose three coter
minous edges OA^ OB, OC are taken along the axes of x, y and
z respectively. Also OA^a, OB=>b and OC=c. The co-ordinates
of the different vertices are as follows ;
O(0. 0. 0), A (a, 0, 0). B(0. b, 0), C(0. 0. c), D(0, b, c),
: E(a, 0. c), F(a. b, 0), P(a, b, c).
Consider a diagonal AD and the edge OB not meeting this
diagonal. Now we shall find the length of S. D. between AD
and OB.
The.equations of^jP are
or x-a y z
0-0 6—0 c—0 ■
b c
The equations of OB are
X y—b
0
1
z
0
...(1)
●..(2)
Let /, m, n be the d.c.*s of the line of S.D. between the lines
(1) and (2). Since the line of S D. is perpendicular to both the
lines (1) and (2). we have
/ (—a)-|-m.6-l-».c«=0 and /.0-fm.l4'n*0=0.
1
Solving,
*●
c
0
a V(c*+0+a*)
V(«’+o*)
-O)
.% The length of S.D. between AD and OB
«the projection of the join of {a, 0, 0) {a point on (I)}
and (0» b, 0) {a point bn (2)} on the line of S.D.
whose d.c.’s /, m, n are given by (3)
(a—0)+m (0—6)+n (0—0)i=/o
=*oc/v'(c*-|-a*).
...(4)
192
Analytical Geometry 3-/)
Similarly the shortest distance between other pairs of lines
can be found.
Ex. 15. A square ABCD of diagonal 2a is folded along the
diagonal AC So that the planes DAC, BAC are at right angles. Find
the shortest distance between DC and AB,(Agra 1976, 80; M..U. 90)
Sol. ABCD is a square of diagonal 2a, so that AC=BD=2a.
Let O, the centre of the square, be chosen as origin of co-ordi
nates and the diagonal CA be taken along x-axis. Hence the
/ p{OMCt)
/
/
/
C-ctAd)/
c.f
'O
>X
ACaAd)
co-ordinates of the vertices A and C are {a, 0, 0) and (—o,0,0)
respectively.
Now as given in the problem, the square is folded over alongthe diagonal AC so that the planes DAC and BAC are at right
angles. This implies that the lines OB and OD become at right
angles. Also OA is perpendicular to the plane DOB. Hence the
lines OA, OB, OD are mutually orthogonal. . Let us now take OB
and OD as y and z axes respectively.
The co-ordinates of
respectively.
and Z) are (0, a, 0) and (0,0, a)
The equations to AB are X—
a
0_z—0
—a 0
-(I)
The equation to DC are
z—a
®
0
a
...(2)
a
The equation of any plane.through DC and parallel to AB
[f.«. through the line (2) and parallel to the line (1)J is
x-0
z—a
a
0
a
a
—a
0
M)
Shortest Distance
or
or
193
X (a^)-y(—(^)+(z—a)(—fl*)«0
..(3)
x-\-y—z+a=^0.
The S.D. between DC and AB
«=the length of perpendicular from a point (o, 0, 0) on
(l)]to theplane(3)
a+O-O+fl
.2a
V{(i)H(i)*+(-on‘V3
Ex. 16. Find the length and equations of the shortest distance
3x-9y-\-5z^0~x+y-z
between
and
6x+8;^+32-13=0=X+2)'+2-3.
(Meerut 1984 P, 86)
Sol. Here w’e shall use method IV of § 2. The equations of
the planes through the given lines are
(3x-9j'+52)+Ai(x+y-2)«0
and
(6x+8;?+3z-13)+A2(x+2>»+2-3)=>0
or
^(3+Ai) (-9+Ai)+z(5—Ai)«0
X2)
and
x(6+As)-l-j'CS+2As)+z(3+Aa)—(13+SAg) 0.
If the planes (1) and (2) are parallel, then their coefficients
are proportional and so we have
3-|-A| ■—94'A| 5—A|
~-k (say).
...(3)
3+Aa
6-I-A2 84-2Aa
Taking the ratios 1st, 2nd and 3rd with k respectively in (3),
we get
or
3+Ai-6fc-fcAa<=0
(34-A,)=*(6-l-A2)
...(5)
or
—-9-|'Ai—8fc”’2/c Aa“0
(—9-f-Ai)=»A:(84' 2Aa)
. . (6)
5-Ai-3fc-rfe
Aa=:0.
(5-AO=/:(3-|-Aa)
or
Substituting (6) from (4),
.(7)
-2+2Ai-3fc=0.
Subtracting 2 tiroes (6) from (S),
..(8)
-19-f3Ai-2fc«0.
Solving (7) and (8), A =53/5, A:=32/5.
Potting the values of Ai and k in (4), we get A^= — 31/8.
Substituting the values of Ai and Aa in (1) and (2), the equa
tions of the parallel planes through the given lines are
..(9)
17x+2v-72«0
(10)
and
\7x-\2y~7z-nr^0.
The required S.D. is the distance between the parallel plam s
(9)and(i0).
J
J94
Analytical Geometry 3-D
Any point on the plane (9) is (0,0,0).
/. The length of S.D.<=> the length of perpendicular from
(0,0, Q) to the plane (10)
11
0+0-0-11
V{(I7)«+(2)>+(-7)*) V(342)
[Numerically]
The equations of S.D.
The equation of any plane through the first given line is
X(3+AO+y(-?+A,)+z(5-A,)==0 [See (1)] ...(11)
If the plane (11) is perpendicular to (9) or (10), we have
17(3+Ax)+2(-9+A0-7(5-Ai)-0 or Ai=t=l/13.
Putting the value of Ai in (11) the equation of the plane'
through the 1st given line and perpendicular to the plane (9) or
(10) is given by
...(12)
10:e-29y+16z=0.
Again the equation of any plane through the 2nd given line
is [See equation (2)]
*(6+A,)+j;(8+2A3)+z (3+At)-(13+3Aa)c=0. ...(13)
If the plane (3) is perpendicular to (9) or (10), we have
17(6+Aa)+2(8+2Aa)~7(3+Aa)=0 or Aa=--58/7.
Putting the value of Aa in (13),. the equation of the plane
through the 2nd given line and ■ perpendicular to the plane (9) or
(10) is given,by
...(14)
13x+82y+55z-109«0.
The equations (12) and (14) are fhe required equations of the
shortest distance.
Note. We can solve the above problem by reducing both the
lines to symmetrical form and then' using method 1 or 11. The
problem can also be solved^ by reducing only one line to symmetri
cal form and then using-method III.
Ex. 17. Find the shortest distance between the lines
JC=0, iy+iz=l
(Agra 1982)
andy^^^O, Jx—Jz= I.
Sol. The equations of the 1st line in symmetrical form are
x_y—2_ z'
0 2
—3
...(1)
The equations of the 2nd.line are
y=0, Jx-tz=l
. ...(2)
The equation of any^plane through the line(2) is
195 -
Shortest Distance
iix-lz-\)+Xy==0
.. (3)
3x-{-my-4z-\2=0.
If the plane (3) Is parallel to the line (1), then the normal to
(3) is perpendicular to the line (1), and hence we have
0.3+2.12A-3.(-4)=0 or
Putting the value of A in(3),the equation of the plane through
the line (2) and parallel to the line (I) is given by
(4)
3;c~6y-4z-12=0.
The line (1) clearly passes through the point (0, 2, 0).
The length of S.D.=»the perpendicular distance of (0,2,0)
from the plane (4)
24
0-12-0-12
(N”rt'<^ncally).
or
v/{(3)*+(-6)»+(-4)>} vm '
Ex. 18. Prove that the S.D. between thelHnes
ax+by+cz+d=0=o*x+b’y-\-c*2-\-d*
and
ax+/5y+yz+8=0=sa'x+j8'’j>+/r+8' is
d
d'
8
S'
a
a*
a
a'
b
V
^
]8'
c
where A=^bc'
c'
y
y'
and /4'=j8y'—jS'y etc.
Sol. The equations of the given lines are
...(1)
ax^by’\-cz-\-d<^0==a'x.-^VyA-c'z-\-d*
and
«*+/5;>+y2+S=0=a'x+iS>+y'z+8'.
...(2)
We shall use method IV. The equations of any planes through.‘
.the given lines (1) and (2) are
(ax+Siy+cz+e/)+Ai (fl'x+^>+c'z+dO*=0
●●●(3)
or
x{a+Aia')+y(6+Aih')+z(c+Aic')+(d+Aid')«0
...(3')
and
(aX+i5y+yz+8)+A8(a'x+j8>+/z+8')=6
...(4)
or
X (a+Aga'l+y fjS+AajS'l+z (y+Ai,yO+(8+A,87=0
...(4')
If the planes (3') and (4') [/.a. (3) aiiH (4)] are parallel, then
their corresponding coefficiehls are prbpbirlfohal and so we have
a+-Ato' 6+A|h'
aj
●At (say).
ot+Ajot' jS+AgjS^ y‘l"A2y'
From these relations, we get
fl+Ajd'—Ar«—>A2a'c=»0
.
-(S)
\
196
Analytical Geometry 3-D
\
'=0
●(6)
c-^-Xic'—ky — kX'iY—O,
^
...(7)
Eliminating X —k, —kXi betv^en (3),(5),(6) and (7), the
equation or the plane through (he lineXl) and parallel to (2) is
given by
A
ax^by-\-cz-hci
a'x-\-b’y^c’z^d’
0
^
a
a'
ct ■
a'
b
b'
jS
jS'
c
c’
Y
Y
=0.
Adding (—x) times second, (—y) times third and (—z) times
fourth row to the first row, we get
d
d'
-(»x-\-^y+Y2)
~(a'x+i5>+y'r)
a
a’
b
b’
,
c
a
a'
j3
iS'
V
V
«=*0 ...(8)
Now we shall evaluate the coefficients of x. yandz in the
expansion of the determinant in (8).
The coefficient of X
d
a'
b
d
c
a
d
a’
a'
+
C;
c'
d
b
c
«=»C' (ac
9
d
a
b
d
. p
I c
/
:/
aa
b
a
-r-a/
y.
a
d
«a
b
b'
pd
c
c'
/●
; ota - aa!
Yd
a
d
d
Pd:-a.P' : -.-r:
b
b'
ct
ya'—a/ . !
c
d
dcy:^B’ {ad-dhy
0
■ ■—€'■■■
.
■ .
B': \
/
[ Y : It is given that
^'«=»y«';-r-y'a, C=<mp'—dp]
.
‘.
Shortest Distance
197
=C'
(C)«C5'-5C'.
Similarly the coefficient of y=AC—CA*
and
1
the coefficient of z^^^BA'—AB',,
Now suppose(xu yu 2i) is a point on the line (2), so that we
have
aXt+)3yi+y?i+8t=0 and a%+]8>i+y'zi+S'c=0
or aXi+^j>i+yZi*=>—8 and a'jci+j3'j>i+y'zi=i—S'
...(10)
Now the required S.]^^etween the lines (1) and (2)
«=*The length of the^rpendicular from the point (xi, yu Zi)
on the line (2)from the plane (8)
d
d\ -r(«Xi+M+yzi)
a
a*
b
b*
c
c'
-*(«%+i8>i+y'2i)
oe
a'
V
/
-r'^[{coefficient of ;c)*+(coeflF. ofy)*+(coeflf. ofz)^].
Putting values from (9) and (10), we get the required S.D.
d
d\
S
8'
a
a*
«
a'
b
b'
P
c
€●
y
Ex. 19.
~V[S{BC-B^C)^}.
y‘
Proved.
Two straight lines
A
mi
^yj. X -aa_y—jSa
rtx~ ’ /a
OTa
z-va
Hi
are cut by a third tine whose direction cosines are A, /*, v. Show
that ●d* the length intercepted on the third line is given by
d A
mi
n,
ai—aa
yi—ya
A
Wa
«a
A
/Ml
A
A*
V
'A
/Ma
«8
i)edluce the length of the shortest distance between the first two
lines.
Sol. The equations of the given lines are
(*~*iVA“(y—ft)//Mi«i(2—yi)/«i=in (say)
.(1)
198
Analytical Geometry 3-i>
...(2)
and {x—<x.i)lk={'y—Pi)lm2=‘{z—Yi)lft'<i^'‘i (say).
Any point on the line (1) is J* (/iri+ai, mirj+^i, nifi-j-yi).
Any point on the line (2) is
Q (/2^8"l“*a»
Wa^a+ya).
...(3)
Let the third line with d.c.’s A, n, v meet the line (1) at the
point P and the line (2) at.Q so that PQ=d.
Now the third line with d.c.*s A, fi^ v is passing through the
point P, hence its equations are
x—Uirt^QLj) j>-(Wiri+jSi)
A
.V The co-ordinates of the point fi at a distance *d* from
the point P on the line (4) are
...(5)
rfA-i-/iri+«i. dfi+mifi-^-^u <fv+«iri+yi
Now the co-ordinates given by (3) and (5) are of the same
point Q, Hence comparing (3) and (5), we get
.
A+/iri+=/a^a+«2
i//tt+OTifi
«= mararf-Pa
dv +Jiiri-l-yi=na<^8+ya
or
dA+(ai—ota)+/iri—/ara*=»0
...(6)
d/t+(Pi-Pa)+7Wiri^mara=0
dv+(yi-ya)+niri-wara=0
Eliminating ri and ra from the relations (6), we get
lx
h =0.
dA+(ai—aa)
wi|
d/i+(Pi-Pa)
Wa
t
Ha
«i
dv f(yi—yj)
Spliiing this determinant into two determinants, we have
/i
It «0
d .A
h
/a i+ a^—aa
or d
or
d
mi
ma
(i
mi
m2
Pi-]8a
V
«i
Ha
A
lx
yi—ya
ni
— ai—aa
P
mi-
ma
V
Hi
Ha
lx
Mil
Ml
/a
ma
Ma
/|
HIi
Hi
A
P
V
/a
ma
Ha
Ia
Pi—Pa
=—
»a
A
HJi
/a
ma
Ha
Hi
yi-ya
«i—aa Pi—Pi yi-ya
...(7)
199
Shortest Distance
Since is the distance, hence neglecting the negative sign in
(7), the required result is obtained.
Now if d stands for the S.D; between the given lines(1) and*
(2), then the third line with d.c.*s A, /*, v is perpendicular to both
the given lines (1) and (2) and hence we have ●
/iA+Wi/i+WiV.»0, /2A+wi2/i+Wavs=0.
Solving, we get
V
A
V(A>+f**+v’)
Ixmi^krhi^y/\,S (min2—maWi)*]
(min2—
“wi/a—
1
The value of the distance
given by (7) will become the
length of S.D. between the given lines (1) and (2)if the values of
A, /i, V are substituted in it from (8).
Now the coeflScient of d in (7).
/i
Wl
/a
m2
«2
A
■ V
«A (miW2—m2«i)+M («i/a—Waix)+v (litrit—ltmi)
[putting the values of A, /t, v from (8)]
Using this value for the coefficient of d in (7), the S.D. *d* is
given by
«i —«a jBi—i^a yi-ya
(miHa—WsWi)*].
h
m,
/.8
ma
«i
/la
Exercises
1.
Find the equations of the straight line perpendicular to both
of the lines
x+2
y-5 z+3
y-l z+2
„
,
1 “2
-y-=3-=— and -3
x-2 y-3 z-1
4 “*-5
Ans. ^
2.
Find the length and equations of the common perpendicular
to the two lines
200
Analytical Geometry 3-D j
x+3 y-6 2
-4 ^“*3—™2”
3.
x-\-2 y
2-1
-4 “l ■^'l ' ‘
(Meerat 1984; Andhca 68; Gorakhpur 74; Madras 76)
Ans. The length of common perpendicular (/.c. S.D.)=9.
The ei]uations of S.D. are
32x+34;>+13z~I08=0«4x+Mj;+5r~27.
Show that the length of shortest distance between the lines
x-2 yJf-l 2^
2 "* 3
4 »
4.
2a'+3>^-5z-6«0=3x-2;;-z+3 is 97/(13^6).
(Rajasthan 1975)
Find the length and the equations of the shortest distance
between the lines
5,
5jc—
z=0«=x--2>+z+3
lx-Ay—22^0=>x-~y-\- z-3,
(Meerat 1986 S)
Ans. The length of S. D. is 13/5^2.
The equations of S. D. are
17a+20j^-19z-39«0, 8a+5;;—31z+67=0.
Find the equation of the shortest distance and its length
between the lines
x-3 y—S 2—2
1 “_2
f
x-1 j>+l 2+1
'7
(M. U. 1990)
and
6
Volume of Tetrahedron
§ 1.(A) To find the volume of a tetrahedron, whose three
coterminous edges in the right-handed orientation are a. b, c. where
a, b, c are vectors.
Let OABC be a tetrahedron. Let
er
O be the origin and let position vec
tors of the vertices A, B, C be a, b, c
' f
respectively, so that
—>■
● —►
Oi4c=a. OB«=»b, OC=c.
’ Then the volume V of the tetra- ^
^
hedron is given by
(area of the triangle OBC) X (perpendicular length from
A on the plane OBC),
(2)
Now the area of AOBC=J | bxc 1.
If n be 1 he unit vector perpendicular to the plane of the
triangle OBC such that b, c and n are in right handed orientation,
then
A
bxc
“"^ IDxcr
since b, c, bxc are in right handed orientation.
A the length of the perpendicular from A on the plane
(9BCc=tbe length of the projection of OA on the perpendicular to
the plane OBC in the direction of n
[a, b, c]
a*(bxc)
bxc
eai(?/4»nc3a»
IbxcT “ IbxcI “ I bxcI
Putting the values from (2) and (3) in (1), we get
[»5 b, c]
...(3)
F^filbxcl i bxc 1
...(4)
La* b, c].
This is the required formula for the volume of the tetrahedron.
or
202
Analytical Geometry 3-D
(B) Tofind the volume of the tetrahedron OABC whose one
vertex O is at the origin and the co-ordinates of the remaining three
vertices A,B and C are (Xi., j'l, rx),
and (X3, >-3, Zs)
respectively,
Let a^ b, c be the position vectors of the vertjces A,B^C
respectively w.r.t. O as origin. Since the co-ordinates oi A are .
(^i» yu Zi)» therefore, the position vector a of the point A is given
by
a=:zOA=Xi\-{-yil+Zik.
—>●
Similarly b=a05=Xai-l-y2j-|-z:2k and c=OC«=*jn:ai-l-;>3j-{-Zak.
Now the volume V of the tetrahedron OABC is given by
[a, b, c]
or
[See § 1 (A), for complete proof deduce this
result here]
xi
yi
Xi
;’2
Xs
y^
zi
^(5)
Z3
The formula (5) is the required formula.
(C) To find the volume of the tetrahedron whose vertices have
(Xir.Fi* Zi)» (xg, ^’2, Zi)y (Xs, >'3, Z3) and
y^, z^) as co-ordinates.
Let i4, B, C, D be the vertices of the tetrahedron DABC. Let
(Xi, yn Zi)f (Xn j»2i Z2), (X3, ;^’a, Zs), (x^t ^4, Z4) be the co-ordinates
nf the points A, B,- C, D respectively. Then the position vectors
of A,B,C,D are
Xii+y,j+zxk, jCai-b^'sj+Zgk, Jfai+^'sj+Zak,
x«i+.F4j+Z4k respectively.
We have X>^=(xii-f j^ij+Zxk)-(x4i-|-:t;4j-|-Z4k)
■=CXi~A-4) i-l-(:Fl-:F4) j+(Zi--Z4) k.
.DB^{Xi-Xt) i-h(;’2-:F4) l+iZi-Zt) k.
and
/>C=(^a-rX4) i+(j|;8-;;4) j+(Z8-Z4) k.
A
The volume V of the tetrahedron DABC is given by
Kc=A [DA, DB, DC] [See § 1 (A)]
: xi-x^
.F1--.F4
Zl — Zt
Xi-Xi
yi-y*
Za-^Zi
X3—X4
>'8“J4
Z3 —Z4
203
Volume of Tetrahedron
Xi-Xi
O'
yi-yi
Xa-X4
^a--X“4
>’8->'4
yi
Xa
Za-^4
0
28-^24
0
24
1
Adding 4th row to 1st, 2nd and 3rd rows, we get
1
Xi
yi
2,
X2
y^
2a
\
^8
J’3
23
1
*4
3^4
24
1
...(6)
The formula (6) is the required formula.
Corollary. Condition for four points to be coplanar,.
The four points A, C, D will be coplanar if the voiume of
the tetrahedron formed by them is zero, t.e.
1
2i
^1
yi
X2
yz
2a
1
Xa
3’8
28
1
X4
3'4
24
1
S30●
§2. Tofind the volume V of a tetrahedron, in terms of the
lengths of three concurrent edges and their mutual inclinations.
LQiOABChc the tetrahedron. Take the vertex O as the
origin. Let the lengths of the three edges OA, OB, OC be a, b, c
and the angles jBOC, COi4, OB be A,/4, V respectively. Let any
three perpendicular lines through the origin O be taken as co
ordinate axes.
Again let the direction cosines of the lines OA, OB and OC
be li, mi, »i, /a, ma, and k, m^, ns respectively, so that the co
ordinates of the vertices A, B and C are (/lU. WiO, nia); {Ub, mib,
ttsb) and (/gc, msc, ngc) respectively. We have
—►
OA==a^liai+mia}+niak, OB=b=/a6i+m2&j -j-Wabk,
—
and OC*=* c= hci -h/wscj+wsck.
204
Analytical Geometry 3-D
Again
or
(/i/a+/Wim2+niMa)=fl6 cos v
h*e^bc ikk+mims+nM^bc cos A
c*a<=>cn (/s/i-hmami-f’^anijejca cos (i
1
^i/8+Wima+ni«a=cos v
/a/a+W2ma+«a”a=COS A ’.
4^+w?8/Wx+«a Wi=cos jti
...(1)
Now the volume V of the tetrahedron OABC is given by
y=>o [a, b| c]c=^
lio
miQ
nia
i abc
m^b
nj)
he
m^c
iigc
h
mi
/8
ma
«a
h
ma
»3
1
/. V*a.
36
1
kb
h
mi
«i
/a
ma
«a
h
ma
Wa
/i*+mi^-fwi*
X
h
mi
«i
/a
ma
«a
I3
ma
na
/j/aHi-mima-^WiWa /i/a+mima+niWa
/a/1 H-mami+Wa»i /a*+ma®+Wa®
1
=:rr O^C®C®
36
/a/a+mama+Wa^a
/a/i+mami^ffa'^i /a/a+mama^-Kaffa ZaHmaH^a®
1
cos V
cos fi
COS V
cos fi
y=:ki ubc
1
cos A
[using (be relations (l)j.
I
cos A
1
cos V
cos II
cos V
1
cos A
cos /t
cos A
1
I
The negative sign will be neglected in calculating tbe magni
tude of the volume y.
§ 3. Tofind the volume F of the tetrahedron when equations of
itsfourfaces are.given.
I
205
Volume of Tetrahedron
Let the equations of the. planes representing the four faces of
the tetrahedron be
...(1)
...(2)
a^x -{- b^y
=0
.(3)
fls*4-^8y+C8Z+</s=0
oiK
...(4)
Now a set of any three planes out of the four planes given
above, will intersect in a point, a vertex ofthe tetrahedron. Hence
the four planes, taking three at a time, will intersect in *Ca l.e. A
points, the four vertices of the tetrahedron.
Now solving (2),(3) and (4) by the method of determinants,
we get
X
-y
C2
d2
flg
Ca
d2
^^3
Cz
ds
Oz
Cs
dz
hi
C4
d.
Ot
f4
d.
1
Z
«2
bi
dz
dz
b,
di
Suppose
A=
i^a
Ca
dz
bz
Cz
Oi
b.
Ci
^2
d,
i
di
bi
Cl
di
Qz
bz
Cj
dz
dz
bz
Cz
di
bi
Ci
...(5)
di
Let the capital letters represent the co'factors of the correi
ponding small letters in the determinant A» i c. Aj,
Cu D:
4zt j9a«... etc. represent the co-factors of ni, hi, Cj, du dz, bz,... etc.
respectively in A- The result(5) may be written as
Z
jc - CO : ~y
r.
D1
Ai . ~Bi Cl
di
■(
.
The point of intersect ion of thu planes (2), (3), and (4j| is
(AilDi, BilDitCilDi). Similarly solvin.; (he other three sets of
three planes, the points of intersection i.e. the other three vertices
of the tetrahedron ure
206
Analytical Geometry 3-2)
lAi Bi Ci,\ lA^ ^
/dl —
\Fa*
Fj’ADa’ 2)3* D,V
DJ
The required volume V of the tetrahedron is given by
Ax
_
A
^
●2)j
Di
D\
-Ba
1
Dt
2>3
a
2)2
1
As
D,
J?3 .
Da
Cs
Ds
1
A
Bi
. C4
D4
D*
1
D4
[See § 1 (C), equation (6)]
1
Cx
Dx
Ba
Ca
Da
Bs
Cs
Da
D4
C4
D*
6D1D2D3D4
^a
^4
A
4-1
[since if A'
the determinant of nth order
formed.by the co-factors of the elements of
the determinant A. then A'=A""']*
6DiDaZ)32)4
Aa
ODf 2)42)3 2)4
SOLVED EXAMPLES
Ex. 1. Find the volume of the tetrahedron^ the co-ordinates of
whose vertices are (2, —1, —3)4.(4, .1, 3), (3^ 2, — i) and (1, 4, 2).
(Pilnjab 198i< 82)
Soi. The volume F of the tetrahedron, the co-ordinates of
whose vertices are (2, —I, -r-3), (4, 1, 3), (3* 2, —1) and (14 4, 2)
is given by
1
-3
:i
(using (6) of § 1 (C)]
F=i
2
4
I
I
3';
I
’
■
■
■
I- '
i -
2
-1
:'i
4;
2
!
207
Volume of Tetrahedron
2
~1
-3
1
2
2
6
, adding(—1) times 1st
row to 2nd, 3rd and
4th rows
0
1
3
2
0
-1
5
5
0
2.
2
1
3
6 , expanding the determinant
along the 4th column
2
-I
5
5
0
12
0
8
-1
5
6
12
16
8
7
16 , adding 2 times 3rd row to
1st row and adding 3rd row
to the 2nd row
7
5
, expanding with respect to 1st column
(84-128)«=»~|^~p (Numerically).
Ex. 2. If the volume of the tetrahedron whose vertices are
(fl. 1, 2).(3.0.1).(4, 3, 6) and.(2, 3. 2) is 6Jind the value of*a\
(Agra 1974)
Sol. The volume V of the given tetrahedron is jgiven by
a
1
2
1
[using (6) of § 1 (C)]
3
0
1
1
4
3
6
1
2
or
3
2
1
a-3 1
1
0 . adding (—1) times 2nd
row to each of the other
rows ,
1
‘1
3
0
I
● ●
●1
- 3.
1
0
* ●
\
208
Analytical Geometry 3-D
1
6
1
6
a-2 1
1
1
3
5
-1
3
1
-2
fl~2
6
0
u-2
6
-12
. ixpanding v^ith respect to
the 4tb coiumn
I
,adding 3rd column to
the 1st column and
(—3)times 3rd column
5
to the 2nd
1
0
-2
6 i'
— 12 , expanding w.r.t* 3rd row
1
1
{-12 (a-2)-|-6 2}«i(-12a+36)=(-2fl+6).
But we are given that the volume V=6.
— 2fl+6«=>6 or a=0.
Ex. 3. Prove that the volume of the tetrahedron formed by the
planes my-\-nz=Q, nz+/x=0, ix-\-my^0 and h:+my-\-nz==>p is
8 p^l(lmn).
Hence deduce the area of the triangle formed on the plane
Ix+my-^ nz^p. (Agra 1973, 79; Garhwal 78; Eajasthan 77)
Sol. The equations of the four planes are
(1>,
nz+/x:=0
.. (2)
(3).
...(4)
lX'\-my-\-nz r^p.
The planes (1),(2) and (3) pass through the origin, and so
the point of intersectioo of the planes (1),(2) and (3) is O (0, 0,0).
Now to get the coordinates of one other vertex of the tetrahe*
dron we solve the equations (1),(2) and (4).
Adding 0)and (2), we have
/x+m>^+2nz=0.
...(5)
Subtracting (4)from (5),
nz——p or z=>—pln.
Putting this value of z in (1) and (2), we get
x==p}l,y=plm.
Heiice the planes (1),(2) and (4) intersect at
^ ipll* P/tn, —pin).
Similarly the other two vertices of the tetrahedron are
^(pI^, —pint, pfn), C {—pH, pirn, pjn}.
The volume P'01 the ietrahedron 0,»BC \s given by
209
Volume of Tetrahedron
pit
pfm
-Pin
Pli
-pim
pin
-Pfl
pIm
Pin
1
[Using (5) of § 1 (B)]
1
1
I
-I
-I
1
0
0
2
0
0
2
2
0
0
6lmn
3
6lmn
adding 3rd column to 1st and 2nd
3
.-P
6lmn
6lmn
2 ,expanding along the first row
2p8
(numerically).
Hmn
...(6)
Proved.
To dedoce the area of the triangle. Let the area of the triangle
formed on the plane (4) be A- Also let the length of the per
pendicular from the opposite vertex O to the plane (4) be denoted
by d, so that
d-
-P
...{7)
Also the volume V of the tetrahedron is given by
A
3K„3.2p»
d 3lmn *
P
[Putting the values from (6) and (7)]
Ex. 4. AiB^C are three fixed points and a variable point P
moves so that the volume of the tetrahedron PABC is constant. Fin^
the locus of P and show that i: is a plane parallel to the plane ABd.'
Sol. Let the co-ordinates of the fixed points A, B^ and C be
taken as (a, 0, 0),(0, by 0» and (0, 0, c) respectively. Also let the
210 .
Analytical Geometry 3-Z>
variable point P be (xj, y^ Zi). Now we are given that the volume
of the tetrahedron Pi4JC*=*constaht, say k.
a
0
. 1
0.
1
6
0
b
0
1
0
0
c
1.
1
Xi
or
or
a
b
0
1
0
c
Pt
Zi
0
1
1
b
0
1
1
0
c
1
=6A:
a {b (c-r,)+l.(0-cy,>}-Xi {1.(hc-0)}c=6A:
or
or
0
“^1
abc-abzi-‘acyi-‘bcxii=>6k
bcxiHrcayI
Dividing by abc, we get
/.
6k),
^i/o+7i/^+2i/c«(flhc—6A:V(aAc).
Locus of/*(x„ yj, z,) is
xla-^ylb’^zlci^{abc-~~6k)l{abc)
x/fl+y/d+z/c=constant.
or
..(1)
The equation (I) is the required 'locus of the point P and is a
plane parallel to the plane ABC whose equation is
xja’{‘ylb’^zfctsa I.
, Ex. 5. .1. 5,C are (0, 1. 2). (3,0,1), (2,3,2). Find the
locus of P if the volume of the tetrahedron PABC is 6.
Sol. Let the point P be taken as (;zi,
K of the tetrahedron PABC is given by
d
1
2
1
3
0
1
1
2
3
2
r
Xi
yi
Zi
1
r,). The volume
[Using (6) pf§ 1 (C)J.
,
211
Volume of Tetrahedron
[’●*
or
0
1
3
-1
-1
0
2
2
0
0
^'i—l
Zi-2
0
2
1
f^=6, as given. Also adding (—1) times 1st row to each
of the other rows].
( 3
-I
-1
36=2= —
2
2
Xj
yi-\
3
-4
-1
2
0
0
0
Zi-2
expanding w.r.t. 4th column
or
36=0—
*1
or
36=2
“●*1+^1““ 1
-4
Zi—2
adding (—1) times 1st column to the 2nd
-1
2,-2
expanding w.r.t. 2nd row
or
or
or
18^
4(2,-2)+l.(-xi+;;,-l)
18=—42,4-8—x,+>>, — l
*i-3'i4-42,-f llcaO.
.*.
The locus of /> is the plane jc-^j;4.42+1 1=6.
Ex. 6. A variable plane makes with the co-ordinate planes a
tetrahedron of constant volume 6^h^. Find
(/) the locus of the. centroid of the tetrahedron.
(Meerut 1977; Indore 76)
flnrf (//) the locus of the foot of the perpendicular from the origin to
the plane.
Sol.
Let the equation of the variable plane be
xla^ylb+zlc^\:
●●(1)
The equations of the coordinate planes are x=0, ^=0, 2=0.
Solving the above four equations taking three at a time, the
vertices Of the tetrahedron are given by
0(0, 0. 0); A{a, 0, 0); ^(0. b, 0); C(0, 0, c).
I)
r.
I
Analytical Geometry 3●/)
212
The volume V of the tetrahedron OABC is given by
a
0
0
0
b
0
0
0
c
abc
a6c=384*».
or
[ ●/
ISee(5)of§ 1 (B)].
64/f* as given]
...(2 )
(i) Now let the co-ordinates of the centroid of the tetrahe
dron OABC be (oe, |3, y), so that we have
0-f-0-|-0-l-c
O+fl-l-O+O „ O-f-O+6-i-O ●
4
a=
^—'
* 7
4T-—^—● p*=
or
a—4ot, b=4p^ ce=4y.
Substituting these values of fl, d, c in (2), v.'e get
(4a) (4/5) (4y)==>384 A:® or a^y«=6fc®.
● .5. the locus of the centroid (a, /3, y) is xyzt=6k\
●
(ii) Let ixu yu Zi) be the co-ordinates of the foot of the per
pendicular from the origin O to the plane (I ).
The direction ratios of this perpendicular are Xi—0, j'l—0,
z,—d t.e. Xu yu Zi- Also the d.T.’s of the normal to the plane (1)
are 1/a, 1/6. 1/c.
The two being parallel, we have
...(3)
\fa lib lie
The foot of the perpendicular (Xi.yi, Zi) lies on the plane
(1) and so we have
From (3), we have
2
'
x,!a
or
z>®
Xx* _yx* „zi
y,!b
zxic
(x,/a)-l-(y,/6)+(z,/c)
1
[using (4)1
6yi=czx =» Xi* 4*yi®+Zi*.
a=(Xi*-l-j>i*-4‘Z,®)/Xi, 6=’(xi*+>>i*+zi*)/yi,
c=(x,*-l-yi*-l-zi®)/ri.
Substituting these values of a. b and c in (2), we have
(Xi*+>'iHzi*)»=384 Xx y, z,
The locus of the foot of the'.perpendicular i.e. the locus
of(Xi,yi, z,)is given by
(x*-b/+zY=384 xyz fc®.
are
Ex. 7. The tenths of two opposite edges of a tetrahedron
fl . 6, their shortest dista nce is equal to d and the angle between them
is e. Show that the volume of the tetrahedron is^ abd sin 0.
Hi
Volume of Tetrahedron
Sol. Let ABCD be a tetrahedron. Let the vertex A be taken
as origin and let
Let the direction cosines of the Wnt aB
be /, m, «, so that equations to AB are
X—0 j;—0 2—0
/
m
n
Since the point B is on the line AS and is at a distance *a
from A(0, 0, 0), hence the co-ordinates of B are (la, ma, na).
The edge opposite to AB is CD. Let the co-ordinates of C be
(a, y) and CD=>b. Let the d.c.'s of the line CD be A, /x, v, so
that the equations to CD are
X -a
A
U
Since the point D is on the line CD [i.e.(2)] and is at a dis
tance ‘b’ from C(a, /3, y), hence the co-ordinates of D are
(Ab-fa,
vb-by).
Now B is the angle between the lines AB and CD i.e, between
the lines (1) and (2), so that
(3)
sin B=“y/ [i?(mv—n/i)®].
Also d=the S.D. between the lines (1) and (2)
a-0
y—0 -i-y/iS (mv—nfi)^]
i8-0
A
i
! a
C3|
or
d sin y=>
n
m
/
/
V
P.
Y
m
n
A
a
P
V
.6
y
/
m
n
A
IM
V
-^sin 6
[using (3)]
...(4)
The volume K of the t etrahedron ABCD [one vertex A being
originj is given by
al
am
an
a
i3
Y
bA-i-a ■
bfi+fi
[see § 1 (B))
bv-l-y
■
I
2i4
Analytical Geometry 3-2)
al
am
on
a
/S
y
al
+
bv
1
ab
"”6
● a
P
y
/
m
n
/*
V
A
flW sio ^
i
)
am
a
^
OL
an
y
y
+0
[using (4) and neglecting —ve sign]
Ex.8. A point P moves so that three mutually perpendicular
lines PA,PB,PC may be drawn cutting the axes OX, OY, OZ at
i4*AXlattd4he-t^hsms-€ftk€-^ahed^^ OABC is constant and
equal to k^jb Prove that P lies on the surface „ ^
^
Sol.
Let the co-ordinates of ihe moving point P be
(●^1. J’l* Zi): Since is a point on the x-axis {i.e. on OX), there
fore, let the co-ordinates of A be {a, 0, 0).
Similarly let the co
ordinates oi B and C be (0, b, 0) and (0, 0, c) respectively.
Now the direction ratios of the lines PA, PB and PC are
Xi—a, yi, zii Xi, yi—b, Zi; and Xi, yi, Zi—c respectively.
It is given that the lines PA, PB and PC are mutually perpt ndicular, so using the formula *0102+^1^2+CiCa«0*, we have
(Xi—U]. Xi +
or
—- h)-|-
s= 0
[PA and PB being perpendicular]
.
...(1)
Similarly PB and PC; PC and P^ are perpendicular, so we have
^1*+Pi*+Zi^ '^byi -i- czi
...(2]
...(3)
The volume of the tetrahedron OABC is given to be 4^/6.
a
0
0
[using § 1 (B)]
0.
2K
0
^1*+>'1* +
0
or
0
eaflXi
c
abcsak^.
...(4)
Now in order to find the locus of the
m
“If
(i)» (2), (3) and (4).
qualities «.
fhe'^^taiionf
215.
Volume of Tetrahedron
Adding (1),(2) and (3), we get
3(Xi»+7i*+z,*)=2 (axi+6^i+crt).
Subtracting 2 times (2)from (5), we have
xi*+yi^+2i^=2axi.
Similarly (3) and (5) give
...(5)
...(7)
and 0)and (5) give
...(8)
Multiplying (6).(7) and (8), we have
(^x**+3'i*+Zi*)®=8 abc XiyiZi
or
W +3'i*+Zi“)* =»8** XiyiZu
[using (4)].
Hence the locus of the point Pfoci, y^ ^) is given by
Proved.
{x^-\-y^-\-z^f=%k^xyz.
Ex. 9. Show that the volume of a tetrahedron of which a pair
of opposite edges isformed by lengths r and r* on the straight lines
Whose equations are
{x—a)ll^iy-b)lm^{z-c)ln;
ix-a')fr={y-b')lm'^iz^c/)ln* is
a—a*
b-b'
c—c'
/
*
m
n
mr
n'
[Rajastban 1975]
Sol. The equations of a pair of opposite edges of a. tetra>
hedroo are given by the equations
(1)
(x-a)ll^(y-b)lm=iz-c)fn^r
and
.»(2)
(x-ayr=(y~-b')lm*^{z-c')ln'^r\
Let A(a, b, c) be one vertex of the tetrahedron on the line (1),
then the other vertex B on the line (1) at a distance r from
i4(a, b/c) has co-ordinates (ifr+a» mr+6,fir+c)>
Similarly other two vertices C and D on the line (2) are
(o', b', c'J and (/'r'+fl', m'r'+bV«V+c') respectively.
Hence the required volume V of the tetrahedron ABC is given
by
1
c
a
b
V*=»^
/r + a
mr-fb
nr+c
1
fl
'
b'
c'
1
/V'-i-fl' mV'+b' nY+c' 1
216
Analytical Geometry 3-D
a
b
c
1
Ir
mr
nr
0
a'
b'
c'
.1
mY
nY
0
I'r'
adding (—1)times 1st row to 2nd and (—1)times
3rd row to 4th
=|rr'
a
b
c
]
I
m
n
0
a'
b'
c'
1
V
m'
n'
0
c—c'
fl—a'
Ok
/
m
n
0
a'
b'
c'
1
r
»r
a —a!
0
adding(—1)times 3rd row to 1st
b-V
c c' ,
I
m
r
m’
n
expanding along the 4th column.
Prored.
Exercises
1.
2.
Find the volume ol' the tetrahedron, the co-ordinates of
whose vertices are (1, 0, ●)), (0,0, 1), (0, 0, 2) and (1, 2, 3).
Ans.
Find the volume of the tetrahedron formed by the planes
y+z—0, 2+x=0,
and x-t-y+z«l
Hence deduce the area of the triangle formed on the plane
x-\-y-\-z^\.
Ans. Volume-=2/3, area—2/3.
7
Skew Lines
§ 1. The equations of two skew lines.
To show that by a proper choice of axes the equations of two
skew lines can be given by the equations
yzs>X tan a,z=c ; y -- —x tan a, z=—c.
^Calcotta 1974)
In the adjoining figure let AB
A
and A'B' be the two given sktw
(non-interseciing) lines, and let CD
oflength 2c be the shortest distance
between them.
Take the axis ot z along DC
and O, the middle point of DC,as
origin. Draw OK ard OL parallel
to AB and A'B' respectively and
take the plane KOL as the plane
2=0. Take the internal and exter
nal bisectors of the angle KOL as ,
the axes of X and y respectively. In the
2.
C
B
V
B'
o
\-z:
figure. OX and OY
represent the axes of x and y respectively. Let the angle between
the lines OK and OL {i.e. between the given lines AB and A'B’)
be 2a.
As explained above the line OK(which is parallel to AB) is
inclined at angles a, Jw —a,|jt with x, y, z axes respectively and
therefore, the d. c.’s of OK (i.e. of AB) are cos a, cos (Iw -a),
cos ^TT i e. cos a, sin a,0.
Again the line OL (which is parallel to A'B') is inclined at
angles -a,|7r + a.
with the co-ordinate axes respectively and,
therefore, the d. c.’s of OL (i.e. of A'B*) are cos(-a), cos (4reH-a),
cos iw or cos a, — sin a, 0.
Now it is required to find the equations of ihe given skew
lines AB and A'B'.
218
Analytical Geometry 3-D
Since CZ)=2c and O is the middle point of CD, therefore
OC^OD=c. Hence the co-ordinates of the points C and D
(on the z-axis) are (0,0, c) and (0, 0, — c) respectively.
Thus we sec that the given line AB has d.c.*s cos a, sin 0
and passes through the point C(0,0, c) and hence its equations
are
jc—0
0_z— c
cos a sin «
0- or y=x tan a, z=c.
(1)
Also, the line
has d.c.*s cos a, — sin
0 and passes
^ through the point D (0, 0. - c) and hence its equations are
X—0 _ y^ z-fc ory= X tan a, z=»—c.
cos a —sin a
...(2)
The equations(1) and (2) are the required equations of the
given skew lines AB and A'B'.
If we put tan «=OT, the equations(1) and (2) become
y=mx,z=c and y=—mx,z*= —c.
The equations(3) may be written as
z-l-c
X y z—c . X
X
7=
1 m=-n0 and -*=T—
1
m
0
(3)
...(4)
SOLVED EXAMPLES
Ex. 1. Prove that the locus of a variable line which inter^ects
the three given lines y^mx,z—c; y==—mx,z-=»—c; y~z, mxt==—c
is the surface y'^~-m'^x^c=z^—c^.
(Agra 1978; Jodhpnr 76; Meerut 74, 89)
Soi The equations of the given lines are
y=mx,zt==c;
...(1)
mx,2c=»—c;
...(2)
and
y‘=^z, mx=—c.
...(3)
We know that any line intersecting the lines (1) and (2) is
given by two planes, one through each line.
The equation of any plane through the line (1) is
(j^—mx)-l-A (z—c)=0.
...(4) .
Also, the equation of any plane through the line (2) is
()*-i-mx)-l-/it (z-l-c)=0.
...(5)
The planes (4) and (5) intersect in a line and if this line
meets the line (3), then putting mx»—c and z=y in (4) and (5),
we have
(j'+cI+A (y-c)«0 and (y-c)-^fi(y+c)=^0
y-c
or .
A= y+c and
y+c
y-c
219
Skew Lines
...(6)
Multiplying these relations, we get
The required locus is obtained by eliminating A and /t between
(4),(5) and (6), and so eliminating A and between these equa
tions, the required locus is given by
_(y- mjc)'l ^
2+c
or
y-^mx\^ 1
z-c J
; m*jc*e=z*—c*.
Hence proved.
Ex. 2. Find the surface generated by the lines which intersect
the lines y=^mx, z ~c; y= —mx,z= —c and x-axis.
(Meerut 1973; Garbwal 81)
Sol. As in Ex. 1 above a line intersecting the first two lines
namely
z- c; y—~mx,z=—c is given by the planes
...(1)
iy-mx)-\-X (z-c)=0,
and
...(2)
(y-{‘mx)+fi (z-{-c)=0.
The planes (1) and (2) intersect in a line and if this line meets
the x-axis i.e. y=^0, z=0,then putting j;=0=»z in (1)and (2), we
have
(0—/wx)+A (0—c)=0 and (0+mx)+/tt(0+c)=0
or
A=—mx/c and /i=»--mx/c.
..(3)
From these two relations, we get A=/t4.
The required locus is given by eliminating A and fi between
(1),(2) and (3) and is given by
y—mx
y-^mx
z~c
z+c
or
{y-^-mx)(z-c)=(y—«ix)(z—c) or cy—mzx.
Ex. 3. A variable line intersects the x-axis and the curve x^=y,
y*t=cz and is parallel to the plane x=0. Prove that it generates the
paraboloid xy --cz
(Meerut 1972, 76, 89S; Kanpur 77; Kuruksbetra 74)
Sol. The equation of any plane through x-axis i.e. y=0=z is
...(1)
y^Xz.
The equation of any plane parallel to the plane x*=0 is
...(2)
X^(l.
The planes (1) and (2) intersect in a line which intersects the
x*axis and is parallel to the plane x=0. If this line meets the
curve x=y, y^=cz, we have by putting x=y in (2),
...(3)
y^cy^c2
From (1), A=_ cz y5 [V
cz=/3
220
or
or
Analytical Geometry 3-D
Xz=c/y
or X^cjfiy using (3)
Xft=^c.
...(4)
The required locus is given by eliminating A and n between
(I),(2) and (4;, and is
(yfz) x<=-c or xy—cz.
Ex. 4. Show that the straight lines which intersect the three
lines y - z-==ly x~0;z~x^]y y=0 and x—y=l,z~{) lie on the
surface whose equation is
+/+2*—2V2 - 2zx—2xy—1—0.
(Kanpur 1981)
Sol. The equations of the given lines are
0, x-0
-.(1)
z- x—1 ^--0, >'=0
...(2)
and
X—
1=0, 2=0,
..(3)
The equations of any planes through the lines (1) and (2) are
(j;—2—l)-j-Ax=0
...(4).
ann
(2-X-l)+/X.V=0,
The planes (4) and (5) intersect in a line and If this line is:to
meet the line (3) ttier we shall eliminate x, y^ z between the
equations(3),(4) and (5). From (3), we have z=0, x=;;-l-l.
Putting these values in the equations (4) and (5), we have
(>'-0-l)-|-A (>-f-lj=0and (0-;^-i-lj+/ij’=0
or
1—A;--f A=0 and —y—2i-ixy=Q
or
y (14-A)=1-A and y {-l+n)=2
l-A
or
and v«=»“”
H-A
li~l
or
Equating the two values of y, vie get
1~A
2
or A*-i-/iA4-A=2-l-2A
1+A II—[
/iA-i-A - /a+3=0.
(6)
1 he required locus is obtained by eliminating A and fi between
(4),(5) and i5>; and hence is given by
_2-x-l
y-z-l y- -2—1 2-X—1
-—+3=0
y
X
X
or (2-x-l)(y-z-1)
:y-z-l)+ x (2-x -l)+3x>;=0
or yz—z^—z—xy--xz-i-x-y-{-Z’\-l-~y^-{.yz
■\-y+xz-x"-xi-3xy=0
or X* +y^ -I- 2^—2yz - 2zx ™ 2xy - i =0.
Proved.
Ex. 5. Prove that the locus of a line which meets two lines
y^-izmXy zsa^c and the circle x^-\-y^=ia^, 2=0 is
cW icy-mzxf+c^ oz-cmxy-^a^m^
221
Skew Lines
Sol. The equations of the given lines are
y—mx<=‘Q, z—c=0
...(1)
...(2)
y+mx=0,z+c=0.
and
Also the equations of the given circle are ;
z - 0.
...(3)
The equations of any planes through the lihes (1) and (2) are
(j;—mx)4-A (z—c)—0
—
...(5)
and
l,y+mx)-[-fi (z-fe)=0.
The planes (4) and(5) intersect in a line and if this line meets
the circle (3) then we are to eliminate, x, y, z between the equations (3),(4) and (5).
From (3), we have z-=0. Putting z=»0 in (4) and (5), we get
(y—mx)—c\<=0,(y-\-mx)-^iic^^0.
c(A—ft)
Solving for x and y, v. e have
■
2m '
2
Putting these values -of x and
in x^-\-y*—a*9 thex, y, z
elimina-t of(3),(4) and (5) is given by
2
c* (A-f/i)*
(A -f*)® .
u4
4m»
...(6)
or
(A +ft)*4-c2/«a (A~/t)»=4a*m*.
The required locus is obtained by eliminating A and ft between
(4),(5), and (6), and hence is given by
c»
or
c» {f;'-mx)(z+c)+(z- e)(j'+mx)}*
2
-C*j'
{(y+mx)(z-c)-(y-mx)(z t-c)}*-4aVri* (z
or c» (yz-cmx)*+c*m^ (cy mzx)^-(r* -c*)^.
Ex. 6. Prove that the surface generated by a straight line winch
intersects the lines x+>»-0-z; x - v-=z. x+y-^2a
f
bola x*=2flz, y=^0 is x*
2qz.
(Kanpur 1983)
Sol. The equations of the given lines are
.. (1)
X-1-;^:: 0, z-0
...(2)
z=0, x+j;-- 2.i=0.
and
Also the equations of the given parabola are
...(3)
x*=2flz, >’=--0.
The equations oi any planes through the lines (1) and (2) are
x47+Az—0
(4)
...(5)
and
(x-;;-z)4-fA (X-1-.V—2fl)=0.
These planes intersect in a line and if this line meets 3he
parabola (3) then we are to eliminate x, y, z between (3),(4)
and (5).
222
Analytical Geometry 3-D
We hav6 3;«=!C from (3). Putting y=»0 in (4) and (5), we get
ac+Az=.0 and x—z-\-fA{x—2a)=0.
Solving for x and z, we have
2a/xX
— 2afx
1 “i* A fJtX ’
i-|-A+/ttA
Putting these values of :r and z in the equation x^*=>2dz
belonging to (3), the x, y^ z eliminant of(3),(4) and (5) is given
4flV*A*
—4a^H
by
(1+A +/ttA)^“(1-I-A+/uA;
or
/*A®= —(1+A+fiA)
or
/iA»+ittA +14*A=0. or fiX (A+l)+l.(A4-I)=0
or
(f*A4-l)(A+l)c=0 or /iA4-I=0 [as A# —1]
...(6)
The required locus is obtained by eliminating A and /u between
(4),(3) and (6) and hence is given by
/x—y—z
or
or
(x-y-z)(x-\-y)+z(x+y-2o)=>0
x^—y^^2az.
Ex. 7. Find.the locus of the variable line which cuts the three
lines y=bt zi=»—c ;z^c, x= a ; x<=a, y=—b.
[Garhwal 1978 (S); Kanpur 73]
Sol. The equations of the given lines are
y-b=>0, z+c«=0 ...(1), z-c=0, x+fl=0 ...(2)
and
● (3)
The equations of any planes through the lines (1) and (2) are
(3'-^)~A (z+c)=0
...(4)
and
(z~c)~fi (x+fl)=0.
■●(5)
The planes(4) and (5) intersect in'a line and if this line meets
the line (3) then we are to eliminate x, z between the equations
(3),(4) and (5).
From (3), x
a, y==^b. Putting these in (4) and (5), we get
—26—-A (z-f*c)«aO and z—c—2a/t=0
or
2*=(—26/A)—c and z^tf+2fl/tc.
Equating the two values of z. the x, y, z eliminant of (3), (4)
and (5) is given by
(—26/A)—c«=.c+2a/it or fl/u+6/A-l-c==0.
...(6)
The required locus is obtained by putting the values of A and
fi from (4) and (5) in (6) and is
Skew Lines
223
a(z—c) , b (z+c)
x+q
y-b
or
a iy-b)(z-c)+&(x+o)(z+c)+c(x+a){y-b)^0
or
ayz-]-b2x+cxy+abc^0.
Ex. 8. How many lines can be drawn from a point to intersect
two non^coplanar lines neither of which passes through the point ?
Find the equations of the lines or line which can be drawn from
the point(2, —1,3) to intersect the lines
.(x-l)/2=(j^-2)/3=(z-3)/4.;
(x-4)/4=:f/5c=(z+3)/3.
Sol. Let(a,)3, y) be the co>ordinates of a given point. Let
the equations of the two non*coplanar lines be
...(1)
^
y—mx=% z—c=0
and
>»+wjc^0,z+c=0.
(2)
The e^n.ations of any planes through the lines (1) and (2) are
\(y—mx)—A (z—c)=0
.. (3)
and
(y+mx)-fi(z+e)=0
...(4)
Since the planes(3) and (4) pass through the point (a, jS, y),
so we have
wa)—A (y—c)==0 and (/3+.»na) —p.(y-l-c)*=»0
A=»(^-ma)/(y-c) and/*=(jS+m«)/(y+c).
Putting the values of A and p in (3) and (4), the equations of
the line through the given point (a, ]8, y) and intersecting the
given lines (1) and (2) are given by
(y-mx)(y-c)--(z-c)(j8—ma)=0, I
and
(y+wx)(y-|-c)-(z+c)(/3-f ma)=0.
Thus only one line can be drawn from a point to intersect
two noo'Coplanar lines neither of which passes through the point.
Second part. The equations of the given lines are
^—l . y—2 z-3 x—4 y z+3
2
3 "" 4 * 4 “5
These lines can be reduced to general form as
3x^2y+l^0=2x—z+1
(1)
and
5x—4y--20=0=3x-4z—24
...(2)
The equations of any planes through the lines (1) and (2)are
(3x-2y+l:)+A (2x-z-l-l)=0
(3)
and
(5x-4y-.20)+/4(3x-4z-24):;=0.
●●(4)
The planes (3) and (4) pass through the. point (2, —1, 3), so
we have
or
224
Analytical Geometry 3-D
(6+24 n+A (4-3-1-1)''--0 or A=-9/2;
-1/5.
(I0+4-. 20)'+/i(6--12-24)=0 or
Putting the values of X and fi in (3) and (4), the equations of
the line through the point (2, —1,3) and intersecting the given
lines (I) pnd (2) are given by
12;c+4j;-92-1-7-0, 1 Kx-10v-l-2z-28-0.
Ex. 9. A line of constant length has its extremities on two
fixed straight Urns. Prove that the locus of its middle point is an
ellipse'whose axe j are equally inclined to the lints.
Sol. Let the equations of the two fixed straight lines be
given by
and
...(1)
x/1 =;>/m—(z-c)/0=ri
...(2)
xJ]—yj(—m)=(z+c)IO^^ri.
The co-ordinates of any two points on (I) and (2) are
P (ri, wri, c) and Q(r^, —rhr^t —0 respectively.
Let fa, Py y) be the middle point of PQ,so that we have
...(3)
a— J (ri+r2),/3- A m (ri~ra),
(c-c)=0.
Also the length of PQ is given to be const ant, say 2d.
or
Then 2d= PQ -dht&nce between P and Q
4d*-(r,(»'a f'■2)H(c+c)®.
...(4)
Let locus of (a. j3, y) is obtained by eliminating the parameters
ri and r* between (3) and (4). So putting the values of f|—r2and
ri + ra from (3) in (4); we get
or
4d“-^ (iPfmf-\-nP (2a)*-f-(2c)^ y=>0
J>,y=0..
m*a*-l-i3*/m»-l-c
Hence the locus of the middle point (a, p, y) is given by
...(5)
- c*. z 0.
The equations (5^ are the equations of an ellipse in the
xy-plane, whose axes lie along the axes of x and y. But we know
(see § 1) that the axes of X and y bisect the angles between the
lines (I) and (2) Hence the axes of the ellipse fin the present case
axes of X and y) are equally inclined to the given lines.
Ex. 10. A point moves so that the line joining the feet of the
perpendiculars from it to two given siraight lines subtends aright
angle at the middle point of their shortest distance {S.D.). Prove
that its locus is a hyperbolic cylinder.
Sol. htt AB and A'B' he the two given lines. Let the
middle point f? of the S.D. between these lines be taken as origin.
225
Skew Lines
The equations of the lines
and 4'5' are respectively given by
[see § 1]
...(1)
xl\ =;p/m=(z-c)/0=ri
and
x/l=W(-m)=fz+c)/0=r2.
, r
Let P(a, y) be the moving point and M,N be the teet oi
perpendiculars from P to the lines (1) find (2) so that we have
M (Tj, mri, c) and N (r^, -mr^y -c)The d.r.’s of PM are a fj, fi-mri, y-c and the d.r.’s of
PiVare a-ra,/J+mra. 7+c.
Now PM is perpendicular to(1)[/.e. AB]and PN is perpendi
cular to (2)[i.e. A'£'\, therefore, we have
l.(a-ri)-|-m (fi' mri)+0.(y—c)=0
...(3)
or
and
1.(a—rg)--m (/?+wra)+0.(y-h c)=0
...(4)
or
According to the problem MN subtends a right angle at the
middle point 0 i.e. OM is perpendicular to ON,so that we‘ have
...(5)
ri-rgH-mri( wral+c(-r)=0 or
(I —m*)—c®=0
Eliminating r, and
between (3),(4) and (5), we have
ia+mfi \ /« -mfi \(1—m*)
or
c^=>0
\H-m2
1-fm* /
(a2(1 +m2)2/(l-m^).
The locus of P (a, fi, y) is
fl -I m2)2'(l_m2).
This is clearly the equation of a hyperbolic cylinder.
Ex. 11. Tn the above Ex. 10, if MN is equal to the distance of
Pfrom the origin O, then prove that P lies on the surface
(1 -m^y fxH.v®)=n +m*)2(4c2- z«).
Sol. We have evaluated in Ex. 10 above that
M is the point (ri, mr^, c); N is the point (r^, —mr^, - c)
ri=(a4 w/5)/(l +m^). r2=(a »*)9j/(l +m2)
and the moving point P is (a, /5, y).
.. .(1)
Now OP^~a^-}~3^+y^—(MNy (given).
But (MNy:=(ri~ r^y-t-m^ ('●i+r.,y+(c-{-cy
a4-m/? . ci~mfi -f4c*.
a+m/i
a m[i
1-i-m® '^l -l- w*
=( l+m* " l+m2
.*. (1 +mV (MNy=Am^^Hm^.4ofi+4e^ (1 ^m^y
or (14-m2)2 (a^+/i^+y^)=4m^
(1 +/w«)2,.using (I)
or (1
(«2+^2^=4m* (a^+j3*)=4c* (l-)-m®)®—y* (14-m*)*
T
226
Analytical Geometry 3-D
or
or
/.
{(I+m*)2-4m*}(a24-y?2)c=(4c2-72)(1
(1 -m»)2(a3+.^®)=(4c2“-'/2j (l+/w2).
The locus of P(a, /?, y) is
(1-mY (xHj'*)=(l+m2)2(4c2-2*).
Proved.
Ejf! 12. F/«rf //re surface generated by a straight line which
meets two line y—mx^ z==c; y= —mx z= —c at the same angle.
(Kanpur 1982)
Sol. The equations of the given lines may be written as
...(1)
xf 1 =y!m .(^ c)/0=fi (say)
and
...(2)
xf\ =yl(—m)=(z-^c)IO=ri(say).
Let the moving line meet the given lines (1) and (2) in the
points C (r^, mri, c) and D (rs, —mr^t—c) respectively. The
d.r.’s of CD are rj—rg, m (fj+ra), c+c. The equations to CD
are
z—c
x-ri _ y—mrj
●■●(3)
ri—rt rH (ri+rg) 2c
Let the line (3) make angles a and p with the lines(1) and
(2) respectively, so that we have
l.iri-ri)-\-m.m (ri+ra)+0.2c
cos a=
●●●(4).
V( I + m»+0) V((ri-ra)2+m‘^
fa)'*+(2c)2}
«
1 .(ri ~fg)-m.m (rg+r,) -f 0.2c
and cos d-^(i+„.+0)y!l(i-,-r,)s+m» (ri+r0*+(2c)'}
...(5)
But according to the given problem, cos a=± cos
Hence
from (4) and (5), we get
ri-ra+m* (ri+ra^=±[(ri-r2)—
the denominators beine same have been cancelled.
Taking-i-ve sign, wc get
● 2m* (ri+ra)==0 or rg=—r,
and ^king —ye sign, we get
2
or ri=/aNow when ri^—rx, the ;^ne (3) becomes
x—ri y—fafi
“ 2c
2ri,~
i.e.
i.e.
i.e.
2rj
X
rx
X
ri
and y—mrx=0
^
z
1* — 1 and ri' y_
m
.
c
z
L.
— and ri= m
c
●
Skew Lines
227
Eliminating ri, we get
xm
or mcx=yz.
..(6)
y
c
Again, when rj=ri, the line (3) becomes
x—ri y—mri z—c
0
2mri ~~ 2c
z.e.,
X ●● rj=0 nnd ylmri=zlc; i.e., ti—x and c;;s=mzri.
..(7)
Eliminating ri, we get
cy=mzx.
Multiplying (6) and (7) the required surface is given by
{mcx^yz){cy-mzx)=0.
Ex. 13. Show that the locus oflines which meet the lines
z
x+a
y _ z
x—a
y
0 sin
-rcos a* 0 ~~sin oi cos tt
at the same angle is
(xy cos a—az sin a)(zx sin d—ay cos «)eaO.
Sol. The equations of the giv^n lines are
...(I)
(jc+a)/0=y/sin «=r/(-^cos a)i=ri
and
...(2)
(x—fl)/0=y/si|i a=z/CQS a=r,.
Let the moving line meet the given lines (1) and (2) in the
points C
sin a, -ti cos «) and P (p» r* sin a, r, cos a)
respectively. The d.r.’s of the line CP are a+ia,(r^ - ri) sin a,
cos a. The equations to CD are
x+a y-ri sin g . z-f-ricosa
2a “(fa-ri) sin g “(rj+ri) cos g*
Let the line (3) ake angles and ^ with the lines (1) and
(2)respectively, so that we have
cos 9=_2a.0H-(ra—Tj) sin a.sin «+(ra+ri) cos g,(—cos g)
"V((2fl)*+(r8-ri)=» sin*
cos* g}
^{O-rf-sin^ g-j-cos* a}
...(4)
2a.0+(rt—ri) sin g.sin «+(fa+,ri) cos «. cos g
sin* «+(^,4-r^* cos* a)
..(5)
V;{9l|-Stin® a+cos* a}
But according to the giyep prohjeiP, cQsi?== ±Qqs
lienee
from (4) and (5),.we get
tra—^i) sin*
^
>
sin* a4-(ra+ri) cos* a}.
Taking rfve sign, we get
2(I’a+r,) cos* a=0 or r,+rj—0 or r*— —ri
rAQ I
■ ®
228
Analytical Geometry 3-Z)
and taking — ve sign, we get
2
sin® a=0 or r2=r^.
Now when ra=-ri, the line (3) becomes
sin g z-i- Tt cos g
0
2n
-2rj sin a
z.e.,
:^+I
a
4—+1 and z-j-ri cos a=0
rj bin oc
rjxsin a= —
and ri=—z/cos a.
Elithinating>1, we get
zx sin
cos a=0.
Again when ra=ri, the line (3) becomes
x+a V—r, sin a z+r^ cos a
2fl
0
2rj cos Of ●
...(6)
or
or
rix cos a=<7z, ri=;;/sin a.
Eliminating r,, we get xy cos a ~az sin a=0.
...(7)
Multiplying (6) and (7), the required locus is given by
Proved.
{xy cos a—nz sin a){zx sin a—ay cos a)=0.
Ex. 14.(a) Find the locus ofa point which moves so that the
ratio ofits distancesfrom two given lines is constant.
Sol. Let P (X|,
Zi) be the moving point and the equa
tions of the given lines be
xl\=yfm={z- c)fO
.. (I)
and
xfl^yl(^m)={z-\-c)/0.
...(2)
Let />! and/;a be the lengths of the perpendiculars from P to
the lines (1) and (2) respectively, so that we have
I
J'l * 1 .1^1
c 2
Zj-C Xif
T
1
m
1+w®
'm
0
0
i ;j
1
or
{(wxi
'’
●● (3)
Replacing w by’ —m and c by —c and pi by pj,[or proceeding
as above for /?,], we have
1
P-/
I-hm®
1
"1+m® {rmxi-f;^i)*+(zi+c)2{m-+ 1)}.
...(4)
I
According to the given problem /?i//?a=A (say), where A is a
\.
constant.
229
Skew Lines
Pi=Xp3 or Pi^^XW
or
+(m*+l)(Zi+c)>}
[using (3) and (4)].
The required locus of P (Xi, j’l, Zi) is given by
(wx-y)2+(w24-l)(z- c)2=A® {(mx4-y)*+K4-l)(z+c^)}
...(5)
Ex. 14(b). Find the locus ofa point which is equidistant from
two given lines.
Sol, Proceeding exactly as in Ex. 14 (a) above and putting
A=sl [since here pi=pal» the equation (5) of the required locus is
given by
mxy+c(l+m*)z=0.
Ex. 15. Py P* are two variable points on two given non-intersecting lines and PP* is of constant length
Find the surface
generated by PP\
Sol. Let the equations of the given lines be
. (1)
x/l=y/m=(z-c)/0=ri
and
...(2)
x/l=;;/(-m)=(?+c)/0=ra.
The co-ordinates of the variable points P and P* on (1) and
(2) are (ri, mr^ c) and (rg, —mr^y —c) respectively.
It is given that PP'=constant, say 2A.
(PPy=4X^
or
...(3)
(ri-roY+m^ (^i+^2)^+4c*=4A*.
Also d.r.’s of PP' are ri-rg, m (ri+rg), 2c.
The equations to PP' are
z~c
x-ri
^
fi-r* in(ri-\-r3) 2c
...(4)
The locus of PP'is obtained by eliniinating fj and r^ between
(3) and (4).
From (4), we have
z-c
z—c
and y -mrx
w(riH-ra) 2c
1*1-ra“ 2c
or
2cx-2cri=ri (z-c)~ra (z-c);
and
lcy—2mcri=mri(z—c)+mra(z—c)
or
»'i (2+c)—ra(z—c)=2cx
...(5)
...(6)
and
wri (z-f c)+mra (z-c)=2cy.
Solving (5) and (6) for r, and rg, we get
.ri-c (y-|-mjc)/{w (z-fc)), r.^c
(z-c)>. ...(7)
230
Analytical Geometry “i-D
'y-i-mx , y—mx} 2c(yz-cmx)
ri+raH=*—
z-c
m
z^c
~~ m (z*—c*)
c y-\-mx y-mx \ 2c imzx—cy)
and
m
z+c
m(z*—c®).
z~c f
Substituting the values of rx-\- r^ and ri—rg in (3), the required
locus is given by
4c^ (yz -mcx)^
4c^ (mzx—cy)^
f4c2=4A2
*
{z^ -c^
.m*(z2-c*)“
or
c® {mzx—cyY+c^m^ {yz- mcxY=m^(A^—c^)(z*-c")*.
Ex. 16. AP and A*P' are two given lines, A and A beingfixed
arid P and P'' variable points on them such that AP.AP'is constant.
Find the surface generated by PP'.
Sol. Let the equations of the lines AP and AP' be respecti
vely given by
.V./1
=(z~ c)/0=ri
...(I)
and
...(2)
^l\^yl(-m)={z+c)IO=rz.
The co-ordinates of the fixed points yl and A' on the lines (1)
tuid (2) are,(0, 0, c)and (Oi 0, - c) respectively.
Let the variable points on the lines (1)and (2) be P(ri, mrj, c)
and P'{r^, mr-i, —c) respectively.
According to the problem /4/»./4'P'=constant, say k.
V{(/‘i--0)H(mri~0)H(c-c)2}.V{(r,-0)H(-wr,-.C)a
M-c+c)*}=k
or
(r2*-fmV)=fc
or
^1^9(1+^”*)=^ or riro=A:/(l-|-w®).
...(3)
The equations to PP' are
x—r, __ y ~mrj _z- c
(4)
ri -r2""m(ri+ r^)^'W
The locus of PP' is obtained by eliminating I'l and rj between
(3) and (4). Hence proceeding as in Ex, 15 above and putting the
values of r^ and r^ from (7) of Ex. 15 in (3), the required locus is
given by
k
c (y-h/wx) c(y-mx)
m {z-^c)' m {z-c) ~ l+m^
or
c2(1 -i-m?)
(z^ -c^).
Ex. 17. Find the locus of a straight line that intersects two
given lines and makes a right angle with one of them,
. Sol. Let the equations of the two given lines be
(1) and y=~‘mx,z= — c.
y-~mXi z=c
...(2)
the equations of any line intersecting the lines (I) and (2) are
given by the planes'
231
Skew Lines
...(3)
mx)—Ai(z-c)=0 or —mx+y’-‘XiZ+chi^0
..(4)
and'(;^+mx) - A2(z+c)=0 or
A2Z—cAa=0;
Let A, fJii V be the direction ratios of the variable line given
the intersection of the planes (3) and (4). Then
—mA+1./»—Ai.v=0,/wA+1.#t-Aav«=;0.
V
A
/*
Solving, we have
32^..(5)
Let the line with d.r.’s A, ft, v be perpendicular to the line (1)
Le. to
so that, wc have
m
0
l.(Ai-Aa)+»i { m (Ai+Aa)}+0.( -2/;i)=0
. (6)
or
(Ai-Aa)-AMMAi+A2)=0.
The required locus is obtained by eliminating Ai and Aa bet
ween (3),(4) and (6). Hence it is given by
jy—mx y+mx —m^ \y~mx y+mx
z—c
z-|-c
z+c
\ z -c
or
-2mzx-{-2cy—m^(2yz-2mcx)=0
...(/)
or
mzx—cy=m^ {cmx—yz).
If the line would have been-perpendicular to the line (2), then
the required locus on replacing m by - m and c by —c in (7) is
given by
(mzx~-cy)=-‘m^(cmx -yz).
1
Ex. 18. Find the surface generated by d straight line- which
intersects two given lines and is parallel to a given plane.
Sol. L^he^uarf6nrbTffie~glveirlines-b
...(1)
and.
z=~c
y=mXt z=c
The equations of any line intersecting the lines(I) and (2) are
given by the planes
...(3)
(;;—mx)—Ai(z—c)=0 or
mx-fj'—AiZ-fcAi==0
...(4)
or (y+mx)—Aa(z+c)=0 or
AjZ -cAa=0.
Let A, jii, V be. the d.r.*s of the variable iine given by the inter
section of the planes (3) and (4). Then
—wA+l ft—Aiv=0, wA-M./tt-Aav=0.
V
Solving, we have
.:(§)
-■/M (Ai”t" Aal ■ —2tn
Ai — ^3
Let the line with d.r.’s A, ft, v be parallel to the given plane
^
say. Ax+By-^Cz-\^D=0y so that we have^
.. (6)
,
+Aa)+C2.(—2w)=0.
The required locus is obtained by eliminating Ai and A2 belween (3), (4) and (6). Hence it is given by
232
Analytical Geometry
A> y-mx
z—c
y-^mx -Bm y-mx , y-^mx
.-2wC«=0
z -c
z+c
z-i-c ^
or
A (2cy -2mzx)—Bm {2yz ~2mcx)—2mC(z*—c*)=0
or
A {cy—mzx)-Bm {yz—mcx)—mC(z*—c®)=0.
Ex. 19. Find the surface generated by a line which intersects
the lines y=z=a;x+3z=a,y-^z=a and is parallel to the plane
x-^y=0.
Sol. The equations of the given lines are
y—a~0tZ-a—0
...(1)
x+3z—a—0yy+z~a=0.
...(2)
The equations of uiy planes through the lines (1) and (2) are
(y—a)—Ai(z-a)=0 or y—AiZ--a+aAi=0
...(3)
and
fx+3z~a)-A2
a)=0
or
x-Ag^+fS—Aa)z-fl+flAa=0.
...(4)
Any line intersecting the lines(1) and (2) is given by the
intersection of the planes(3)and (4). Let A, #*, v be its d.r.’s, then
0.A+1.#*-Ai.v=0, l.A-Aaf*+(3-A,)v=0.
A
V
** 3-Aa-AiAa “^“-1 ●
Now the line with d.r.’s A, p, v is parallel to the plane
x+;>=0 i.e., this line is perpendicular to the normal to the plane
X’\-y<=0 whose d.r.’s are 1, 1, 0. So we have
l.(3-A2-AiAa)+l.(-Ai)+0.(-l)=0
3—Ai—Aa““AiAa=0.
.
...(5)
The reqcifcuiGCus oi the line is obtained by eliminating Ai
^ and Aa between (3),(4) and (5) hence is given by
g
or
or
or
or
;>-a .Y+3z-fl
fl x+3z-<M 0
z—a
y+z-a ~z—a y+z—a °
3(y+z-a)(z-a)~-(y-a)(y+z~a)-(z—a)(x+3z~a)
-(y-a)(x+3z-a)=^0
—yz -jA-{-2az -xz+2ax—xy=0
yz+y^+xz+xy=‘2az+2ax
(y+z)(x+y)=2a(x+z).
Ex.20. Find the locus ofthe lines which move parallel to the
zx^plane and meet.the curves.
xy=<^, z^O; >®=4cz, xo»0.
"~soK—Lgt *h£-ggu§tions of any line be
,A
(x—a)//=(y—
According to the problem!,-the line (1) is parallel tothe-zx-—
plane r.e. the plane y=0,so that we have
1#-.
m
Skew Line^
/.0<^m.l+n.0=0 or m=0.
Putting m=0 in (1), the line becomes
...(2)
{x-a.)li=iy-fi)IO=(z-y)ln.
Now the line (2) meets xy=>c^y z=0,so that we have
(x-a)/lMy-p)IO=^(0-y)ln
or
(x -a)ll=~ylny y=fi or x-^a.—{lyin),y~p.
Putting the values of x and y in xy—d^^ we have
/
O''
or
c2\
1
/
V fi }y~n'
...(3)
Again the line (2) meets y^=Acz^ x=0, so that we have
(0-a)//=0; -|3)/0=(z-y)/n
-a!l=(z-y)jn, y=fi or z«y-(na//), y=fi.
Putting the values of y and z in /=4cz, we have
^=4c(r-^)or
...(4)
Multiplying (3) and (4), to eliminate / and n, we get
or
or
or
(4c7—j8*)«4ca)?y
4cajffy -/S® (a^ -c*)-4c*y=4«;ffy
(a^—c*)+4c®7=0.
Now the locus of the line (1) is the locus of the point (a, fi, y)
and is given by
/(xy-c*)+4c3z=0.
8
Change of Axes
§ 1. Transformation of co-ordinates.
In the study of analytical geometry, the co-ordinates of a
point, the equations of a curve or the equation of a surface are
always considered with regard to a fixed origin and a set of co
ordinate axes. The co-ordinates of a point, the equations of a
curve or the equation of a surface change when either the origin
is changed or the directions of axes are changed or both are
changed. The process of changing the co-ordinates ofa point, the
equations of a curve or the equation of a surface is said to be the
transformation of co-ordinates. This process of transformation of
co-ordinates will be of great advantage to tackle most of the
problems.
§ 2. Change of origin (Translation of axes).
To change the origin of
co-ordinates to another point
(«» Pi y)i whereas the directions
of axes remain unaltered.
Let O be the origin of co
ordinates and OX, OY, OZ be
the original co-ordinate axes
Let O* be the new origin and
(«j Pi
co-ordinates referred
to theoriginal axes.
Draw three lines O'X', 0'Y\
/
Af
/V
L
0‘
<5?
X
V
and O'Z' through O’ parallel to and in the same directions as OX,
OT and OZ respectively. Let
be a point in the space whose
co-ordinates are {x, y, z)referred to the original axes OX,OY, OZ.
Again suppose that the co-ordinates of the same point P referred
to the new axes O’X’, O’Y’, O'Z’ are {x’, y', z’).
Change of Axes
235
From O'draw O'L perpendicular to OZ. Frgm Pdraw PM
perpendicular to ;;z-plane to meet v'z'-plane in N. Then
0’L=y.^NM, PM=x,PN=x\
We have PM=PN-\-NM
i.e.
X=JC'+a.
Similarly .
y=y!+P,z=z’+y.
Also,
x'=.\-'-uy'=y-P, z’=z—y.
Thus: we have the following rule:
Rule. To change the origin to another point (a,
y) replace
X by .'v+a, y by y-^ fi and z by z-^y. The transformed equation is
then obtained with respect to (a,
y) as the new origin and in the
transformed equation also the current co-ordinates are (j:, z).
Remark. Shifting the origin back. Sometimes it is required
to shift the new origin back. Then replace x by x— a, y by y^fi
and z by 2 y in any equation referred to the new origin to get
the corresponding equation referred to the old origin.
§ 3. Change of directions of axes. (Rotation of axes)
Tofind the change in the co-ordinates ofa point when the direc
tions of axes are changed, the origin beingfixed.
Let OX, OY. OZ be the ori
ginal system of co-ordinate axes.
Let OX\ OY\ OZ' be the new
axes through the same origin O.
Let /i, nil, I
wig*
and
/a» Wa, rta be the direction cosines
of OX', OY' and OZ' respecti
vely referred to OX, OY and OZ.
Therefore, the direction cosines
of OX, OY and OZ referred to
OX', OY' and OZ' are li, l^, l^; Wj,
ni^ and
respectively.
Let P be a point in the space whose co-ordinates are
(x, z) and (x', v', z')
referred to old and new axes respectively. Draw PM perpendi
cular;to OX.
●
OM—x ov x=OA/-"the projection of OP on C?Z.
Now referred to new axes the co-ordinates of P are (x', y', z')
and the d.c.’s of OX are /„
/j. Hence we have
x=OM=(x'-0)/i-Hy_0)i^+(z'~0) I3
236
Analytical Geometry 3-D
or
Similarly y^Wix'-^-may'+m^z' V
..,(1)
z=Wi^'+«ay'+WaZ' J
Now multiplying the relations (1) by 4, mi, ni respectively
and adding,.we have
and
or
or
liX-\-miy+niZ=(li^-^mi*+ni^) ^'+(/i/2+WiWj+/iiff2) y'
+(/i/3+/MiMa+niWa) z'
hX'\-miy-\rn,z=\.x‘+0.y+Q.z' *■
[See § 4, Next]
X =liX’\-miy-\-ni7,
}
Similarly y=kx-\-m^-\-n^z
...(2)
and
z'=kx+m^-\rn^z.
Hence an equation referred to OX, OY and OZ as axes is
transformed to an equation referred \o OX\ OY' and OZ' as axes
by replacing x: by kx'+ky’\-hz' y by mi.x'+Way'+Waz'and rby
HiX*+n2y'+n^z'. [See relations (1)]. In practice we leave the dashes
and hence from {\) the rule to obtain the transformed equation
referred to the new axes OX', OY', OZ' is to replace
X by IiX+Igy+lgZ y by miX+may+maZ
and
z by HiX+nay+naZ.
In the transformed equation also the current co-ordiiuites are
{x,y,z).
Again if we want to shift the co-ordinate axes back to their
original^ positions, then replace x {i.e. x') by /jX-f Wiy-f«iZ,
y {i.e. y') by igX+niiy+n^z and z {i.e. z') by liX-^m^y-bnaZ in any
equation referred to the new axes OX', OY' and OZ' to get the
corresponding equation referred to the old axes OX, OY and OZ.
[See relations (2)].
Remark. The results given by (1) and (2) can be conveniently
remembered by the following scheme of transformation.
x'
y
z'
X
lx
h
I:
y
mi
z
Wg
«2
«8
237
Change of Axes
In the above scheme to get the value of jc, first multiply each
element of the row of x with the corresponding elements of the
first row and then add, i.e. we have x^lix'-^rky'-^-hz'- Similarly
the values of and 2 are obtained.
To get the value of x', first multiply each element of the
column of x* with the corresponding elements of the first column
and then add, i.e. we have x‘=liX+mxy^niz. Similarly the
values of y' and z' are obtained.
§ 4(A) Relations between the direction cosines of three matoally
perpendicular lines.
Let OX, OY, OZ be a given set of mutually orthogonal axes,
and OX\ OY\ OZ' be three given mutually perpendicular lines.
Let /j, Wj, Wj ; /g,
and /a, rwa, «a be the direction cosines of
OX', OY' and .OZ' respectively referred to OX, OY and OZ.
Therefore, we have the following six relation :
W2^=l V
...(1)
J
/j/.,4"n/1Wa-f" Wj^2“®
y3+m-2W8+ «2n3=0
/8/i4-m3/Wi-}-W3«i=0.
...(2)
Again the direction cosines of OX, OY, and OZ referred to
OX', OY' and OZ' are /„ 1^, A,; Wp Wj, w, and n^, n^, «a respecti
vely. Therefore, as above, we have the following six more
relations :
/i^+V-4-/a2=l
...(3)
/aWa-l- /a^Mg —0
mxn,_-\~nhini-^m^n^—0
Wt/i+^2/2+Wa^3 —
...(4)
■ (B) The direction cosines of one line in terms of the direction
cosines of other two lines.
From relations (2) above, we have
■ /i/a+Wi'”2 f
A/s l-"h'’^*3 4-Wina=0Solving these relations for /i, m^, /?i, we get
/.
ni^n^ ■ m^n.i
«i
n^k ● ml^
V(i)
c^=±ll\=±{rryh - nhJh), Wi=±(«a/a -rtsL), /»i=
(Lm, -/amj),
...(5)
Similarly other relations can be delei mined.
238
Analytical Geometry 3-D
Let us suppose
D= it
mi
U
Wa
«2
/:
/W3
«3
We have
The CO-factor of /i=(m^n^-ma^a), etc.
Dwce the relations (5) show that each constituent in D
= ±iits co-factor).
(C) To show that D=±\.
We have D*= it
mi .
or
/g
ma
«2
/a
m3
/I3
D?=
D?=
A=
mi
/a
ma
4
/Wg
.
»i
2/./»
A/a
1
0
0
0
1
0
0
1
0
or
X
it
A/a
A/.
or .
ffi
D*=l, or
^/./a
^/a»
[Using relations (1) and (2)]
D=±l.
SOLVEip EXAMPLES
Ex. 1. 04, QB, OC are three rnutually perpendicular lines
through the origin with direction cosines l^, mu «z I /a» **h, w* and
h* »*8, ffa etpd if04=pB—0C^ay then show that the equation of the
plane
is (Ij^f-jg+hi) x+X/nj+m^-l-fna)
(«i+fa+«a)z—o.
SftL Ifthe jinps 04,0% OC be taken as pOTrordinate axes,
thep the aquation pf jtite plapc which makes egual ipercepts (each
equal to a) on these axes is Xla-\- Yla+Zla—\
i.e. \ '
.(I)
'r+r-i-^=?.
But it is required to find the equation of the plane referred
to the origiparl axes. Hepce using the relations (2), § 3, we put
X=liXr\-mjy-\-hiZ, y=/2X-b.Way+«fZ apd Z=l^x-^p!iay+naz jn the
equatipn (i). Thep the equation of the plane a]^C referred to the
original axes is given by
Change of Axes
239
(/io:+Wi3;+Wiz)+(/aX+m,>?+naz)\-{Ux-\-m^-\-nzz)=»a
(/1+/8+/3) x+Cwi+ma+ma):H+(ni+ «*+/«8)z=a.
Ex. 2. The equations referred to rectangular axes of three
mutually perpendicular planes are Pr—
«r^=0, r=l, 2, 3.
Prove,that if(a, y9, y)ds at a distance dfrom each ofthemt then
/aP2~l~ IzPz) P—im\Px'^mzPi’¥m^P^
</= « —
mi+mj+Ws
/1+/2+/3
y-(WiPi+«aP2+«aPa)
ni+»2+»8 ■
or
Sol. The equations of the given planes are
..■(0
/iX+miy+«iZ-pi=0,
(2)
kx+
+Wgz -P^=0>
...(3)
and
/sX+may+nj^-P3=0.
' where /i, nii, Wi etc. are direction cosines of the normals to the
given planes. According to the given condition, we have
rf=fhe. distance of(a, y)from the plane (1)
and
or
Similarly d=l^-\-m,p-^n^y—pt
rf=/3a+m3/?+nay-pa.
...(4)
...(5)
...(6)
Multiplying (4) by /i, (5) by k and (6) by /j and adding, we get
(/1+/2+/*)
(2
2? (/iw,)+y S (/i«i)
(Pl^+P2^2+P|/t)
=«.l+j8.0+y.0-“(pi/i+Pa/9+P8/3)
[using (3) and (4) of § 4]
«—^Pi/i+Pa/f+PsM
...(7)
li+k-hU
Similarly multiplying <4), (5) and (6) by pi,, wia and /«i respec
tively and adding, we get
...(8)
d={fi - (pim,+P2P*8+Pa«?»)}/f«l+?”2+»*8).
And multiplying (4), (5) and (16) by Wi. «, and n, respectively
and adding, we get
...(?)
</= { y — (pi«i -l-P2e2’^^^8®3)V(?*l
The. relations (7), (8) arid (9) |ive t^e re^riked result.
Ex. 3. Show that for any lra^prmati^/i of rectqnj^Jqr axes
without change of origin, a+^+c, (i0 p+g^-^if—bc—cd—qb and
abc-\-2fgh—ap—bg^—ch^ are iqvfiritmts for tij^e expression
T 2/yz+2gzx-i-2ftxy.
240
Analytical Geometry 3-Z)
Sol. Suppose the expression
(U
ax^-rby^-\-cz^-\-lfyz-\-2gzx-{-lhxy
after any transformation of axes without change of origin becomes
...(2)
-l-2G2;c+2^jr;;.
We know that x^-\-y^-\-z^ is ihe square of the distance of any
point {x, y^ z) from the origin. Hence A
does not
change by change of axes without change of origin Hence the
equation
ay^’\‘by^-\-cz^-\-2fyz’{‘lgzx-{-2hxy - A {x^+y^-\rz^)’=^ - (S)
transforms to
Ax^-^By^-\-Cz'^^2Fyz+2Gzx-\-2Hxy-\{x^^y’^^-z^)^i). ...(4)
The equations (3) and (4) may be written as
(a~X) x^i {b~X) >-24 (c-A)z^-\-2fyz+2gzx+2hxy=0 ...(3')
M-A)x^+{B A) y^MC A) z^-'r2Fyz-\-2Gzx+2Hxy=0 ...(40
Now if(3') represents a pair of planes non (4') will also
represent a pair of planes for the same value of A, i.e. A is the
same quantity in both the equations.
The condition for (3') to represent a pair of planes is
h
a A
g
h
b-X ■
/
c-A
8
/
or (a~X){(b-X){c \)P}-h {h (c - X)-fg} j-g {hf g(A-A)}=0
or
}?—{a-\-b+c)X'-{-(hc-\-ca\-ab P g~ //-) A
(abc \-2fgli ap bg- ch^)=0. ...(5)
Similarly the condition for (4)to represent a pair of planes is
A® -iA+B+C)X--h(BC+CA hAB-F^ -G^ WO X
-{ABC+2FGH-AF^ BG^ CH^)^0
.. (6)
Since the roots of (5) and (6) are same i.e. the values of A
given by (5) and (6) are same therefore comparing the coefficients
of like powers of A in 15) and (6j, we get
a-\-b+c=-A+B+C
/24.g2+/,2
and
abc-\ 2fgh op—bg^—chABC+2FGH
BC CA
AB
AF^-BG^ CHU
...(7)
Relations (7) show the required results.
Ex. 4. Show that if
dx-A-by^+cz^-\-2fyz+2gzx-\-2hxyA-2ux+2\y-\-2wz-{-d
he transformed by change of co-ordinatesfrom one set of rectangular
^
241
Change of Axes
axes to another with the same origin, the expressions a-^b+cami
a^-{-v^-\-w^ remain tmaltered.
Sol. The given expression is
ax^-{-by^+cz^+7fyz-\-2^zx-\-2hxy-{-2ux-]-2vy-\-2wz-^(i. ...(1)
Let the given set of axes OX,OX, OZ be transformed to the
new set of axes OX', OY', OZ'. Let d.c.’s of OX', OY', OZ' be
kf trix. Hi; /a, m^, ; k, m^, m referred to OX, OY, OZ. Now
replacing x by liX+lzy+kz etc. [See § 3 relations (1)], the ex
pression (1) transforms into
a {iiX+liy+i3z)^+b {miX-^m2y-\-m3z)^-\-c {niX-\-n.yyjrni=f
+2f(miX-\-miy-hm^z)(/»iX ^n^y+n^z)
^2g(nxX-\-niy+niZ)(liX+l^y -|-/gz)^2h{lix+lzy+hz){mxX-\-m^-\- m^z)
-f2w (/iX-h/iV-f/8z)+2v (mjx-l-mjj'+maz)
-|-2w (nix+n2y-{-n^z)+d.
..●(2)
Suppose the expression (2) is written as
Ax^-^By^+Cz^-\-2Fyz+2Gzx-\-2Hxy+2Ux+2Vy+2iVz+D
where A, B, C, F, G, H, U, V, W and D are clearly given by
A
cn^.-Jc2fmxn^+2gnxlx-t 2hixmx,
B=
bm^ -1- ewa*+^fm^n^ -|- Ign^h 2A/ama>
C—al^^+bnh^-\rcn^^-\-2fm^n^-{-2gn^k Ar2hi^m^.
I/=M/i-fvmi+ivni, K=tt/24-vma-fu’Wa,
W—ul^-\-vmi-\-wn^, D~d.
A
or
A+B+C=a T V+fc -V mx^+c S nx^+2fS mi«i
+ 2gr «i/i-f-2//
5-|-C=n4 ^-1 c
and C/2_|_ j/2^
Hence the
unaltered.
/iWj
[using relations (3) and (4), § 4]
(n/a-j- v/Wa+»t’«a)^
+(M/3+vma +Jv/ia)=M« S /ia+v2 Z mx^-\-w^ S V+2ttv Z ^mx
+2wuZ nxlx-\r2vw ZmxHx
=--m8+v2+u’2 [using relations (3) and (4) of § 4].
expressions a-Vb^c and
remain
vmx+
9
The Sphere
§ 1. Defioition. A sphere is the locus of a point which moves in
: space such that its distancefrom afixed point is always constant.
The fixed point is called the centre of the sphere and the.
constant distance is called the radius of the sphere.
§ 2. i^aatioD of a sphere.
(A) Central Form. Tofind the equation of a sphere when cen
tre and radius are given.
Let O be the origin of co-ordinates. Let
the co-ordinates of the centre C of the sphere
<r
be (a, h, c) and r be the radius of the sphere.
j
Let P(>r, z) be any point on the surface of >o
the sphere. Then
"0C=a\-^b\’\-c}L,
and
OP=3ci+pj+2h-
i-f-(;p-h).j-l-{z-c) k.
We have
o
€P=\ CP J=r.
I (x-a)i+O-h)}+(z-c)|
k =.r
or
or
-m=r
CJc-fl)*+(:P-h)®+(2~c)8=r*.
...(1)
The equation (1)is the. required equation of a sphere whose
centre is (a, h, c) and.radius r.
Partieolar Case. If the centre of a sphere is origin and
radius is r, then we |iave ! OP I =r,'
where O is the centre and OP(==r) is the radius.
Hence we have .
I xi+yl-\-2k \ =r i.e.
or
;e*+^^-f-z*=£=r*
which is the requiredjequation.
(2)
The Sphere
243
Remark.- It can be easily proved that every diameter of a
sphere subtends a right angle at any point of its surface.
(B) General form. To find the general equation of a sphere.
Tofind also the centre and radius of the sphere in the generalform,
[Garhwal 1979]
The equation of the sphere with centre (o, bf c) and radius r
is given by [See equation (1) above]
(x-aY-{-iy-bY+iz-cY=r^
(3)
or
3C*-fy® 4-z®—2ax -2by-2cz+(a^-f
c*-r*)=0.
The equation (3) is of the form
...[4)
The equation (4) is called the general form of the equation of
a sphere.
Comparing (3) and (4), we get
2<7=~2t<, 26=-2v, 2c= -2w,
or
M, b=-~Vy c=—w,
rf.
the centre of the sphere (4) given in general form is the
point(—«, —V, — w) and the radios is
(C) Conditions for a sphere. The most general equation of
second degree in Xy y and z is
ax*+6y*+cz*+2/yz-l-2gzx+2/ixy+2«x+2vy4-2wz+(/=0. ...(5) .
Comparing (5) with the general equation (4) of the sphere,
we have:
(i) The etfuation of a sphere is ofsecond degree in x, y and z.
(ii) The coeflScients of x*, y® and z* are equal in magnitude
and sign both, and
(iii) The coefficients of the product terms involving yz, zx
and xy are zero i,e, in other words, in the equation of a sphere
the terms involving yz, zx and xy must be absent.
Remark!. The radius of the sphere (4) is
which is real when
d > 0.
If!i*4-v®4w*—t/=0, the radius is zero and the sphere reduces
to a point and is called a point sphere. If
< 0,
the radius is imaginary whereas the centre(-u, -v, — w) is real
and in this case the sphere is called a virtual sphere or pseudosphere.
Remark 2. The general equation (4) of the sphere contains
four arbitrary independent constants namely u, v, w and d and
244
Analytical Geometry 3-D
hence the equation to a sphere can be determined if foui condi
tions are given.
(D) Four-point form. Tofind the equation ofa sphere which
passes through four given points A. B, C and D whose co-ordinates
^*)»
yz* ^3) ond
y^, Z4) respectively.
Let the general equation of the sphere be
^®+v*+^*+2tt;r-i-2vy-f-2»v2-frf=0.
.. (6)
If it passes through the four given points A,B, C and D,
then their co-ordinates will satisfy the equation (6) and therefore
we have
’
“+iVi*+*+2uxi-{-2 -f.2m'Zi-f </=0'
^8*+;^9*+
2f/X8+2v7j.-j-2wz8-f</=0
^8®+;V3j+23®+2MAr3-i-2v;;8-f-2wz3+</=0 ‘
...(7)
V+V+^i*+2ttor4-f-2v;^4-h2wz4-f </= 0.
Eliminating the constants «, v, w and d between the equations
(6)and (7), the required equation of the sphere is given by
X
z
1
y
■^'1}
^i®4-^i*+rr
■ V+V+^4^
yi
zx
1
y%
z,
1
yz
zz
I
^4
=0.
...(8)
1
^Remark. The equation (8) has* a determinant of 5th
and hence in numerical problems, the evaluation of 5th
determinant will require much labour and therefore, it is
sable, to solve four equations given by (7) for n. v, »v and 4/,
order
order
advrthen
the required equation of the sphere is obtained by substituting
the values of w, v, w and d in equation (6),
(E) Diameter form. To
the equation of a sphere on the
join of two given points A {x^, y,, z^) and B {x^, y„ 2,) as diameter.
(Berahampur 1981 (S); Indore 79; Lucknow 78; M. Dayanand 79)
Let P with co-ordinates (x, y, z)
p
be a current point on the surface of a
sphere drawn on the line joining the
given pojnts
^ (xu yu Zi) and B U3, y^, Zg)
as diameter. Then clearly the lines AP
ai;d BP are perpendicular to each other.
e
The Sphere
245
The direction ratios of AP are
y-yi, z~zi;
and the direction ratios of BP are x—x^^ y—yz* z—z^.
Since the lines AP and BP are perpendicular to each other,
therefore using the formula
we have.
ix-Xy)(x-~Xa)+(.v -yi)(y -y^-^(z-Zi)(2-za)=0.
...(9)
The equation (9) is the required equation of the sphere.
SOLVED EXAMPLES(A)
Ex. 1. Find the equation of the sphere whose
(i) centre is(- - 3, 4, 5) and radius 7,
(i7) centre is (2, —3,4)and radius 5.
[Meerut 1972]
Sol. Let P(x, yy z) be a. general point on the sphere, then
using § 2(A), equation (1), we have
(i) The equation of the sphere with centre (—3,4, 5) and
radius 7 is given by
{x-(-3))»+(.V-4)2+(2-5)2=(7)2
or
x2+6x+9
-8j;+16+2=-iOz+25=49
or
●v*+J’®+2®+6x“8j>—10z+} =0.
Ans.
(ii) The equation of the sphere with centre (2, —3, 4) and
radius 5 is given by
.
(x-2)=+{:V-(-3)}=+(2-4)==(5)=
or
x= - 4x+4+>»=+6;;+9+2= - 8z+16=25
or
A:=+y=+z= -- 4x+ 6>>—8z+4=0.
Aps.
Ex. 2. Find the equation of the sphere whose centre is
(2, —3, 4) and which passes through the point (1, 2, ~ 1).
Sol. Clearly the radius of the sphere.
= the distance between the points (2, —3,4) and
(1,2, -1)
.
=V{(2- l)“+(-3-2)=+(4+l)=}=V(H25+25)=V(51).
Hence the equation of the sphere whose centre is (2, —3, 4)
and radius ^(-**1) given by [see § 2 (A)]
(x-2)=+(;;-(-3)}=+(2-4)=={v'(5l)}=
or
x=-4x+4+;p=+6;;+9+22-82+16=51
or
Ans.
x’^+y^+z^—4x+6y—8z—22=0.
Ex. 3. Find the centre and radius of the following given
spheres :
(/) x=4 .y2-l-2=—2x+4j>—6z=il,
and Hi) 2x=+2j;2+222-2x+4>^-62=15.
[Burdwan 1976]
246
Analytical Geometry 3-D
Sol. (i) The equation of the given sphere is
...(1)
x^+y^+z^-2x+4y—6z-l 1 =0.
The most general equation of the sphere is
...(2)
j>®+z*+2m:«+2v;>+2wz+rf=0.
Comparing (1) and (2), we get
2u=-2,2v=4, 2w=-6, </=-ll
or
fi=i —1, v=2, w=-3, rf= —11.
The centre is(—«, —v, — w) i.e. (1, -2, 3)
and radius=
^)«\/{(-l)*+(2)*+(-3)a-(-11)}
=V(1+4+9+11)=5.
(ji) The equation of the given sphere is
2x“+2;>*-|-2z®—2x-|-4:»>—6z—15=0.
Dividing throughout by *2% we have
...(3)
x^+y^+z^-x+2y-3z-{l5l2)=0.
Comparing (3) with the most general equation (2) of the
sphere, we have
U-.
J..v=l,
3/2, «?=:-15/2.
The centre is(—a,-V, —w)i.c. (i, — 1,1)
and radius=v^{«Hv«+w*-rf}=V{a)*+(-l)*+(l)* ■( - V)}
= V(i+l+l+¥)=\/(ll).
Ex. 4. Find the equation of the sphere on the join of (2, --3, 1)
and (3, — 1, 2) as diameter,
Sol. The equation of the required sphere is given by [See
§ 4 (E), equation (9)]
(x-2) (x-3)+{;;-(^3)}
(z-2)=0
or
X*—5x+6-{4;V-1-3-i-z*—3z+2 « 0
Ans.
or
X*
z^—5x+4;V - 3z +11 = 0.
Ex. 5. Find the equation to the sphere through the points
(0, 0, 0), (0, I ,-1), (--1, 2, 0) W (1. 2, 3).
(Meerut 1975, 85 ; Madras 74)
Sol. Let the equation of the sphere be given by
x^+j'*+z*-H 2ttx-l-2vy+2wz+=0.
...(1)
If (1) passes through (0, 0, 0), then d=0.
Putting
in (1), the equation of the sphere becomes
JC*+^*-l-z*+2ttx+2v>'4-2wz=0.
-.(2) .
If (2) passes through (0, 1, -1), we get
0-HI + 1+0+2V—2w=0 or V—w-}-l=0.
.(3)
If (2) passes through (—1, 2, 0), we get .
l-|-4-i-0-2«-|-4v+0=o or —2«-|-4v-|-5=0.
...(4)
If (2) passes through (1, 2, 3), we get
H-4-1-9-1-2m-1-4v4*6w=P or 2«-f-4v+6w-i-14=0.
...(5)
The Sphere
247
Adding (4) and (5), we have
..(6)
8v-}-6w4-19s»0.
Multiplying(3) by 6 and adding to (6), we get
14v+25=0 or v=-25/14.
Putting the value of v in (3), we get
—25/14-w+l=0 or w=-11/14.
Putting the value of v in (4), we get
-2tt-50/7+5=0 or «= —15/14.
Putting the values of «, v, w in (2), the equation of the requi>
red sphere is given by
;c2+:v"+z“-(!5/7);c~(25/7) (I 1/7)2=0
or
7(;c*+/
-1 -25;^—11z=0.
Ex.6(a). Find the equation of the sphere passing through
(0,0, 0),(fl, 0, 0),(0, b,0) and {0,0, c). (Agra 11>75 ; Meerut 85S)
Soi: Let the equation of the sphere be given by
. ●(1)
x^+y^'+z* l-2ttX+2vj>+2wz+^/=0.
If(1) passes through (0,0, 0), then J=0.
Putting d=0 in (1), the equation of the sphere becomes
...(2)
jc*+>*+z*+2ttX+2v;;+2wz=0.
If(2) passes through (o,0, 0), we get
aiT40+0+2Mfl+0+0=0 or 2i<=—a or k=—
Similarly if(2) passes through (0, A, 0) and (0, 0,c), we get
v=— and w=—Jc.
Putting the values of «, v, w and/? in (1), the equation bf the
required sphere is given by
...(3)
x^+y^+z^—ax—by—cz^O.
Note. The co ordinates of the centre of the sphere (3) are
(-11,—V,—w) i.e. {\a,\bAc)
and the radius=>/(«*+v®+w*= V{(-i«)H(-● W+(-i<^)*-0}=iV(^+*Hc2).
Ex, 6 (b) Find the equation of a sphere which passes through
the origin and intercepts lengths a, b and c on the x, y and z axes
(Ranchi 1976)
respectively.
Sol. The sphere intercepts a length a on the x-axis and hence it passes through the point (o, 0, 0). Similarly the given sphere
passes through the points (0, b, 0) and (0, 0, c); Hence we are
to find the equation of a sphere passing through the four points
(0,0, 0), (fl, 0. 0), (0, d, 0) and (0, 0, c).
Now sec Ex. 6 (a) above.
248
Analytical Geometry 3-D
_ Ex.6(c) The plane ABC whose equation is
xla-\-ylb^-zlc=^l
.meets, the axes of x, y and z in A, B and C respectively. If O is the
origWyfind the equation of the sphere OABC.
[Aera 1974 ; Rajasthan 78]
Sol. The equation of the given plane is
...(1)
xlaA-ylb+zlc= 1.
The plane (1) meets the x-axis in the point A where j^=0, z=0,
and so (1) given x/fl+0+0= 1 or x=a. Therefore, the co-ordinates
of the point are (fl, 0, 0).
Similarly the co-ordinates of the points j? and C are (0, b,0)
and (0, 0, r) respectively.
Hence the required sphere OABC is passing through the four
points O (0, 0, 0), Ala,(\ 0), B(0, b, 0) and C(0,0, c). Now pro
ceed as in Ex. 6(a) above.
Ex. 6(d) Find the equation of the sphere circumscribing the
tetrahedron whose faces are
x=0, y=0,z=0, xlaA-ylb+zi'c—l.
[Punjab 1976 ; Meerut 80j
Sol. The equations of the plaiies [i,e. faces of tetrahedron]
arc
x=0
...(2)
...0) y=0
2=0
●●.(4)
...(3)
x/a+y/b-\-zlc=^\.
Any three faces (/.e. planes) meet at a vertex of the tetra¬
hedron.
Solving (1), (2) and (3), we have the vertex O (Oj 0, 0).
Solving (2), (3) and (4), we have the vertex
(fl, 0, 0).
Similarly solving (1), (3), (4) and (1), (2), (4) we have the
vertices B (0, b, 0) and C (0, ^ c) re.speciively.
Therefore the sphere circumscribing.the tetrahedron OABC is
the sphere passing through the four points O (0, 0, 0), A (a, 0, 0),
B (0, b, 0) and C (0, 0, c). Now proceed as in Ex. 6(a) above.
Ex. 7. Find the equation uf the sphere circumscribing Ihe
tetrahedron wliose faces are
arc
J'/^+^/c=0, 2/c-l-x/a=C, .x/a-l-y/h=0, x',a-\-y!b-\-zlc^-- 1.
Sol. The equations of the planes {i.c. faces of the tetrahedron]
...(1)
z/c-l-x/fl=0,
...(2)
...(-3)
x'ia^-ylb+zlc=\.
. ..(4)
Any three faces {i.e. planes) meet at a vertex of the tetrahed●«/a+.v;h=o,
249
The Sphere
ron. Hence, solving (1), (2),(3), we get the vertex O (0, 0, 0)and
solving (1),(2),(4), we get the vertex (a, b, —c).
Similarly solving other two triplets of the equations of the
planes the remaining two vertices are {a, - b, c) and (- a, b, c).
Hence the required sphere passes through the four points
(0, 0, 0),(a,
c),(a, -A,c) and (-a, b, cj.
Let the equation of the sphere be
...(5)
.v2 -h y^+z^+lux+2V v+Iwz4- rf-=0.
If(5) passes through (0, 0, 0), we have d=0. Hence (5) becomes
..(6)
2vj;+2wz=0.
If(6) passes through (a, b, —c), (a, b, c) and (—c, by c),
we have
...(7)
a®-!+2t/a+2r/>—2wc=0,
...(8)
+b^+c*H-2ua -2vb+2wc—0,
and
...(9)
a^-\- b^+c^ - 2«a-|-2>'^+ 2ivc=0.
Adding (7) and (8),
2(a*-f6Hc2)+4M«=0 or 2m=-'(a*+6H-c2).a.
Similarlv, we have
2v= -~{aHb^-^c^)lh. 2w=-^(a^+l^+c^)lc.
Substituting the values of //, v, h» and d in (5), the equation of
the required sphere is given by
(a2
|
6*+c2) x-i\!b)(a^-^b^+c^)y
~(!/c)
z=-0
or
:V“+^4-2*
d'‘+b^+c
.V
-_^-L=o.
a
b c
Ex. 8. Find the equation of the sphere which passes through the
points(1, - 3, 4),(1, —5, 2),(1,-3,0) and whose centre lies on
[Jodhpur
19781
the plane x+;y+z=0.
1
Sol. Let the equation of the sphere be given by
z^+2ux+2vy -1- 2H’r+ rf—0.
■..(1)
If(1) passes through (1, —3,4), then we have
1+9+16-b 2w—6v4-8h-+«/-=0.
-(2)
If(1) passes through (1,—5, 2), then we have
l+25+4+2tt- l0v4-4iv4</=0.
-(3)
If(1) passes through (1, --3, 0), then we have
1 +9-{.o+2M-6v4-0+rf=0.
.. (4)
The co-ordinates of the centre of the sphere (1) are (—m, — v,
—w) and if it lies on the plane .y+j'4 z-~Qy then we have
—M—V—»t’ —0.
● (5)
.250
Analytical Geometry 3-i)
Subtracting (4) from (2),
164-8m»=0 or w=—2.
Subtracting (3) from (2),
--4+4v+4»v=0. or -l+v+w—O.
...(6)
Putting the value of w in (6),
--1+V—2=0 or v=0.
Adding (5) and (6), — 1 —m=0 or «=—1.
Putting the values of u and v in (4), </=10.
Substituting the values of u, v, w and d in (1), the equation
of the required sphere is given by
-2x+6;;—4r+10=0.
Ex. 9. Find the equation of the sphere having its centre on the
line 2x—.Zy=0=5y-{-2z and passing through the points (0, - 2, —4)
n«rf(2, — 1, — 1).
[Agra 1977 ; Lucknow 81]
Sol. Let the equation of the sphere be given by
^*+.v*+^*+2mjc4-2v;>+2wzH-rf=0.
...0)
If(1) passes through (0, -2, —4), we get
0+44-16+0—4v — 8»v+</=0
...(2)
If(1) passes through (2, — J, —1), we get
4+1+1+4«—2v—2w+</=0.
. ●..(3)
The co-ordinates of the centre of the sphere (1) are(—«, - v,
—w)and if it lies on the line 2x—3>>=0=5j>+2z, then we have
-2«+3v=0
...(4)
and
-5v-2»v=0.
...(5)
Subtracting (2) from (3),
4«+2v+6w—14=0
or
2m+v+3w=7.
...(6)
Putting the values of u and w from (4) and (5) in (6), we get
3v+v-(15/2) v=7 or v=-2.
Putting v= — 2 in (4) and (5), we get u= —3, w=5.
Putting the values of «, v, w in (2), we get d= 12.
Substituting the values of m, v, w and d in (1), the required
equation of the sphere is
Jc2+J'®+z"-6x-4;;+10z+12=0.
Ex. 10. A plane passes through a fixed point {p^ q, r) and cuts
the axes in A, B, C. . Show that the locus of the centre of the sphere
OABC is
Plx+qly+r’iz=2.
[Allahabad 1981; Bundelkhand 78; Kanpur 30, 82;
Liicknow 80; Meerut 83, 83P, 84S, 86, 86S; 89S
Ranchi 78; Rohilkhand 77, 80; Punjab 77j
151
The Sphere
SoK Let the equation of the plane be
...(1)
The piano(1) meets the eo-ordinate axes in A, B, C. Hence
the co-ordinates of B and C are (a, 0, 0),(0, b,0) and (0,0, c)
respectively. The equation of the sphere OABC is
[See Ex. 6(a) above]
x^+y^+z^—ax—by—cz=0.
If its centre be (a, j9, y), then a=^A,
...(2)
/.
b=^2fi, c^2y.
It is given that the plane (1) passes through the fixed point
ip* Q* 0* Hence we have
pla4-qlb+rlc=:\
[Putting the values from (2)]
or
p/(2«)+^/(2i?)+r/(2y)=l
or
pla-\-ql^+rly=2.
The locus of the centre (a, /5, y) \s pjx+qly+r/z=2.
Ex. 11. A sphere of radius k passes through the origin and
meets the axes in A, B, C. Show that the centroid of the triangle
ABC lies on the sphere 9 {x^+y^+z^)=Ak*.
[Allahabad 1980; Avadh 82; Meerut 84P, 86; Indore 78;
Kanpur 79, 83; Nagpur 77]
Sol. Let (a,0, 0),(0, b, 0) and (0, 0, c) be the co-ordinates
of the points <4, 5 and C respectively. The equation of the sphere
OABC is [See Ex. 6(a) above]
...(1)
x^+y^+z^'-ax-by—cz=0.
The radius of the sphere
so that
(given).
Squaring, we get
...(2)
Let(a,
y) be the centroid of the /^ABCy so that we have
(fl+0+0),
(0+6+0), y=i(0+0+c).
AssSa, 6=3p, c=3y.
Substituting the values of a, 6, c in (2), we have
9a2+9^2+9y*=4/:8.
The locus of the centroid (a, /J, y) is 9 {x^-\-y^-\-z^)=4kK
Ex. 12. A sphere of constant, radius 2k passes through origin
and meets the axes in A^ B and C. Prove that the locus of the cen
troid of the tetrahedron OABC is
[Meerut 1984, 86P]
x^+y^-\-z^=k^.
252
Analytical Geometry 3-/)
Sol. The equation of the sphere OABC is [See Ex. 11 above]
x*+y^’\'Z^—ax-^by—cz—Q,
...(1)
The rad,us of
=
(given).
Squarinc, we gel
a^+.b^+C^^\6k^.
...(2)
Let (a, }») be the centroid of the tetrahedron OABC.
Then
i(O+a+0+0) or a=4a.
Similarly
b-=4fi and c=4y.
Substituting the values off/, h, c in (2), we have
^ (4a)"+(4p)H(4y)*--=I6A:2 or
*. The required locus of the centroid («,
y) IiS
Ex. 13. Find the equation of the sphere with centre at (2, 3,
—4)and touching the plane
2x+6y-32+l5=0.
Sol. The point (2,3, -4) is given as the centre of the
sphere. Since the sphere touche.s the plane
2x+6j;-32+15=0,
therefore the radius of the sphere=the length of the perpendicular
from the centre (2, 3, ~A) to the plane (1)
2.2-H6.3 3.(~4)415 49
^7.
■"v/{(2)=*-h(6)S-(—3)2)
■=
The equation of the required sphere is
(X— 2)2 riy-i ;2 -i- (> + 4)2 = (7)2
or
^‘M-v24-2^-4x—6v-f 8r - 20=0.
Ex. 14. Obtain the equation of the sphere which passes through
(4, 1, 0), (2,
3, 4), (1,0. 0) and touches the plane
2x+ly ~z=U. .
. .
.. .
(Avadh 1982; Garhwal 78J
Sol. ,
Let the equation of the sphere be given by
x^Yy^^2^+lux-\-'h>y-\-2ivz+d={).
The sphere (1) passes through the points (4, 1 0)
. > 4’»
and (1, 0, 0), therefore, we have
’
t6+H-u+8«f 2v+04f/=0
4+9-f I6+4m—6v+8M'4rf=0
1-h 0-f 0 4 2m 4-0+0+ 0.
From (4)^
From (2),
or
Front (3),
●●●(2)
(3)
-(4)
«=(<fj. I).
(5)
2v=-8M-f/- 17=4f/-j-4-<f- 17,
using (5)
(3fi( -13).
...(6)
8
— 4m 4 - 6V—f/~ 29=2i/-f 2+9^/~ ^9—d^ 29.
using (5) and (6)
The Sphere
or
253
w=i (5d-33).
Now it is given that the sphere (1) touches the plane
2jc+2j;-z=11.
,..(7)
...(8)
The length of perpendicular from the centre (—«,— w)to the plane (8)—-the radius of the sphere (1)
or
or
or
or
or
or
V,
2(-»)^2(-V) -( -w) —11
-/|(2)2+(2)2+( _ 1)2) -V(M‘4-v +w ~d).
Squaring,
i-2u -2v-\-w^\\Y=9 {u^-\-v^+w^-d). ...(9)
Putting the values oft/, v, ir from (5),(6),(7), we get
{</+I-3</+13-{-i (5</--33)—41}2=.9 {d+\;^+\(3J-13)*
+(1/16)(5t/~33)a-//}
(l/:6){- 3//-21)2=(9/!6){4 (//+l)2-i-4 (3//-!3)2
+{5d-33f-Ud}
(//+7)2=4(//H2//+l;+4(9ti“ -78t/ rl69)
+(25//2-330//+1089)-16rf
. 64J2—664//+1720=0 or 8//*-83</+2|5=0
Sd^-40d . 43//+215--0 or (d—5)(8//-43)=0
</=5, 43/8.
When </=5, the equations (5),(6) and (7) yield t/=
»v=-2.
3, v= 1,
Substituting the values of u, v, u* and d in (1), the equation of
the required sphere is
x2+,v2+z*-6x+2v- 4Z+5-0.
Similarly when t/=43/8, find the corresponding equation of
the sphere as above.
Ex. 15. A sphere ofconstant radius r passes through the origin
O and cuts the axes in A, B, C. Find the locus of the foot of the
perpendicularfrom O to the plane ABC.
[Allahabad 1975, 79; Kanpur 81; Meerut 73, 83 S, 87, 89, 90
Rohilkhand 78, 79, 81; Nagpur 87]
Sol Let (a. 0, 0»,(0, 6, 0) and (0, 0, c) be the co-ordinates
of the points A, B and C respectively. The equation of the sphere
(7/4.BC is [See Sx. 6 (ai) above]
ax-by-cz=^.
...(I)
Clearly the intercepts on the co-ordinate axes arc //, b, c and
so the equation of the plane ABQ is
x/a+>;/6+z/c=l.
The radius r of the sphere (1) is given by
or n“+bHc2=4f".
...(2)
..+3)
254
Analytical Geometry 3-D
Now the equations of the line through the origin and perpen
dicular to the plane (2) are
;xr-0 y-0 z-0 . : .
1/a “
...(4)
The CO ordinates of any point bn the line (4) are (A/a, A/ft,
A/c). If this is the foot, say (a,% y), of the perpendicular from
the origin to the plane (2), then we have
A/a=a,^/b=p^ Xlc=y or a=A/a, ft=A//3, c=X[y, ...(5)
Since the foot of the perpendicular lies on the plane (2), we
have
...(6)
a/a+p/ft-hy/c=l.
Now putting the values of a, ft,c from (5) in (3) and (6),
we get
..(7)
A2(l/a*+l/jS*+l/y^=4r2
and
...(8)
(l/A)(a2+i92+y2)=l.
Eliminating A between(7) and (8), we have
The locus of the foot of the perpendicular fa, jS, y) is
Ans.
(j:H>>®+z2)2(xr^-\-)T^+z~^)=4r\
Ex. 16. Find the equation of the sphere which passes through
the points (1,0,0), (0, 1,0) and (0,0, 1) and has its radius as
small as possible.'
Sol. Let the equation of the sphere be given by
x^+y^+z^+2ux4-2vy-\-2wz+X=0.
...(1)
If(1) passes through the three given points (1,0, 0), (0, 1, 0)
and (0,0, 1), we have
H-2w-fA=0, H-2v+A=0, l+2w-}^A=0.
Solving, we have
m=v=m'=-i(A+l).
...(2)
Let r be the radius of the sphere (I), so.that we have
A=p*(say)
or
/*=r2=^(A+l)a-A.
[Putting the values frbm (2)]
.nr
,^^^Pt1^radius r is minittitim, then r* i.e. p is minimum. The
necessary condition for p to be minimum is
dpfdX=0 or (3/4).2(A+l)-l-0 or A= -i.
^p 3
-j-ve at A=-ii
Hence p is minimum at A=—
Putting A=—i in (2), we have M=y=H*— — J
255
The Sphere
Substituting the values in (I), the equation of the required
sphere is given by
-f(:x+;'+z)-i=0
or
3
(;c+:y+z)-l=0.
Ex. 17. A point moves such that the sum of the squares of its
distancesfrom a given number of points is constant. Prove that the
locus of this point is a sphere and show that its centre is the centroid
of the given points.
Sol. Let (xr, z,)y r=l, 2, ..., n be the co-ordinates of n
given point. Let(x, >>, z) be the moving point. It is given that the
sum of the squares ofthe distances of(x, >>, z)from n givsn points
(●^r, Pn Zf) is constant, say k. so that we have
S {(x-X,)*-f(j>-Vr)" + (z - Zr)*}=fc
or
n (x^+yj+z*)- lx E x,—ly E y, ~2z Z Zr
y,^^-Z Z,*-A:=0.
Divi ing by n, the locus of (x, y, z) is given by
x«+^+z>-2x(%)-2y{?^)-2z (%)
which is clearly the equation of a sphere the co-ordinates of whose
centre are
\ and evidently the centre is the centn
n
n f
Hence proved.
roid of the given points (x„ y,. ^r). ''=1. 2, . .,n.
Ex. 18. A point moves so that the sum of the squares of its
distances from the six faces of a cube is constant. Prove that its
locus is a sphere.
(Meerut 1979)
Sol. Consider a cube with the vertex O as origin and edges
OA^ OB and OC as co-ordinates axes. Let the length of each
edge be a.
The six faces of the cube OBDCy OAECt OB FA, AFHE,
BFHD, DHEC and their equations are
x=0. y=0, z=0, x=o, y=a, z=a respectively.
Let P («, /?, y) be the moving point. It is given that the sum
of the squares of the distances of P (a,
y) from the six faces of
the cube is constant, say k, so that we have
2a
£
+'
V(i*) [
*1
r
+ vn^)i
y-^V-k
256
Analytical Geometry 3-Z)
Z
7T
H
P
P
/
Y
or
2(aH/5H?=*)-2a (a+3+yH-(3a*-fe)=0.
The locus of P(a, p, y) is
2rjc*4-jH2*)-2a{x+.v+z)+(3a*-ifc)=0
which IS clearly the equation of a sphere.
Ex. 19. 0/4, OB, OC are three mutually perpendicular lines
t rough the origin and their direction cosines are ly, nii, «i; /j, m2,
a> m3,/}g. If OA^a, OB=b, OC~c, prove that the equation of
the sphere OABC is
’\-y^-\-z^~x_ (a/i+A/3+c/g) -y (anil+bm2+cwig)
r (a/?i+A«a4-cn3)-0.
Sol. The co-ordinates of the points ,0. A, B and C are
tiye?*
respec-
Since the sphere passes through- the origin O, its equation
can be put as
x^+y^-hz<‘j-2ux+2vy+2wz=0.
or
-.(1)
Substituting the co-ordinates ofv4 (I), we get
a^ (li^4 mi^+ni^)+2a (ulj-\-vmi+wnj)—0
a+2u/i+2vmj-^2wni=d.
...(2)
[V /iHmiHwi^-l].
Similarly substituting the co-ordinates of 5 and C in (1), we
get
and
b -I- 2M/a -I- 2v/Ma-i- 2M7it ■= t K
...(3)
c f-2i/,-|-2i7»,'1h:»v/3«.0.
...(4)
257
The Sphere
Now 7i, /a, /a; fni,
m»; «i, w». "s are the d.c.’s of OX, OY,
OZ respectively with reference to the three mutually perpendicular
given lines.
/ia+/a*+/a?=mi*+ma*+m8*=ni*+«a*+»8*=l.
and /i/Mi+/aWa+/aWa~ WiWa+WaWa+Ws^s-/iWi+/»"*+^8«8=0.
Multiplying (2), (3) and (4) by /i, h
adding and using the above relations, we get
h respectively,
fl/i+b/a+cf3+2tt*='0 or 2u= —(fl/i+Wa+c/a).
Similarly multiplying (2),(3) and (4) by mi. ma and m,. and
then by /ii, fia and Wa, and proceeding as above, we get
2v=»—(flmirffcma+cma), 2w« —(<wii+b»8+^»a)Putting the values of 2«, 2v and 2w in (1), we.get the equation
of the sphere as given.
Ex. 20. A plane passes through a fixed point (a, b, c), show
that the locus of the foot of the perpendicular to it from the origin is
the sphere OABC,
x*+y^+z*—ax’—by-cz>=‘0,
Sol. The equation of any plane through(a, b, c) is
/(x—flH-m
c)=0.
...(1)
The equations of any line through the origin and perpendi
cular to the plane (1) i e. the equations of any line through the
origin and paraild to the normal of the plane (1) whose d.r.’s are
/, m, n are given by
...(2)
xfl—yfm=>zfn.
The line (2) intersects the plane (1) in the.foot of the perpen
dicular and hence the locus of the foot of the perpendicular is
obtained by eliminating /, m, n between (I) and (2), Therefore
the required locus is given by
X ix—a)+y(y~-b)+z(z-c)^0
or
x^-^y*-\-z^-ax by—cz=^
which is the equation of the sphere OABC, [See Ex. 6(a) above].
§3. Plane. sectidh of a sphere. To prove that the sectlhn
of arspher^y a plane is a circle and to find its rad^ and the
centre.
V
258
Analytical Geometry 3^D
Let us consider,the section of
a sphere by a plane and let P be any
point on this section. Let O be the
centre of the sphere. From O draw
OC perpendicular to the plane so
that C is the foot of the perpendi
cular and is a 6xed point. Join C
to P. Clearly CP is a line in the plane and hence CP is perpendi
cular to OC. (Note that OC is perpendicular l(to the plane and
therefore, OC is perpendicular to every line.lying in the plane).
Thus we have
But
and
Opa=»OC*+CP® or CP=>y/(OP^~OC*).
...(1)
OP«=»the radius of the sphere and is constant
OC=the length of the perpendicular from the centre
of the given sphere to the given plane and hence
is constant.
Thus OP and OC are constant and hence from (I) it follows
that CP is constant for all positions of the point P and hence the
locus of the point P is a circle with centre C and radius CP, which Is
given byi\).
Therefore, Ifa sphere is cut by a plane then the plane section of
the sphere is a circle.
Theeqoations of a circle. Let the equation of the sphere be
. ^*+J'*+z*+2MX+2v;>+2wz-|-rf=0
...(2)
and the equation of a plane be lx-\-my-\-nz^p.
...(3)
We have shown above that the section of a sphere by a plane
is a circle. Hence the equations(2)and (3) together represent the
equations of a circle.
Great Circle. A great circle is the section of a sphere ,by a
plane passing through the centre of the sphere. Clearly the centre
and the radius of a great circle are the same as those of the
sphere.
§ 4. Inlerseciion of two spheres.
.2'o prove that the curve of intersection of two sphere is a circle.
Let the equations of the two spheres be
5ia**+J»H2*+2wiX+2viy+2w,z-l-rfi«0
and
St^x*-{-y*’irZ*+2ttiX+2viy-{r2w»zi-d»^0.
The Sphere
259
The equatioD of the plaoe which contains all the common
points of the spheres Si=0 and Sa=0 is given by
(M|-Ma) Jt+2(vi-va) y^2 (w,-^Ws)
Thus the points of intersection of the two spheres are the
same as those of the intersection of any one of the two spheres
and the plane Si-Si=Q and hence these points lie on a circle.
Hence the curve of intersection of the two spheres is a circlet
and the two equations of the spheres together represent the equations
of a circle,
§ 5. The system of spheres through a given circle.
Let the equations of a circle be given by
...(1)
...(2)
and
P=lx+my+nz—p=0,
...(3)
Consider the equation S+AP=0
or
x*+
+2’mx+2vy-1-2wz+rf+A(/JC+wy+«2-p)=0
where. A is a constant. The equation iS-l-Ai*=0 for all values of A
is satisfied by ail those points which satisfy both S=0 and P=0
i.e. which lie on the circle 5=.0*=P.
Also the equation 5+AP«0 is of second degree in which the
coefiScients of x*» y^, z* are equal and the terms containing yz,zx,
xyare absent.
Here the equation 5+AP=0 represents a system of spheres
passing through the given circle whose equations are given by(1)
and (2).
Similarly the equation iSi+ ASs^^O represents a system of
spheres,through the circle of intersection of the two spheres
i9i=0 and ^a^O.
Remark, The value of A in § 5 above is deterniined by an
additional given condition.
i^OLVED EXAMPLES(B)
Ex. 1. Find the radius of the circle given by the equations
3**+3y*+3z*+x-5y-2=0, x+y==2. (Calcutta 1973,81)
Sol. The equations of the given circle are
and
...(2)
x+y=2.
The centre of the sphere (1) is the point O(—i,f, 0)
and radius OP=-\/{(—ii*+(6)*+(0)‘—(—1;)[See figure of § 3J
or
OP« 5/(3V2).
260
Analytical Geometry 3*Z)
Now OC==the leogtb of the perpendicular from the centre
^(—f.0) to the plane (2)
-1/6+5/6-2
4
A
The radius of the circle=C/>«
OC>)
Ex. 2. Find the radius and centre of the circle of Intersection of
the sphere
2;>—4z^=« 11 and the plane x~\‘7y-\-2z:^\5.
(Agra 1976; Bundelkband 78; Madras 74; Meerut 84 S, 85(S)
Sol. The equations of the given circle are
x*-\-y^+z^-2y~^4z-U^O.
● .(1)
and
x-\~2y\-2.z—15«=0.
...(2)
See figure of § 3 above. The centre of the sphere (1) is the
point O (0, 1, 2) and its radius
. O/»=V{(0)>+(l)*+(2)»-(-n)}=4.
Now OC=the length of the perpendicular from the centre
0(0. 1,2) to the plane (2)
“9
04-2.1 4.2.2—I
t=i-j =3(numerically).
.A The radius of the circle
=CP--.V((?P*--OC2)= ^(16-9)V7.
To find (he co ordinates of the centre C of the circle. The line
OC being perpendicular to the plane (2), is parallel to the normal
of the plane (2) and hence d.r.’s of the line OCare 1, 2. 2.
.A The equations of the line OC i.e„ the equations of the
line, passing through O (0, 1. 2) and having d.r.’s 1, 2, 2 are
*-0 .F-1 z-2
, \
—
(say).
1
2
●● (3)
Any point of(3) is (r. 2r-f l, 2r+2). If this is the foot of
the perpendicular U.e. the centre C of the circle) then it must lie
on the plane (2), so that we have
r +2.(2r-M)-}-2(2r+2)-I5=*0or r=J.
Putting the value of r, the co-ordinates of the centre C of the
circle are (1, 3, 4).
Ex. 3. If r is the radius of the circle
x^-\-y^-\-7^4r2ux \-lvy-\-2wz-\-d-t.% lx-\-my:\-nzr=0.
prove that
(RoMikhand 1982; beihi 75;|
261
The Sphere
Sol. The centie of the given sphere is the point
O(—M, —V, —w)and its radius OP
[See 6gure of § 3]
Now OC=the length of the perpendicular from the centre
O(“U, —V, —w)to the plane /x
+«z*=»o
f(~u)+w
(-H>) /M+mv+/iw
(numerically).
The radius r of the circle is given by
; f/i4+mv+nw)*
ra=,OP*-OC®=(M*+v*+w2-d)Of
or
(r*+d)(P+mHn*)«(«‘+vHw*)(/»+»*»+«*) |
—{lu^mv^nwY
(rHd)(/8+mH«*)=’(w»v-«v)*+(nM-/w)*+(/v-/wM)*,
using Lagrange’s identity.
Ex. 4. Find the equation of the sphere through the circle
x^-\-y^+z*=9,
2z+4e=iG
(Lucknow 1978)
and the origin.
Sol. The equation of any sphere through the given circle is
...(1)
+2®-9)+A(
-2z+4)«=»0.
If the sphere (1) also passes through the origin (0, 0, 0), then
we have
(0+0+0-9)+A (0+0-0+4)=0 or;A=9/4.
Putting the value of A in (1), the equation of the required
sphere is given by
x»+y+z*-9+(9/4)(x+3^-2z+4)«0
or
4(x«+>»*+z*)+9x+9;^-18^='0.
Ex. 5 (a). Find the equation of the sphere which passes through .
the point (a, y) and the circle
z«0.
(Agra 1980; Gorakhpur 82; Meerut 71,83,8SP;
Kanpur 77,81; Panjab 78)
Sol. The equations of the given circle are
Jc*+y*t=o*, z=0.
Xhe-above-equatiorts-otHhe-eircle may be rewritten by intro
ducing the term z* as follows
...(1)
x®+)'*+z®=a*, z=0.
The equation of any sphere through the circle (1) is
...(2)
X*+y*+2®—fl®+Az«=*0.
If the sphere (2) passes through the point (a, y), we have
a’^j38+ya«fla^Ay*=0 or A«—
162
Analytical Geometry 3-/)
Putting this value of A in (2), the equation of the required
sphere is given by
(x*'-\-y^+z*-a*) y-(a.^+P\+y^-a^) r=0.
Ex. 5 (b).
Find the equation of the sphere through the circle
and through the centre of the sphere
(Nagpur 1977)
(x-oiy-¥(y-Pf+{z-y)^^r\
Sol. The co-ordinates of the centre of the given sphere are
(«* Pr y)> Hence it is the same problem as Ex. 5(a) above.
Ex. 6. A circle, centre (2. 3, 0) and radius 1, is drawn in the
plane z=0. Find the equation of the sphere which passes through
this circle and the point(1, 1, 1).
(Lucknow 1979)
Sol. The equations of the circle of radius 1, in the z^O
plane aind with centre (2, 3,0) are given by
(x-2)H(>'~3)8=1^zc=0.
These on introducing the term of z* can be rewritten as
(X-2)»+(y-3)*+2*=l.z-0.
...(1)
The equation of any sphere through the circle (1) is
(x-2)H(>'“3)a+2®-1+Az=0.
. .(2)
If (2) passes through (1, 1, 1), we have A=—5.
Putting the value of A in (2), the equation of the required
sphere is
(x-2)8+(j'-3)Hz*-1-5z=0
!or
xH;V»+2V4x-6y-5z-l-12=0.
Ex. 7. Find the equations of the spheres through the circle
2x+4y+5z=6 and touching the plane z=0.
(Avadh 1979; Meerut 77,81)
Sol. The equations of the given circle are
2.«+4><+5z-6=0.
...(1)
The equation of any sphere through the given circle (1) is
given by
(X*+)>*+2*-1)+A (2x+4y+5z-6)=0
or
x“+.y*+2®+2Ax+4A>'+5Az-(6AH-1)=0.
...(2)
The centre of(2) is(—A, —2A, —|A)and radius
=y|A>+4A>+^ +(6A +1)|=§v'(45A»+24A+4).
Now if the sphere (2) touches the plane z=0, then we have
the length of the perpendicular trom the centre (—A, —2A, —|A)
to the plane(z»0)athe radius of the sphere (2) /.e.,
(--5A/2)
iV(45A>+24A+4).
V(i*)
The Sphere
263
25A»«45A*+24A+4
SquMog, we have
20A*+24A+4=0, or 5A*+6A+1=*0, or A«-l,
Since we have two valhes of A and so we shall have two
spheres satisfying the given conditions. Now substituting these
values of A one by one in (2), the equations of the required spheres
are given by
z*-2x—4;>-52+5=0
and
5(x*+j>*+2*)—2x—4j>—5z+1=0.
Ex. 8. Find the equations of the spheres which pass through
the circle x*+>>‘+2*=*5. x+2y+3z=3 and touch the plane
4x+3j'=»i5.
(Agra 1981; Kanpur 79; Lucknow 82; Nagpur 78;
Bundelkband 78; Kurukshetra 76)
or
or
Sol. The equation of any sphere through the given circle *8
x*+;^*+2*-5+A (x+2;y+32-3)«0
...(1)
x*+.V*+2HAx+2A;;+3A2-(3A+5)=0.
The centre of the sphere (1) is (-iA,,-A, -fA)and its radius
“V{(-iA)>+(-A)*+(-tA)*+(3Af5)}-iA/(i4A*+12A+20).
Now if the sphere (I) touches the plane
...(2)
. 4x+3j;—15=0,.
then we have the length of the perpendicular from the centre of
the sphere (1) to the plane (2)=the radius of the sphere(1)
or
or
^^^=^^^'=iV(14A»+I2A+20,
2(-5A-15)=5v'(14A*+12A+20).
Squaring, 100(A*+6A+9)=25(14A*+12A+20)
or
5A*-6A-8=0 or A=2, -4/5.
Substituting these values of A one by one in (1), the equations
of the required spheres are given by
x*+>>*+z*+2x+4;>+62—11«0
and
5 (x*+3^*+2®)—4x-8;>-12z—13=0.
Ex.9. Prove that the circles
**+>*+2“*-2x+3>>+42—5=0,5j>+62+1»0;
jca^_^a^2i_3x—4j>+3z—6=0,x+2>—72=0
lie on the same sphere andfind its equation. Alsofind the value *d*
for which x+y+z=aV3 touches the sphere,
(Meerut I982S; Robilkhand 81; Garbwal 82; Kanpur 78;
Madras.77; Puujab 78);
264
Analytical Geometry 3-D
Sol. The equations of any spheres through the given circles
are respectively given by
(5;^+6z+l)=.0. ...(1)
and
x»+>M-2r*-3*-4;;+5z~6+Aa(*+2;»~7z)c=o. ...(2)
Now if the given circles lie on the same sphere, then the
equations (1) and (2), for some values of Aa and Aa, must represent
the same sphere. Hence comparing the coefficients of oc, y,z and
constant terms in the equations(1) and (2)^ we get
—2«=»—3-f Aa
3+5Aa=-4+2Aa;
...(4)
4+6Aa=5-7Aa
...(6)
●●●(5);
—5-l-Aa=3^—6.
Solving (3) and (6), we get Ai=-1, Aa« l.
Clearly the equations (4) aud (5) are satisfied by these values
of Ai and Aa and hence the given circles lie on the same sphere.
Putting Aaa — 1 in (1) [or Aa*»l in (2)], the. equation of the
required sphere is given by
**'+3'«+z»-2*~2;^-^2z-6=0.
...(7)
The centre of the sphere (7) is (1, 1,1) and its radius=3.
The equation of the given plane is
x+y+z»=^ay/3.
...(8)
The plane (8) will touch the sphere (7>, if the length of the
perpendicular from the centre (1,1,1) to the plane (8)
the
radius of the sphere (7).
. .
l-H + l-flV3
^±3 or fl=\/3±3.
Ana. .
** V{(1)*+(1)*+(1)*}
Ex, 10, Prove that the sphere
5i=^^+)'*+2‘+2tfjX-|-2vij»+2»ViZ+<ii=>0
cuts the sphere
^*=^*+3'*+^*+2tfaX+2v8y+2H>a2+</a=0 in a great circle’ if
2 («a*+v**-f->Va*)—</a=>2 (UiMa+ViVa+WiM>a)—</i
or if
2 («i«a+ViVa+WiW8)=2ra*+di+</a
where r^ is the radius of the second sphere 5a**>0.
Sol. The equation of the plane through the circle of inter
section of the given spheres Sk-sO and 5a«=»0 is 5jr-5a=0,
Le,
2 (Mi-i#a) *+2 (Vi-va) 3'+2 (Wj-Wa)
...(1)
If the sphere
cuts the. sphere 5a >=>0 in a great circle,
then the plaie (1) will pass through the centre (-n*. -v,, ~iv,)
of the sphere 5aa0, so that we have
2 (Ui—«a) (—a*)+2 [Vi—Va) (—Va)-f-2 (Wi—Wg) (—Wa)
rfacsO.
or
.2 (tt8*+Va‘+Wa*)—<^«2 (WiWa+ViVa-l-iViWa)—di.
...(2)
265
The Sphere
Agaio f2 is the radius of the sphere 5a“O, hence
.8
ra"=
using (2)
or
ra*« J {2(UiUa + ViVa-fWiWa)-rfi+^a}
..(3)
or
2(MiMa+ViVa+ H'iH’a)==2ra^'t-</i+</a‘
The relations (2) and (3) are the required conditions.
Ex. 11. Find the equation of a sphere for which the circle
x^-\-y^+z^+7y--2z+2^0, 2;c+3;;+4z-^8
[Garhwal 19811
is a great circle,
Sol. The equation of a sphere through the given circle is
(x»+;;*+2*+7>'-~2z+2)+A(2«+3>'+42-8)«0
or ;c>+>;*+za-i-2AA:+(7+3A) A—2(1--2A)z+2~8Ac=0. ...(1)
The centre of the sphere (1) is
(-A. ^H7+3A). 1-2A).
If the given circle is a great circle of the sphere (i), then the
centre of the sphere (1) will lie on the plane 2x+3y+4z==3 and
hence we have
2(~A)+3{-i(7+3A)}+4(-2A)=8 or A= -l.
Putting this value of A in (1), the equation of the required
sphere is given by
x*+y*+z*-2x-{-Ay—6z+\0^0,
Ex. 12. Prove that the plane jc+2;;—z=4 cuts the sphere
x*^y»-^z*--x+z-2^0 in a circle of radius unity and find the
equation of the sphere which has this circle for one of Its great
circles,
[Gorakhpur 1975; Madras 76]
Sol. The equation of the given sphere is
2*=0
...(1)
and the equation of the given plane is
...(2)
x-i-2y — z=^4.
Refer figure of § 3. The centre of the sphere (1) is the point
O (*. 0, -i)and its radius OP=V{(4)'+0+(-i)*+2} = V(t)Now OC*=the length of the perpendicular from the centre O
to the plane (2)
«__i±2±izi
V{(2)*+(2)®+(-m
A
(1)
, numerically^
vfi >/\2/
The radius of the circle
«=CP=\/(OP»-GC*)
This proves the 1st part of the problem.
For the second part proceed exactly as in Ex. 11 above. The
equation of the required sphere is given by
166
Analytkal Geometry
x*+y^^z*—2(x+y—z—l)t=»0.
Ex. 13. Find the equation of the sphere passing through the
circles
y^-^z^>=9t xv=>4.andy*+z^^36t x=l.
[Ponjab 1975]
Sol. The equations of the given circles are
:v»+2*=9,x=4 ...(1) and y+z«c=.36.:c=l ...(2)
Introducing the terms of x', these equations can be written as
je>+J'*+z*=25, x=4
...(3)
and
**+JV*+2’'=37.x=1.
...(4)
We are to find the equation of a sphere passing through the
circles (3) and (4.), or in other words to find the equation of a
sphere so that the circles (3) and (4) lie on this sphere;
The equations of any spheres through the circles (3)and (4)
are respectively given by
**+yH2*-25+A,(x-4)=0
●●(5)
and
...(6)
x*+;>>+2®-37+Aa(x-1)=0,
Now if the given circles lie on the same sphere, the equations
(5) and (6), for some values of Ai and Aa, must represent the same
sphere. Hence comparing the coeflScients of and constant terms,
in equations(S)and (6), we get
A,=Aaand-4A,~25=-Aa-37.
Solving, we get Ai=Aa^4.
Putting the value of Ai in (5) [or that of Aa in (6)], the requi
red equation is
Ac*+.y®+2H4x--41c=.0.
Ex. 14. Find the equations to the circle whose centre is (a, /3,
y) and which iiea on the sphere x*+j>*+2*e=o*.
Sol. The equation of the given sphere is
...(1)
The centre of the sphere (1) is the point O (0, 0, 0) and the
centre of the circle is given to be the point C (a, j8, y).
.*. The d.r.*s of the line OC are a—0, /3-0, y—0 or a, j8, y.
The equation of the plane through C (a, j3, y) and perpendi
cular to OC is
«(x—a)+j8 (y-/3)+y (z—y)«0
or
ax 4* A)'+y^'=*
+y*»
...(2)
The equations (1) and (2) are'the required equations of the
circle.
The Sphere
26t
Ex. 15. A variable plane Is parallel to the given plane xja
+ylb+z/c=»6 and meets the axes In A, B^C respectively. Prove
that circle ABC lies on the cone
yz {blc-\-cfb)+zx (cla^a/c)-{-xy (a/b+b/a)^0.
(Agra 1975; Allahabad 76. 82; Bandelkhand 79; Kanpnr 83;
Lacknow 76; Madras 77; Meerut 74, 89, 90, 90P; Rajasthan 78)
Sol. The equation of any plane parallel to the given plane
...(1)
xfa-\-ylb-^zjc=0 is x/fl+y/h+z/ci=A.
The plane (1) meets the co-ordinate axes in the points id, R
and C whose co-ordinates are (oA. 0,()),(0, bX, 0)and (0,0,cA)
respectively. The equation of the sphere OABC is
+z® -flAjc—hA;y-cAz=0
[See Ex. 6(a) of Solved Examples (A)]
or
...(2)
x^-\-y^-\-z^—\{ax^by fcz)==0.
The equations (1) and (2) together represent the circle ABC,
Eliminating the variable A between (1) and (2), the required
locus is given by.
(x/n+.y/h +z/c)(ax-\-by-{-cz)=‘0
or
yz(b}c-\-c}b)+zx (c/fl+fl/c)-fxy (a/6+^>/fl)=0.
Ex. 16. P is a variable point on a given line and A, B,C are
its projections on the axes. Show that the sphere OABC passes
through afixed circle. (Allahabad 1982; Kanpur 79; Lucknow 77)
Sol. Let the equations of a given line be
..^(1)
(X-ayi=^(y-P){m=‘(z-y)fn=r (say).
The co-ordinates of any point P on (1) are (/r+«» wr-l-j5,
nr+y). The projections of the point P on the axes are A^B and C
whose co-ordinates are(/r+a, 0,0),(0, mr+/5,0) and (0,0, nr-j-y)
respectively. The equation of the sphere OABC is
x»+y«+z*-(/r+a) x -(iwr-l-)5) y-(#ir-|-y)z=0.
[See Ex. 6 (a) of Solved Examples(A)]
or
..(2)
x^+y^+z^-<xx^Py—yz—r(lx-\-my+nz)—0.
The sphere (2) is of the form 5+AP=0. This sphere for all
values of A passes through the fixed circle 5=0,P=0
i.e.
X*-fJ'*+2^——jSy—yz=0,/3c-|-my+nz=0.
Ex. 17. Find the equation of the sphere through origin and
whose centre lies in positive octant^ and which cuts the planes x=0,
y=0,z=0 in circles of radii a^s/l, by/2, cy/2 respectively.
Sol. The equation of any sphere passing through the origin
is
X*-f-y*-h z*+2ttx+2vy+2wz=0.
...(1)
268
Analytical Geometry 3-D
The intersection of the sphere (i) with the plane xc^O is
...(2)
^8^2*+2vj>+2wz—0, x=*0.
Equations(2) are the equations of a circle lying in the yzplane. The radius of the circle(2) is clearly
But we are given that the radius ot the circle (2) is ay/2.
Hence we have
...(3;
V(v*+ w“)=aV2 or v“+w*«=>2a>.
Again since the sphere (1) meets the planes ^>=>0 and z^O in
circles of radii by/2 and cy/2 respectively^, we have proceeding as
above
...(4)
and
i<a+v*=2c®.
...(5)
Adding (3), (4) and (5), we have
or
Subtracting (3),(4) and (5)from (6), w'e respectively have
a*=s6*4"C®—0*. v*=c*+fl*—6*.
c® ■
v*=±V(c”+fl“—^®),
c®).
Putting the values of u, v, w in (1), the equation of the
sphere is
x*+y®+z*±2V(^>*H-c®-o*) x±2v'(c»+fl»-6*)
. ±2V(fl®+6®-c®)z«0. :.(7)
It is given that the centre of the sphere lies in the positive
octant, hence the co-ordinates of the centre of the sphere are. all
positive. The co-ordinates of the centre of the sphere (1) are
(—u, — V,. — w). Therefore taking everywhere the—ve sign in
equation (7), the equation of the required sphere is given by
x^-{-y^+z^-2y/(b^-\-c^-a^);c-2v'(c®+fl»-6*) y
^2V(fl®+6®-c®)z«0.
Ex. 18. POP' is a variable diameter of the ellipse z=0,
x®/fl®-fy'*/6®=l and a circle is described in the plane PP'zz' on PP'
as diameter, prove that os PP'varies, the circle generates the surface
{x®-|-y®-fz®)(x®/a®+y®/6«)c=:x®-l-y'®.
Sol. Recall the analytical geometry of two dimensions.
The parametric co-ordinates of the extremities of any diameter of
the ellipse
are (a cos 6, b sin d) and(—a cos d.
—b sin 6).
Here in the present problein (of three dimensions POP' is a
variable diameter of the ellipse x*/a®>|-y>®/d®«=l, z=0. Hence, the
co-ordinat.:8 of the extremities P and P' may be taken as (a cos 6.
b sin 6, 0) and(—a cos 6, —b sin 0) respectively.
269
The Sphere
The equation of the sphere on PP* as diameter is given by
(jc—fl Cos 6)(Jc-}-fl cos d)-hiy—b sin 6)(y-{^b sin 6)
+(z-0)(z- 0)=0[See § 2(E))
...(1)
or
cos- d+b^ sin^ 0.
To find the equation of the plane PP'zz'.
The equations of zz' f,e. z-axis are x=0,>>'=»0.
The equation of any plane through zz' is
...(2)
If the plane.(2) also passes through P(a cos 0^ b sin 0t 0)then
a cos 04-Afebin 0==iO or A*=—(acos 0)l(b sin 0),
Putting the value of A in (2), we get
X
x—{(a cos 0)l{b sin 0)}
or a cos 0
Z..-,
b sin 0
...(3)
The plane (3) clearly passes through
P'(—a cos 0, —b sin 0,0).
Hence (3) is the equation of the plane PP'zz\ Thus the
equation^ of the circle described in the plane PP'zz' on PP'as
diameter are given by the equations (1) and (3).
The required,locus is obtained by eliminating the variable 0
between the equations (I) and (3) From
we have
(xjaljtyip) _ V{(xla)^Mylb)^}
cos 0'"'4ih^0 " -v/tcos'** 0+sia* 0)
A/(fx/a)«-Ky/^>*}.
1
a cos 0=x/y/{{xla)^-t(ylbf}, 6 sin 0=-ylV{ix!a)*+iylb)*}^
Putting the values of u cos 0 and h sin 0 in (1), the required
locus is given by
JC2
y
{{xla)r+{ylb)^J’^Maf+{ym
or
{x^ )^.y^-\-z^f (x^la^-\-y^lb^}-x^+y^.
Ex. 19. A Is a poini on OX and B on OY, so that the angle
OAB is constant qnd equal to a. On AB as diameter a circle Is
draw^i whose plane is parallel to OZ. Prove that as AB varies, the
circle ^eperates (he cone 2xy~z^ sin 2oc.^O.
Sol. Since .4 is a point on OX, we may lake A as (a, 0, 0).
Similarly B being ou OY may be taken as (0, b, Oj. Given
Z_OAB
therefore from right angled triangled AOB, we have
ian (X.—OBIOA=‘bla.
The equation of the sphere on AB as diameter is
(x- a)(x-0)+t>^-0) iy-h)+iz-Q)(z->0)=0 [See § 2(E)]
or
X* 4-^2 4- z* «=-● ax 4- by.
■■●(2)
270
Analytical Geometry 3-D
Also the equation of the plane passing through the points A
and B and parallel to OZ is
(4)
xla-\-y(b==\.
Thus the equations(2) ahd (4) are the equations of the circle
on AB as diameter and having its plane parallel to OZ.
The locus of this circle is obtained by eliminating the varia
bles a and b between (1),(2) and (4).
From (2) and (4), we get
x^+y^-{-z^^iax+by)(xla-^y/b)
or
x^+y^+z^=x^+y^-\-xy (alb-\-bla)
or
x^+y^-{-z^=x^-{-y^+xy (cot a+tan a), using (1)
or
z^==‘Xy (cos a/sin a+sin a/cos a)
or
za=.Tj»/(sin a cos
or
sin a cos a z^—xy or 2xy—z^ sin 2a=0,
which is the required locus.
Ex. 20. Find the equations of the clrcumcircle of the triangle
ABC, whose vertices are A (a, 0, 0), B(0, b,0) and (0,0, c). Find
also (i) the co-ordinates of its centre and (//) its diameter.
(Meernt 1984)
Sol. The three given points are A {a, 0, 0,), B(0, b, 0)' and
C(0,0, c). Let the fourth point be O (0, 0, 0). The equation of
the sphere
is [See Ex. 6 (a), set (A)]
—ox——CZ=0.
...(1)
The equation of the plane ABC is given by
...(2)
x/a+y/6+z/c.= l.
The equations (1) and (2) together are the- equations of the
circumcircle of the A^5C.
To Olid the co-ordinates of the centre of the circle.
A line perpendicular to the plane(2)of the circle and passing
through the centre (^a, ^6,|c) of the sphere (1) will pass through
the centre of the circle. The equations of this line are given by
x-ld_y-lb_z-\c
r (say).
1/a
\jb
1/c'
..,(3)
Any point on the line (3) is
(ifl-f-r/a, ih+r/h, ic+r/c).
...(4)
If this is the centre of the circle, then it will lie on the plane
(2) and hence, we have
(2+p)+(?+^)+(5+?)=‘
271
The Sphere
or r
-1
'●
and
2"'■=
(S5+P+?j
I
2a (a-»+6-*+c-*)
2 a “2 2a (a-»+^>"“+c-*)
_ g»
fl (fr-»+c-«)
2a (a-»H-6-2+c-*)“2(a-»+^»-*+c-“)
_
c (a‘*+6“*)
c ""2 (a-^+i^-Hc-^*) ‘
Putting these values in (4), the co*ordinates of the centra of
the c^cle are given by
c (a-^+h-*)
h (c~”4-g~^)
a(6-Hc-»)
(\2 (a-2+h-M^)* 2(a-a+h-»+c-») » 2 (a-a+h“*+c"*) )●
To find the diameter of the circle. Let r be the radius of the
circle and A the radius of the sphere. If p be the length of the
perpendicular from the centre of the sphere to the plane (2), then
...(5)
rs=,j?8-pa
Now /{<=>radius of th^iphere (l)=iy(a*+6*+c*)
and
‘ ]^=>the length of the perpendicular from
}c)
the centre of the sphere (1) to the plane (2)
1
—
l+i+i-t
...
V{( m*+(1/^)*+(1 m
2V(fl-*+^’*+o
Putting the values of R and p in (S)» we get
1
r* = o*+6*4-c*
4
4 (a-*+h“»+c-«;
or
#0
4r*=a*H-hHc‘-
h»c»+c*a*+a*h»
„(y+g^) (c«+o^) (q^+h»)
h-cs-t-c^g*+a*h»
The required diameter=2r
Ib^+c^)
(fl^+h*)l
(h*c*+c*a®+a*h*j
j’
§ 6 (A). The intersection of a straight line and a sphere.
Let the equation of the sphere be
x>+;»*+z*+.2i/;c+2v;>+2w2+</=0,
and that of the.straight line be
(X—jci)//=»(y-j>i)/ffi«(z-zi)/n«=»r (say).
The co-ordinates of any point on the line (2) are
Or+Xi, mr+J'i.
..(I)
...(2)
...(3)
272
Analytical Geometry 3-D
If this point lies on the sphere (1), we have
(lr-\-Xi)^-h(fnr+yiY \-{nr-{‘Zjy-\-2u (/r+.Vi)+2v (mr-fj>i)
or
(/*+TO*+«®)+2r {/(x,+ «)+/» (>>i+v)4-« (zi+ w)}
+(^i®-h
H-^i® i 2i/x,+2vj>i+2wzi+ rf)=0.
...(4)
The equation (4) is a quadratic in r and hence gives us two
values of r which may be real and distinct, real and equal or
imaginary and, therefore, accordingly the line (2) will meet the
sphere (1) in two points which may be real and distinct, real and
coiucident or imaginary.
The co-ordinates of the points of intersection are obtained by
putting the values of r from (4) in (3).
(B) Power of a point. If a straight line In any direction and
through a given point A meets the sphere in two points P and Q then
AP.AQ is constant and is said to be the power of the point A.
Let the given point A be taken as (Xi, yi, Zi) and let /, m, n
be the actual direction cosines of the line (2) through A so that
l*-{-m^+n^=l. The line (2) meets the sphere (I) in two point P
and Q and hence ri and r^, the roots of the equation (4), are the
actual distances of the points P and Q from the point A
t.e.
AP—riaixdAQ-r^.
AP,AQ^fir^^Xi^A‘yY +zYi-2uxi+ 2vyi+2wzi-\-d ...(5)
or AP.AQ=-constant, being independent of /, m,n.
(C) To find the condition that the given line (2) may touch the
sphere (\).
The line (2) will touch the sphere (1) if the roots of the equa
tion (4) are equal and so the required condition is given by
{/(Xi-f
(>^i+v)-hn (z,+H»)}*
==^{P-\-m^Arn*)(Xx^-\-yY+Zi*+2uxi-h2vyt-\-2wzi+d). .,.(6)
§ 7. The equation of the tangent plane.
(A) To find the equation of the tangent plane to the sphere
x^-\-y^ftz^+2ux-\r2.vy-\-2wZ'\-.d=0 at any point (Xj, yi, z,) on it.
(Agra 1973; Avadh 80; Madras 73; Kanpur 80, 82; Ranchi 77, 80)
The equation of any sphere is .
+3'“+z*+2mx+2 4-2wz+
0.
...(1)
The equations of any line through (Xi, y,, Zj) are
.(2)
(x-x,)//=(y—yi)/m«=(z- z,)/n—r (say).
The points of intersection of the sphere (1) with the line (2)
are given by [See § 6(A), equation (4). Give complete proof
here]
The Sphere
273
r*(/Hm*+n*)+2r {/(x,+«)+m (;F,+ v)+n (zi+w)}
(^1*-\-y*+2i^+2t/xi 4-2 +2wzi+</)«=0. ...(3)
The point (xi,
2,) lies on the sphere (1), and so we have
...(4)
JCiHTi*+«i*+2mx,+2v;;i+2wzi 4d=»0.
Using (4), the equation (3) reduces to
r« C/*+w*+n*)+2r{/(Xt+M)+m (>^,+ v)+n(2,+w)}=0. ..,{5)
From (5), we get r=0
and r=—2{/(Xi+i/)+m f>i+v)+» (Zi 4 w)}/(/*+»»*+«●). ...(6)
Now one root of the equation (5) is zero i.e., one value of r
given by (3) is zero. Butr is the distance from the point (Xj, j>i, zi)
of a point of intersection of the sphere (1) with the line (2) and
so one point of intersection coincides with (Xi, yu Zj). Therefore^
if the line (2) is a tangent line to the sphere (1), the other root of
(5) [i.e. the value of r given by (6)) must also be zero and for tliis
we have
...(7)
/ (Zi4tt)+w (yi4*v)4« (zi+w)«*0.
Hence the line (2) is a tangent line to the sphere (1) at the
point (Xi, yi, Zt) on it if the direction cosines /, m, n of the line
(2) satisfy the condition (7). . The locus of all such.tangent lines for
different values of /, m, n is called the tangent plane at the point
(*i» yu zi) to the sphere (1).
Therefore, the equation of the tangent plane at (Xi, yu Zi) to
the sphere (1) is obtained by eliminating /, m, n between (2) aiid
(7), and is given by
(X—Xi) (Xi+tt)4(T“tO (Ti+v)+(z—Zi) (Zi+w)«i0
or xxi-^-xu+yyi-^-yv+zzi+zw
- (Xi*+>>if 4 Zi*+
+wzi)-= 0
or xxi 4 yyi 4 zzj 4 «x 4 vjy 4 wz 4 (nxi 4 v;>i 4 wzi 4 d)=0
fusing (4)]
or xxi43'3'i4zzi4« (^4^1)4V (3'4Ti)+w fz4Zi)4</=0. ..(8)
(B) To find the equation of the tangent plane to the sphere
x*-\-y*-{-z*=>a^ at the point {xu yu Z\) on it,
(Kanpur 1976: Madras 77; Guru Nanak 75)
This is a particular case of § 7 (A). Pr'oceeding exactly as in
§ 7 (A), the equation of the tangent plane et (^1. yu Zi) to the
sphere x*4;/* 4 z*=fl* is given by xxi4j/j>i4?z, = a*.
Rule. The equation of the tangent plane at any point
(JCi» yu Zi) to a sphere can conveniently be written by replacing
X* by xxi, y^ by yyu z® by zzu 2x by x4^i» 2;> by )'4)'i. 2z by
z4zi and retaining the constant term.
274
Analptlcal Geometry 3-D
(C) The tangent line at any point ofa sphere is perpendicular
to the radius through that point.
Letthe equation of the sphere be
...(1)
Let the equations of any line through the point (xi, yu Zt)on
the sphere (1) be
y-yy
/
m
Z-Zx
n
...(2)
The line (2) will be a tangent line to the sphere (1) at the
point (Xi, yi, zi) on it if[See.equation (7), § 7(A)]
/(Xi+u)+in (jyi+v)4-n [Zi+w)=0.
...(3.)
The co-ordinates of the centre of the sphere (1) are(—«, — p,
—w)and so the direction ratios of the radius through the point
(Xif yu Zi) are Xj—(—li), y,-(—v), Zi-~(—w) i.e. are Xi-|-m, yi+v,
Zi+w. Hence the condition (3) shows that t^ tangent line (2) at
the point (Xj, y,, Zj) is perpendicular to the radius through the
point(X„
zi).
(D) The tangent plane at any point is perpendicular to the
radius through that point.
The equation of the tangent plane at any. point (Xi, yi, Zi) of
the sphere
x*+ya+z*4-2MX+2vy-|-2wz+d=0 is given, by
(JC+x,)+v (y+y,)+w (z-fz,)+d=0
or ^(x,+ii)+;;(y,4-v)+z(zi+w)+ MX,+v.y,+ wzi+d=0. ...(1)
The d.r.’s of the normal to the tangent plane (1) are Xi+m,.
3>i+p» Zi+ H».. But these are also the d.r.’s of the radius through
the point (x,,
z,) (See § 7(C) above].
This shows that the normal to the tangent plane (1) is parallel
to the radius through the point (xi, y^ zO and hence the tangent
plane at any point (Xi, pu z,) of the sphere is perpendicular to the
radius through the point (xi, pu zO.
(E) Condition of fangency. Tofind the condition that the
plane Ix-^my+nz ■. p touches the sphere.
'
J'*+z“+2wx+2ry 4-2»PZ+d=0.
(Calcutta 1981; Gorakhpur 82; Madras 77; Kanpur 78)
The equation of the given sphere is
^*+3'*+z*-|-2M*+2vy-l-2wzH-d=0.
(1)
...(2)
The equation of the given plape is /x+my+nz=p.
The Sphere
275
If the plane p)touches the given sphere (I), then the length
of the perpendicular from the centre(— m, —v, — w) of the sphere
(1)to the plane (2)=tbe radius
of the sphere (1)
l(--u)+m (-v)-\-n {~w)-p
».«●,
V(/“+m»+/|2)
Squaring, the required condition is given by
(/M+mv-f nw+p)®=(u®+v2+w*—</) {/*+/»*+»*)●
Corollary.
The condition that the plane lx-\-my+nz=p
touches the sphere x^-\-y^-\-z^==a^ is given by
p^=a^
[Proceed as in § 7 (E) above].
SOLVED EXAMPLES (C)
Ex. 1. (a). Find the equation of the sphere described on the line
joining A (2, —1, 4) a/iflf fi (—2, 2, —2) as diameter arid find also
the equation of the tangent plane at A.
Sol. The equation of the sphere described on the line
joining the points (2, — 1, 4) and .B (~2, 2, —2) as diameter
is given by
Cc-2)
(3'-2)+(z-4) {z-(-2)}=0
Ans.
Or
'
>'~2z—14=0.
The equation of the tangent plane at A{2~—\, 4) of the
above sphere is given by
jc.2+;;.(-l)+z.4-i (;;-I)-l.(z+4)-14=0 [See § 7 (A)]
or
ilUS.
4jc-3;;+6z-35=0
Remark. The tangent plane at A can also be found by writing
the equation of the plane passing through A and perpendicular
to AB.
Ex. 1. (b). Find the equation of the sphere described on the line
joining the points [3, 4, 1) and (—1, 0, 5) as diameter and find also
the equation of the tangent plane or (—1,0, 5).
(Madras 1976)
Sol. Proceeding as in Ex. 1 (a) above we have the required
equations as
x^-\-y^+z^—2x—4y—6z+2==G,
z+6=0.
Ex. 2. (a). Find the equations of the tangent planes of the sphere
4y+6z—7^0 which intersect in the line
6x—3y^23=0=3z-\-2. .
(Robilkband 1977)
Sol. The equation of the given sphere is
+3'®+z*+2x - 4>>+6z—7=0.
...(1)
The equation of any plane through the given line
6x-3;;-23=0=3z+2
276
Analytical Geometry 3~D
is
or
6x-3y-^23-\-X(3z+2)=0
6x-3j>+3Az+(2A-23)=0.
...(2)
If the plane (2) is the tangent plane to the sphere (1), then
the length of the perpendicular from the centre (—1, 2, —3)of
the sphere (i) to the plane (2)=the radius of the sphere (1)
6(-l)-3.2-l-3A (-3)+2A-23
i.e.,
V{(l)»+(-2)*+(3)»-(-7)}
V{(6)*+(3)«+(3A)>}
or
-7A-35= v'(21)v'(45+9A«).
Squaring and simplifying, we get 2A*—7A—4=0
or
2A*-8A+A~4=0 or A=4,
Putting the values of A in (2), the required equations of the
tangent planes are given by
2x-3>4-42—5=0and 4ic—2j?—2—16=0.
7" Ex. 2.(b). Find the equations of the tangent planes to the sphere
x*4-j>*4-z*+6x—2z^l=0 which pass through the line .
3(16-x)=32=2y+30.
(Punjab 1977)
Sol. 'J'he equations of the given line may be re-written as
(*1i?)-3z=0=2;»-3z+30.
Now prbceeid as in Ex. 2(a) above, the equations of the tan
gent planes are given by
x+2y-2z+l4=0 and 2x-4-2y-z^2.
Ex. 3. Find theTequatfons of the tangent planes to the sphere
x*+y*+z*=24 which are parallel to the plane 2x+;>—2=0.
Sol. The equation of the given sphere is
x*+;^*+z*=24.
...(1)
Let the equation of the tangent plane parallel to the plane
2x+;>—2=0 be 2x+j>—2+A=0.
●■(2)
If the plane (3) touches the sphere (1), then the length of the
perpendicular from the centre (0, 0, 0) of the sphere (1) to the
plane (2)=±the radius of the sphere (2)
,
2.0+0--0+A
/nA\
V{(2)*+(I)H(~1)*}
A = ±V(24).v'6=±12.
Putting the values of A in (2), the equations of the required
tangent planes are given by
2x-l-j»^z±12=0.
^
Ex. 4, (a). Show that the plane 2x—2y+z+\2*=>0 touches the
2x—4j'+2z—3=0 and find the point of contact.
(Avadh 1981; Agra 82; Berahmpur 81; Gauhati 11)
* *
or
The Sphere
277
Sol.
The equation of the given sphere is
...(1)
jc*
+z*—2;c—
2z—3=0.
Its centre C is (I, 2, — 1) and its radius
=V{(-l)*+(-2)»+(l)’*--(-3)}=3.
The equation of the given plane is
...(2)
2x-2;»+z+12=.0.
The length of the perpendicular from the centre C(1» 2, — 1)
2.1-2.2-1 +12
9_,
to the plane (2)'
::the radius of the sphere (1).
Since the length of the perpendicular from the centre
C(I, 2, —1)of the sphere (1) to the plane (2) is equal to the
radius of the sphere^ therefore,(he plane (2) touches the given
sphere(J).
Tilt.
ui* tin. uviui'ul—to tlwTo find the
taugemTiirSucTzf^e 2;-2, 1» and, therefore, the d.r.’s of the
[See § 7(D)].
radius CP are 2, —2, 1.
Thus the equations of the line (radius) CP are
.. (3)
● (x-l)/2=(;^-2)/(-2)=(z+l)/l=r (say).
The co-ordinates of any point on (3) are (2r+l, —2r+2, r—1).
If this is the point of contact P, then it will lie on the tangent
plane (2) and so we have
2(2r+i)-2(-2r+2)+(r-l)+12c:.0 orrc=-l.
Putting the value of r, the co-ordinates of the point of contact
Pare(2( -1)+ 1, -2(-l)+2, -l-l)or (-1,4, -2).
Ex. 4.(b). Show that the plane 2x+y-z>^ 12 touches the sphere
x^+y*+z^==24 and find its point of contact,
Sol. Proceeding as in Ex. 4(a) above the point of contact
is(4, 2, -2).
Ex. 5.(a). Find the equation of the sphere which touches the
sphere x^-{-y^+z^-{-2x—6y+l^0 at(1, 2, —2)and passes through
(Indore 1979; Meernt 82)
the point(1, —1,0).
Sol. The equation,of the given sphere is
...(1)
x^+y^-\-z^+2x—6y-\-1=0.
The equation of the tangent plane to the sphere(1) at the
point (1, 2,—2)is
x.l+j^.2+z(-2)+(x+I)-3(>»+2)+l«0
or
...(2)
2x—y-2z-4=0.
Therefore, the equation of the sphere which touches the
sphere (1) at (1, 2, —2)is given by
278
Analytical Geometry 3-D
x^+y^+z^-h2x-6y-h]+X{2x-y-2z-4)=0.
...(3)
If(3) passes through the point (1, — I, 0), then we have
1H-1+0+2+6+H-A (2+l -0-4)=0 or A=ll.
Putting ihe value of A in (3), the equation c*" the required
sphere is given by
;c2+ya+28+24jc~17y-222-43=0.
Ex. 5 (b). Prove that the equatton of the sphere which touches
4 (x®-|-y*+z*)+10jc--25y—2z=0 at (1, 2, —2)and passes through
the point
is
JC*+/+z2-1-2jc-6y+1=0.
(Meerut 1978)
Sol. Proceed as in Ex. 5(a) above.
Ex. 5 (c),
Find the equation of the sphere which touches the
through
spnere X-+y--fz--tia — \jy -r i iC
the origin.
Sol. Proceed upto equation (3) of Ex. 5 (a) above.
If (3) passes through (0,0, 0), then we have
O+O+O+O-O-l-H-A(0-0-0-4)=0 or A=l/4.
Putting the value of A in (3), the equation of the required
sphere is given by
(A*+ya+z24-2x—6y+l)+i.(2x—y—2z-4)=0
or
4(x*H-.V®+z®)+10x-25y-2z=0.
Ex,6. Find the equations of the spheres through the circle
x“4-/+z*=l,2x+4y+5z=6
and touching the plane z=0.
(Gorakhpur 1974; Meerut 85S)
or
Sol. The equation of any sphere through the given circle is
x*-ry*+2^—1+A (2x-f4y+5z—6)=0
X®
+z«+2Ax+4Ay+5Az-(H-6A)=0.
...(1)
If the sphere (1) touches the plane z=0, then the length of
the perpendicular from the centre (-A, — 2A, —|.5A) of the
sphere (1) to the plane z=0=±the radius of the sphere (1)
-i.5A
i.e.
V(1«>
Squaring 25A®
4
or
45A®
4
6A+1 or 5AH 6A +1=0
5AH5A-f-A+l=0 or (5A+1)(A-f-l)=0 or A
-1/5, - I
Putting the values of A in (1), the equations of the required
spheres are given by
5 (x®+.FHz’*)-2x-4y-5zH-l=0
and
x’HyHz^—2x—4y—5z+5=0.
The Sphere
279
Ex. 7(a). Find the equations of the spheres which pass through
the circle
2x+j^+z«=4 and touch
the plane 3x+4;>—14=0.
Sol. The equation of any sphere through the given circle is
;c84./+z*-2x+2>>+4z-3+A (2x+;y+2-4)=0
or
;c2+j,a+28-2x(l-A)+X2+A)+z(4+A)-(3+4A)=0....(l)
If the sphere (1) touches the plane
..(2)
3x+4;?—14=0,
then the length of the perpendicular from the centre
(l-A. -(2+A)/2. -(4+A)/2)
to the plane (2)=±the radius of the sphere (1)
3.n-A)-4.j(24-A)-14
i.e.
or
-(A+3)=±iv^(6A*+20A+36).
Squaring and sinaplifying, A^—2A=0 or A=0, 2.
Puttingthe values of Ain (1), the equations of the required
-I-z8^2x+2;^+4z—3=>0
spheres are
and
x®+>'*+z*+2x+4>>+6z —11=0.
Ex 7(b). Find the equations of the spheres which pass through
the circle x*^+/+z»=5. 2x+3;'+z=3 and touch the plane
3;c+4;,_15=0.
(Punjab 1982)
Sol. Proceed as in Ex. 7(a) above.
Ex. 8. Find the equations of the tangent line to the circle
3x^-\‘3y^+3z^—2x—3y—4z—22t=0,3x+4;>+5z—26«=0
at the point (I, 2» 3).
Sol. The tangent line to a circle at a point is the line of
intersection of the tangent plane to the sphere at the point and
the plane of the circle.
The equations of the given circle are given by the sphere
(1)
(2)
3x+4>»+5z—26=0.
and the plane
The equation of the tangent plane to the sphere (1)at(1, 2, 3)
x.l+;;.2+z.3-J +
(.V+2)-S(z+3)-\^=0
4x+9>'+14z—64=0.
●●●(3)
Thus the non-symmetrical equations of the required tangent
line are given by (2) and (3) together. To find its symmetrical
form, let /, m, n be its d.r.’s, so that we have
3/+.4m+3«=0 and 4/+9m+14=0.
is
or
280
Analytical Geometry 3-D
Solving, we get
/
-11
m
n
/
or — = m _n
22 -II
1
-2 f
The equations of the tangent line at (1, 2. 3) and having
‘i.r/s 1, -2, 1 are
(x-l)/l=.(;;-2)/(-2)=(z-l)/l.
£x. 9 fa). If three mutually perpendicular chords of lengths
du di, da be drawn through the point (Xi, y^, Zi) to the sphere
x*+y^-hz^*=a\ prove that .
+<4^=120*—8
Sol. The equation of the given sphere is
...(1)
n^ be the actual d.c.*s of
the three nsutualiy perpendicular chords drawn through the same
point A{Xi, yi, Zi). The equations of the first chord are given by
Let /i, /Ml, //i; 4, /Wa, n% and /a,
(^—
)//«!<=(z—2i)/«i=r (say).
...(2)
Any point on (2) is(/ir+Xi, m,r+y,, /iir+z,).
Substituiiog the co-ordinates of this point in (1), the points
of intersection of the chord (2) with the sphere(1) are given by
(Ar+Xi)2+(#»ir-|-;;,)2+(«,r-fz,)a=a*
...(2)
or
r- (li^+
+2r (liXi-hmiyt +njZi)
or
Let the chord (2) meet the sphere in the points P and
that AP=ru AQ=^r% where rx and ra are the roots of(3),
di^PQ^AP-AQ rx-ra
or
+f3)*=4nra
or
rf,*^{-2 {IxXx+miyx ^-/iiZ,)}*-4
or
^i*=4 (4x1+/W,
zj8_ 4
Sinailarly </a*=-4 (/aX,+/W3:Fi+«8^i)*--4
and
</a-=4 (/aJfi-|-ffi8;'i+/i3ri)2-4 (JfiH.l'iH V-a®).
Adding (4),(5)and (6) and using /iH/a*+/3*=l etc.,
(i^x+ZaWa-f-./a/Ms—0 etc., we get
Q so
...(4)
●●(5)
...(6)
<4’+rfaH<4*'=4 (Xi^+;;,«+
2 (x,2+;;i5<-i-z,8„fla)
=I2aa-8 (Xi»+:Fi*+V).
Ex. 9(b). Show that the sum of the squares of the intercepts
made by a given sphere on any three mutually perpendicular straight
lines through thefixed point is constant.
\
\
Sol. Let the equation of the given sphere be x^-\-y^-\-z^ = aand the fixed point be (Xj, yi, Zx).
28t
The Sphere
Now this problem is nothing but a different* statement of the
Ex.9(a) above.
Ex. 10. Ifaiiy tangent plane to the
makes
Intercepts a, b and c on the co-ordinate axes^ prove that
(Agra 1975, 79; Kanpur 73, 81; Punjab 82)
Sol. The equation of the sphere is given to be
...(1)
Let (Xi, yu zi) be a point on the sphere (1), so that we have
Xi^-{-yi^-\-zi^=r^.
—(2)
The equation of the tangent plane to the sphere (1) at the
xxi -{-yyi+zzi^f*
point (xi. yu zi) is
z
X
y
1.
or
(3)
r*/xi r^/yi r^[zi
The intercepts made by the tangent plane on the axes are
r^/xu
and r*/2i; but these intercepts on the axes are given to
be a, b and c respectively.
a=^r^lxu 6=r^lyu c«=» r^jzi
or
yi~f*lhf Zi=r*/c.
Substituting the values of Xi, yu Zi in (2), we get
Ex. 11. Find the equation of the sphere touching the three co
ordinate axes. How many such spheres can be drawn ?
Sol. Let the equation of the sphere be
...(1)
.X*+4-z*-|-2ttx+2
2wz+d=»0.
If the sphere (1) meets the x-axis i.e., y^O,z=0. then from
(1), we have
...(2)
x*-l-2ttx+d=0.
Since the sphere(1) is touching y=0=z, the roots of(2)
must be equal and so we have
[using B^—4AC^0]
or
Similarly if the sphere touches y and z axes, we have
v*=d and w**=d.
/. tt*<=v*=w*=></=A*(say).
tte=sV=W=dbA.
Putting the values in (1), the equation of the sphere(t)
becomes
...(3)
xH/+z®±2A (x-h;;+z)+A*==0.
282
Analytical Geometry 3-D
Since A may have any real value, an infinite number of spheres
can be drawn satisfying the given conditions.
In case the radius of the sphere be given, say r, we shall have
only eight spheres as shown'below.
Now r«=*the radius of the sphere=-\/(«®+v*+ w*-d).
r^^d-hd+d-d
[V y8_y2_.
or
r^=2d.
u^T.--v^ :r.w^==>d--r~^r^ ox u—v — w~±,rjy/{l).
Hence the equation (1; of the sphere becomes
...(4)
x^-\-y^-\-z^±rxy/2±ryy/2±rz\^l-y\r^==Q.
The equation (4) represents only eight spheres satisfying the
given conditions.
Ex. 12(a). Find the equation of a sphere touching the three
co-ordinate planes. How many such spheres can be drawn ?
(Robiikhand 1979)
Sol. Let the equation ot the sphere be
...(1)
x^-\-}^+z^+2ux-\-2vy-\-2wz-\‘d=0.
In the sphere (1) touches the >>z-plane J.e., x<=>0, then the
length of the perpendicular from the centre (—w, —v, —w)of the
sphere (1) to the plane.(x=0)«=>the radius of the sphere (I)
l.e.
—ull= y/{u^i-v^+w^—d),
-(2)
o2 _ ^ v2^ w^—d or v*+
d.
Squaring,
Similarly if the sphere (1) touches the planes ^=0 and z=0,
then we have
tt*-}- v®=ci.
...(4)
Adding (2),(3) and (4). we get
...(5)
2(u^-f
-3t/or m*4=>§</.
Subtracting (2), (3) and (4) from (5) in turn, we get
=
=
r- (say), or M=v=w=±^*
Also radius of the sphere=V(*<“+’'^4 w*—d)
-v'('’H^*+r>-2r8)=r.
Substituting the values in (1), the equation of the required
sphere is given by
...(6)
x^-\-y^-{-Z‘±2rx±2ry±2rz-\-2r^=‘Q.
Since r may have any real value, an infinite number of such
spheres are possible. In case the radius, r of the sphere be given,
then only eight such spheres are possible.
Ex.12(b). Find the equation of a sphere which lies in the octant
OXYZ and touches the co-ordinate planes.
The Sphere
2S3
Sol. Proceed upto equation (6) of Ex. 12(a)above.
Now if the centre of the sphere lies in the octant OXYZ,then
the centre of the sphere is (r, r, r) and hence the equation of the
required sphere is of the form
...(7)
+2^-2rx-2ry-2r2+2r*=0.
Ex. 12 (c). Shovf that in general two spheres can be drawn
through a given point to touch the co-ordinate planes and find for
what positions of the point the spheres are (i)real and{ii) coincident.
Sol. Let the given point lie in the positive octant. Obviously
the centre of a sphere passing through this given point and touch
ing the co-ordinate planes will also lie in the positive octant.
From Ex. 12(b) above, the equation of any such sphere touching
the co-ordinate planes is given^y
,
If this sphere passes through the given point (xi, yu ?i)
then we have
+:j;i3+2,*-2r (Xi
-f zi)+2r*=»0
...(2)
2r2-2r Ui-l-;i+z,)-Izi^)=0.
or
This equation (2) is a quadratic in r and hence gives two
values of r and accordingly two spheres are given by (1) passing
through a given point (Xi, jI, Zi).
.
The two spheres will be real or coincident according as the
roots of(2) are real or equal and the condition for the same is
{- 2 (xi-l-;;,-f2x)}2-4.2.(Xi2+>i"-l-2i") > 0
[using 5®—4^C ^ 0]
or
or
{Xi+yi+ziY-2(x,H;V+Zi'‘)
P
2
z,x,-)-Xi>»i) ^ Xi*+;;iH2i*.
Note, li the co-ordinates of a point are given, the point will
lie in any one of the eight octants. Any sphere touching the co
ordinate planes and passing through this given point will have its
centre in the same octant.
Ex. 13. A sphere touches the three co-ordinate planes and passes
through the point (2, I, t). Find its equation.
(Burdwan 1978)
Sol. The given point (2. I, 5) lies in the positive octant
OXYZ. Hence the equation of the required sphere is of «the form
[See Ex. 12(b) above J
x-i-y‘^+z^-2r (x+y+z)-\-2r^=0.
...(1),
[Deduce thi.s result, here]
If the sphere (I) passes through the point (2, 1, 5), we have
<l+ l +25~2f (2-H-i-5j-i-2f*c=0
284
or
or
Analytical Geometry 3-Z)
2r>-16r+30=0. or r2-8r+15=0,
(r—3)(r—5)=0 or r=3,5.
Substituting the values of r in turn in (1), the equations of
the two required spheres are given by
x^-{-y^+z^~6x-6y~-6z-\-U=0,
and
x*+;;a+z2-l0x -10;;-10z+50=.0.
Ex. 14 (a). Find the equation of the sphere inscribed in the
tetrahedron whosefaces are
x=0, y=0, z«=() and 2x+6;;—3x+9—0,
Sol. The equation of the fourth plane is ;2x+6;;—3z+9<=>0.
This plane clearly meets the x,;; and z axes in the — ve, — ve and
+ve directions. Hence the sphere touching the three co-ordinate
planes and the plane 2x+6;;—3z+9=0 lies in OJTT'Z octant and
so the centre of such a sphere is of the form (—r, —r, r) where r
~Ma-the^ra^us of the spHweT^' ~~——Since the sphere touches the plane
2x+6;;-3z+9=0,
...(1)
theiefore the length of the perpendicular from the centre(—r, — r,
r) to the plane (l)=tbe radius r of the sphere
—2r—6r—3r+9
i.e.
*=»r
“V(4+36+^
or
— lir+9=7r or r«=|.
/. The centre of the required sphere is (—i, —i,i) and
radius is
Therefore its equation is given by
(^+i)*+Cl'+
or
xH/+z®+x+>—Z+J=0.
Ans.
Ex. 14(b). Find the equation of the sphere inscribed in the
tetrahedron whoie faces are
x=>0, .v=^-0, z~0 artrf2x+6;;-f3z—14=0.
Sol. Here the sphere lies in OXYZ octant. Proceeding
as in Ex. 14 (a) above, the equation of the required sphere is
given by
81 (xa+/+2*)- 126(x+;;+z)+98=0.
Ex. 15. Provt: that the centres of the spheres which touch the
lines y-mx,z=c; y= — mx, z=>—c lie upon the conicoid
mxy-ircz
(Bundelkband 1979; Lucknow 74, 81; Kanpur 79; Meerut 89S)
Sol. Let the equation ui the sphere be
JC*+>*+
2mx+2v;;-h 2h*z+</=0.
The Sphere
285
The points of intersection of the sphere (1) with the line
are given by
x^-{-m^x^‘hc^+2ux+2vmx+2wc+d—0
.(2)
or
X*(l+m*)+2(u+vm) x-^{c*-{-2wc+d)=0.
If the sphere (1) touches the line y—mx,z*=>c then the roots
of the equation (2) will be equal and so we have
{2(m+v/»i)}2-4(l+m«)(c8+2wc+d)c=0
...(3)
or
(tt+vw)*«=(l+m*)(c^+2wc-{-d).
Again the sphere (1) also touches the line
y=»—mx,z=—c.
So putting — m for m and — c for c in (3), the required condition is given by
...(5)
(u—v/7i)2=»(14-«*)(c**—2wc+d).
Subtracting (4)from (3). we get
4ttv»i=(H-m*)(4wc)»
or
m«v—cw (l4-m*j=0.
or
The locus of the centre (-i/, -v, -w)is given by
m(-X){-y)-c(-Z)(1 +OT*)«0
mxy+cz(l+m*)=>0.
§ 8(A). Plane of contact.
The plane of contact is the locus of the points of contact of the
tangent planes which pass through a given point (not on the sphere).
(B) Tofind the equation of the plane ofcontact.
Let the equation of the sphere be
...(1)
x*+y*+z*+2«x+2vy+2wz+d=0.
Let(a, p, y) be a given point external to the sphere (1). The
equation of the tangent plane at any point (Xi,yi, Zi) of the
sphere is
xxi+yyi-^-zz^-^-u (x-f-Jfi)+i'
(z+^i)+^=®*
If it passes through (a, p, y), we have
a^i+jSj'i+yZi+w (a+x,)+v (^+yi)+w (y+Zi)+d==0.
The locus of the point of contact (Xi, yu Zi) is
<xx+/5y+yz+M (x+a)+v (y+/3)-|-w (z+y)+d=0
which is the equation of the plane ot contact of the point (a, Pt y)
with respect to the sphere (1).
§ 9. Pole and polar plane.
Definition. Consider a line through a fixed point A to inter
sect a given sphere in ihe points P and Q. Take a point R on
286
Analytical Geometry 3-D
this line such that
1
1
2
Ar
A^
AjK
{i.e. the
distance AR is the
harmonic mean of the distances AP and AQ), then the locus of the
point is a plane called the polar plane of the point ^4 with
respect to the given sphere and the .fixed point ^ is called the
pole of the polar plane.
The equation of the polar plane.
Let the equation of the given sphere be
.. (1)
x^+y^+z^+2ux-t2vy-i-2wzi-d=0.
Let the co-ordinates of the fixed point A be
Zi) and let
/, m, n be the actual direction cosines of the variable line through
^(*i» Pu 2i) and so its equations are given by
X—Xi
y~y% z—Zx
. .
—*=r(say).
m
..(2)
The co-ordinates of any point on the line (2) are
{lr-\-Xx, mr^yu nr+zj).
The distances AP and AQ of the points of intersection P and
Q of the variable line (2) and the sphere (1) are the roots of the
equation
(/r-l-x,)2+(mr+};,)H(wr-|-Zi)2+2M (/r-|-A:,)-l-2v (mr-l-y,)
-f2w («r-i-Zi)+</=0
or
and
●
or
r*+2r {/(xri-l-M)-l-m (;>i4-v)-l-/i (zi-fw)}
●\-{Xx^+yi^+Zx^-\-2uxx+2vyx-{-2wzx-\-d)=0
[V
...(3)
fi-f r2=^P-t-/4g=—2 {/ (XiA-u)+m (>>r+v)-|-n (zi+w)}
ri.ra==AP.AQ=:Xi^+yp+Zi^+2uxi ’\-2vyi-{- 2wzx -1- d,
1 , 1
APA-AQ — 2 {/r.3fi 4-M)+m fvi 4-v)-I-« (z,-l-w))
AP^AQ^AP.AQ
+Zx^ ^ luxx -H 2v;;i -J- 2 »vz, -|- rf
2
[See definition above]
""AR
Xx^-\-yx^-\-Zx^-\-2uxx-\-2vyx-\-2wzx-\-d
= -{/ AR
(z,-fw)} ...(4)
Now let the co-ordinates of the point R be {x,y^ z). The
distance of R from A{Xx,yx. Zx) is AR i.e. putting r=AR in (2),
we get
x-Xi=l.AR, y-yx=>m.AR, z-Zx=h.AR.
...(5)
The locus of the point R is obtained by eliminating the
variables
between. (4) and (5). Hencethe locus of^P i.e.
287
The Sphere
the polar plane of A (Xi, yi, Zx) with respect to the sphere (1) is
given by
+:Fi"+Zi*+2i«t+2vyi 4-2wzi+
= —{(x-Xi)(Xi+«)-l-(y-yi)(yi+v)+(z-zi)(zi+w)>
or
xxi-\-yyi+zzi+u (x+x,)+v(y+yi)+»v (2+zi)+<^=0 .- (fi)
To find the pole of a given plane. Let the pole of the given
plane be (jCi, yi, Zi). Write the equation of the polar plane of
Zi) with respect to the given sphere. Now compare the
given plane with this polar plane and then obtain Xi,yi, Zi,
§ 10. The properties of the pole and the polar plane.
Property I. Salmon’s Theorem. The distances of two points
from the centre of a sphere are proportional to the distance of each
from the polar plane of the other.
Let the equation of the sphere with centre O(0,0,0) be
...(1)
x^+y*+z^=a^.
Let P(xuPu Zi) and Q (X2, yt* Z3) be the two points whose
polar planes with respect to the sphere (1) are
...(2)
xxi’^yyi+zzi^a'^
and
...(3)
xXi-{-yyz+2Zi^a^.
Now the distance of7* from (3)
the distance of Q troth (2)
(●yi^2+.VlJ'2+ZiZ8—0^)/\/(^2^ + V2*+Z2^)
^{X2Xx+yiyx+ZiZi—a^)lyfW+yx^-\-Zx'^)
V(Xil±£lLt£l!)_
Proved.
“V(^a‘+V+Z2*)"(?0'
Property II. The polar plane of a point P with respect to a
sphere is perpendicular to the line joining the point P to the centre of
the sphere.
Let the equation of the sphere with centre O (0, 0, 0) be
...(I)
Let the co-ordinates of the point P be (xi, yu Zi) and so the
equation of the point plane of P (Xi, yi, Zi) w.r.t, the sphere (1) is
● (2)
xxi+yyt+zzi^a\
The direction ratios of the normal to the polar plane (2) are
Xi, yu zi. Also the d.r.’s of the line OP are Xi—0, yi—0, Zi—0
. i.e. Xu yu zi. Hence the normal to the polar plane (2) is parallel
to the line QP l.e,, OP is perpendicular to the polar plane (2).
Property III. If the line Joining the centre O of the sphere and
a point P meets the polar Of P in the point Q, then
OP.OQ = (radius)^.
288
Analytical Geometry 3-2)
Let the equation of the sphere be
...(1)
The centre of sphere is O (0, 0, 0)and radius=>a
Let the co-ordinates of the point P be (Xi, y^ Zi)> The
equation of the polar plane of F{Xu yi» Zi) with respect to the
sphere (I) is
...(2)
xxx-\‘yyx+zzx^a\
Now OP is perpendicular to the plane (2) and meets it at Q,
03»the perpendicular distance of the centre O (0,0,0)
from the polar plane (2)
Q‘
Property IV. If the polar plane of a point P(xi,
Zi)passes,
through another point Q (x,,
^2)1 then the polar plane of Q will
pass through P. The two points P and Q satisfying this property are
called conjugate points.
Let the equation of the sphere be x*-|-y*-f-z*=a*.
The polar plane of P (xi, yi, Zi) is xXf\-yyi~\‘ZZi^a*
If it passes through Q (xa, yg, ^2), then
XiXi-^yiyi+ZiZi=aK
The polar plane of Q (Xa, ya» z^) is xxi-\‘yyi+zzi=a*.
If it passes through P (xi, yu Zi), then
...(1)
...(2)
XiXi-)ryxy%-\-ZiZ^^a*.
.●(3)
The relation (3) is true since the relation (2) is true, Hence
the result.
Property V. LeM/i=0 and «a=0 be the equations of two
planes. If the pole of the plane Ma=0 lies on the plane
then
the pole of the plane tta*=0 will lie on the plane Mi=0.
The two
planes
and fia=>0 satisfying this property are called conjugate
planes.
The proof is similar to property IV above.
§ It. The polar line.
●The polar line of a given line AB is another line CD such that
the polar planes of all points on the line AB pass through the line
CD.
Let the equation of the sphere be x*-l-y*4-z*=a*.
...(1)
Let the equations of the given line AB be
{x—a.)ll^{y’-^)lm^{z-y)ln^r (say).
...(2)
The co-ordinates of any point on the line (2) are(/r+a, mr+jS,
nr+y). Its polar plane with respect to the sphere (1) is
X ilr+a)-\ry (mr+j8)-f r (nr+Y)^a^
or
{a.x+py-{-YZ—a-)-{-r (/r+my+wz)=0.
289
The Sphere
This plane for all values of r passes through the line
...(3)
ax+Py-\-YZ-a^=Oi=lx+my-\-m,
Thus the equations (3) are the equations of the polar line CD
of the given line AB[given by (2)].
§ 12. The angle of intersection of two spheres.
If the two spheres intersect then the angle of intersection' of
these two spheres is defined as the angle between the tangent
planes to them at their common point of intersection. Since the
radii of the spheres through the common point are perpendicular
to the tangent planes at that point, therefore the angle of inter
section Is equal to the.angle between the radii of the two spheres
through their common point.
Let Pbe the point of intersection of the two spheres whose .
centres are Ci and Cz and radii fi and rg. Let 6 be the angle of
intersection so that Z.CiPCa=^* Also let CxCz—d.
Applying cosine formula for the
we get
or d=cos-1
\ 2rira
/
Two spheres are said to intersect orthogonally if their angle
of intersection is a right angle. In this case the spheres are called
orthogonal spheres.
To find the condition for orthogonality of two spheres.
(Avadh 1981; Kanpor 78)
Let the equations of the two spheres be
+z*+2«i*4-2Viy+2W|Z+dx «=0,
(1)
and
(2)
x*+y*+z* f2tfa^+2Pay+2waZ+rfas=0.
Let Cl and Ca be the centres and ri and fa the radii of the
spheres(I) and (2) respectively. Then
Cl is(-«„ -Vi, -Wi), Cz is (-Ma, -Vg. -Wa)
and
ri=<\/(«i®+Vi®+Wi*-di), ra« VW+Va^+Wa*—rfa).
Let P be a common point
of intersection of the spheres
(1)and (2). If the spheres (1)
and (2) intersect orthogonally
then the angle between the
radii through the common
point P is a right angle and
hence we have
CiP*+CaP*=CiCa* or ri*-|-ra*«CiCa»
290
Analytical Geometry 3-D
or
+(M'i-lVa)*
...(3)
or
2i/iWa+2viVa+2H’i
+</a,
which is the required condition.
§13. Tonching spheres.
If the distance between the centres ofthe two spheres is equal
to the dijfference of their radii, then the two spheres touch
internally.
If the distance between the centres of the two spheres is equal
to the sum of their radii, then the two spheres touch externally.
SOLVED EXAMPLES(D)
Ex. 1. Prove that the equation of the polar plane of(Xu yu Zi)
with respect to the sphere x^-\-y^+
a* is
xxi-{-yyi+zzi=>Q\
(Delhi 1974)
Sol. proceed as in § 9 above.
Ex. 2. Find the pole of the plane lx+my-{-nz=p with respect
to the sphere x^-{-y^Jrz^=a^.
(Gorakhpur 1976; Rajasthan 78)
Sol. The equation of the sphere is
(1)
(2)
The equation of the plane is lx-{-my-\^nz=p.
Let the pole of the plane (2) w.r.t. the sphere (1) be (Xi, yi,
Zi). The equation of the polar plane of(Xi, yi, Xj) with respect to
the sphere(1) is
...(3)
The Plaines (2) and (3) are the same, so comparing the coeffi'
cients of like terms, we get
/ m n
xi
or
yi
zi
a»
a»/
arn
Xi=.
The pole of the plane (2) is
a^mjpt a*nlp).
Ex. 3. Show that the polar of
respect
to the sphere x*+y*+z*c=9 is given by
x+2y+3z-9=0, 2x+3y+4z=0.
Sol. This problem is based on § 11.~ The co-ordinates of
any point on the line
are (2r^\, 3r+2.4r^3).
Its polar plane w.r.t. the sphere x*+y*^2®=9 is
291
The Sphere
(2r+l) x+(3r+2)>>+(4r+3)z=9
(x+2y-{-3z-9)+r(2x+3y+4r)=»0.
This plane for all values ofr passes through the line
*+2:F+3z—9«0,2x+3j>+4z«0.
Thus these are the equations of the polar line . of the given
line.
or
Ex.4. Prove that the polar plane of any point on the\llne
(y-1)=J(z+3) wlth respect to the sphere
1
passes through the line
(2*+3)/13=(;—i)/(-3)=.*/(-I). (Allababad 1OT6)
Sol. This problem is based on § 11.
The co-ordinates of any point on the line
(y-l)=J(z+3) are (2r, 3r-l-l, 4r—3).
Its polar plane w.r.t. the sphere x*-l->>*4-z*=l
jc.2r-|-;>.(3r+l)+z.(4r-3)=l
or
(;,-.3z-l)+r (2x4.3yH-4z)=0.
This plane for all values of r passes through the line
...0)
3z-l=0, 2x4*3.v+4z«0.
The symmetrical form of the line(1) is
*+3/2^>-1_z^
13
-6 -2
or
2x4-3 _;^-l_^ z or 2x4-3 _ji»—l_^ z
26
-6 -2
13 “ -3“-!*
Ex. 5. Find the angle of intersection ofthe spheres
x*4-j>*4’Z*—2x—4jk—6z4-10=0
♦o«d
*8+;,8+z8_6x-2;»4-2z4-2«0.
Sol. The equations of the given spheres are
0)
x*4-J'*4-z*—2x—4>—6z4-10=0,
and
(2)
x*-l“J'*4-z*—6x—2>>4-2z4-2=0. .
The centre of(1)is Ci (I, 2, 3), its radios
n=\/a*+2*4-3»-10)=2.
The centre of(2) is Cg (3, I, —1),its radius
= V{3»4-1*4-(-1)‘'-2}=3.
Now C|C2=the distance between the centres Ci and C%
={(3-l)*4-(l-2)»+(-l-3)>)=^/(21).
Let d be the angle of intersection of the spheres (1) and (2),
then
292
Analytical Geometry 3-Z)
cos
(2)»+r3t«-2l
2rira
: 2.2.3^
(=>I (taking the acute angle).
^=co8-» (f).
-it. 2
lT““
Ex. 6. Show that the two spheres x*+y^-\-2*^6y-{-2z-{-Z=>0
4.g>
+Sy'\-4z-{-20-=i0 are orthogonal,
Sol, The equations of the two spheres are
**+>'*+2*+6;;+2z+8=0
...(1)
**+7*+2*+6x+8;^+4z+20=i0.
...(2)
For the sphere (I), we have «,<=0, v,=3, w,«l,
and
for the sphere(2), we have 1/9=,3.
Wa«2, </a«20.
We know that the condition of orthogonality of two spheres'
is [See § 12. equation (3)]
2«i«a+2viV8-|-2»ViM'9«=» </i
...(3)
Putting the values in (3).. we have
2.0.3+2.3.4+2.1.2=8+20
or
0+24+4=28 which is true.
Hence the given spheres are orthogonal.
Ex. 7, Two points P and Q are conjugate with respect to a
sphere S; prove that the sphere on PQ as diameter cuts S ortho
gonally.
and
Sol, Let the eqi^tion of tl^e sphere S be given by
fex*+y*+z®+2t/x/+2vy+2M»z+</=0. .
...(1)
Suppose the co-ordinates of the given points P a»d Q are
(^1.
Zi) and (xa, ya, Za) respectively,
sphe^fin”**^*^”
w.r.t. the
xx^+yy,+ZZ1+II(x+xi)+v(y+y,)+w (z+z,)+</=0.
The point Q (xa, ya. ^a). will lie on this.plane, since P and Q
are conjugate points and hence, we have
*i*a+yiya+ZiZ8+tt(^1+Xa)+V (yi+ya)+w(Zi—Za)+d=0
The equation of the sphere on PQ as diameter, is
(x~xi)(x-xa)+(y-yi)(y-7«)+(z-^i)(r-Za)«0
or
x*+y*+z*-x (X,+xa)~y(yi+ya)-z(z,+Za)
+(XiX8+yiya+ZiZa)=0.
...(3)
. Now the sphere(3) will cut the sphere (1) orthogonally, if
(^1+Jfa): jr
(yi+ya).
^ (^i+^a),w
«d+(x,Xa+yiya+ZiZa)
py *i^8+yiya+^iXt+tt(:«i+JCa)+v (yi+ya)+w (Zi+Za)+d=0
which is true by virtue of(2). Hence proved.
293
The Sphere
Ex. 8(a). Obtain the condition that the spheres
a {x^-^y^-^z*)-\‘2lx-{‘2my-\-2nz-\-p<=Q
and
may cut orthogonally.
(Paojab 1975. 81)
Sol. The equations of the given spheres may be re-written
as
and
or
+2*+2 (//o) x+2(w/a) y-\-2(nia) z+(/>/fl)=0, ...(I)
(*“/*)=0.
...(2)
The spheres (1) and (2) will cut orthogonally if
2 {iJa).0+2(m/fl).0+2 {nla).0=^(p(a)+{^k^lb))
[See § 12]
a
b
or
pb^ak*
which is the required condition.
Ex.8(b). Find the condition that the spheres
and
+>*+2*+2«x+2v;;+2WZ+</=0
may cut orthogonally.
(Punjab 1979)
Sol. Proceeding as in Ex. 8 (a) above; the required condition
is a^-d.
Ex. 9. Two spheres of radii r% and r^ cut orthogonally. Prove
that the radius of the common circle is
rxr2jy/{r^’\-r^),
(Kanpur 1977. 78; Rajasthan 77; Meerut 85. 87P)
Sol. Let the equations of the common circle be taken as
z~ 0.
...(1)
The radius of this circle is clearly a .and we are required to
find it.
The equations of the two spheres through the circle (1) are
given by
2AiZ=0,
...(2)
and
.(3)
(.X* -\-y^ -f 2*- a®)4-2AaZ—0.
The radii r^ and r^ of the spheres (1) and(2) are given by
■
or ri®=Ai*+a*
and
r2*-Aa®+aa.
Again the spheres (2) and (3) intersect orthogonally, so apply
ing the condition
we have
●2MiMa+2vi Vg+2wiWa == </i+d^,
0H-0-i-2AtA2=—a®-a* or AiAa*=—a*.
A,*Aa®=a*
Squaring,
or
(fa*—a*) -a« or fi*/*a*—a* (ri*+ra*)=0
294
Analytical Geometry 3-D
or
Ex. 10. Prove that every sphere through the circle
x*+>»*--2ax+r*«=(), z=0
cuts orthogonally every sphere through the circle x^-\-z*=r\ y=>0.
(Punjob 1977(S))
Sol. The equations of any spheres through the given circles,
are given by
...(1)
-1-2*—2fljic+r*-f2AiZ*=0.
...(2)
and
**+>»*+z*—»‘*-1-2A8^«0.
If the spheres(1)and (2)cut orthogonally, then applying the
condition
*iuiUi’{-2vtV2+2wiWi^di-\-di, we have
2.(-fl).0+2.0.Aa-|-2.A,.0=r*-f(-r*)
0=0
or
which is true for all values of Ai and Ag. Hence the result.
Ex. 11. Find the egupfton of the sphere that passes through the
circle x*+y*-{-2*—2x+3y-^42+6^—0t3x—4y+5z-‘l5c^0 and cuts
the sphere x*-|-;>*-|-2*+2x-l-4y—^z-Hl=0 orthogonally,
Sol. The equation of any sphere through the given circle
is
2x-14z-f6H-A (3x—4j/-|-5z—15)=0
*®+>*+2*+(3A-2) X-H3-4A)j»+(5A-4) z-K6-15A)=0.
...(1)
The equation of the other given sphere is
...(2)
**+A'*+2®+2xH-4^-6z-f11 =0.
If the spheres (1)and (2)cut orthogonally, then applying the
condition ●
*2«i«i-|-2yiVg+2»ViWg=</i-l-<fa*. we have
(3A-2) ,. ,, (3-4A)
, 15A-4)
2.
1-1.2, -5 ,.2-1-2. ^'
.(-3)=(6-15A)-l-n
or
3A_2Hr6-8A-M2-15A=17-15A
or
16-5A=17 or A=-I/5.
Putting the value of A in (1), the equation of the required
sphere is given by
5(x*-l-3»*-fz*)-13x-|-19j»-25z-f45=0.
Ex. 12. Find the equation of a sphere which touches the plane
3x^2y~-2+2*^Q at the point (I, -2,1) and also jcuts orthogonally
the sphere x*-fj>*-l-2®—4x-l-6r;>-l-4=0.
/■
(Avadh 1978,80; Agra 80; Punjab 76; Jodhpur 76)
The Sphere
295. ,
Sol.
The equation of the given plane is
...(1)
3x-|-2^—2+2=0.
It is given that the plane (1)is a tangent plane to the required
sphere at the point i4(l, —2« 1) and hence the line joining the
centre Ci of the required sphere and the point A is . perpendicular
to the plane (1). The d.r.’s of the normal to the plane (1)(/.c. of
the line ACi) are 3, 2, -1 and so the equations of the line ACi are
given by
...(2)
Since the centre Ci lies on the line (2). therefore the co-ordi
nates of the centre Ci.may be taken as
(3r+l,2r-2, -r+1).
The radius of this sphere^Ciil
= V{(3r+1--l)”+(2r-2+2)*-f(-r+1-!)»>« \/(14r*).
The equation of the other given sphere is
(3)
x*+y^+z^-4x-\-6y+4=0.
The co-ordinates of the centre C2 of the sphere (3) are (2, —3,
0)and its radius= V(2*+3“-|-0>—4)=3.
Now the two spheres will cut orthogonally if
(radius of one sphere)*+(radius of other sphere)*=(CiC2)*
i.e. rV(14r*)]*+(3)*=(3r-l-l-2)*+(2r-2-l-3)>+(-r+l-0)*
or
14r*+9=9r*-6r+l-l-4f*-f4r-hl+r*-2r-M
or
6=—4r . or . r=s=—3/2.
Putting the value of r in the co-ordinates of Ci, the centre of
of the required sphere is
(-t-H.-3-2,1+1) or (-i,-5,f)
and its radiu8=‘\/04r*)=V'(14.S)i=f V(i4).
.*. the equation of the required sphere is'given by
(x+7/2)«+(>^+5)*+(z-5/2)*«lt.V(14)]*
or
;t*+;>»+z*+7*+10^-52+12«0.
Ex. 13. Prove that the general equation of all spheres through
the points ia,0,0),{0,b,0) and {0,0,c) is
x*+>'*-l-2*—
C2—A (x/fl+y/6+2/c—1)«0.
Find also the value of A so that this sphere may cut orthogonally
the sphere
x2+;»>+2*—2<ix—26;>—2c2=0.
Sol. The three given points are A{a, 0, 0), B(0, b, 0) and
C(0,0, c).‘ Let us take the fourth point as 0(0, 0, 0).
296
Analytical Geometry 3-D
The equation of the sphere OABC is given by
S^x^-{-y^-j-z^—ax~by—cZ‘=^0.
...(1)
[See Ex. 6 (a) of the set(A)]
The equation ot the plane ABC is given by
Psxla-\-ylb^zJc— 1 *= 0.
...(2)
The general equation oflhe sphere through the intersection
S’~AP«0
of(1) and (2) is
or
(x/fl+>'//>+2/c-l)'=0.
...(3)
This proves the first part of the problem.
The equation (3) of the sphere may be rewritten as
X»+j^»+2*-(fl+A/fl) x~(fr+A/h);;-(c+A/c) 2-)-A=0. ...(4)
The equation of the other sphere is given as
x^-\-y^+z^—2ax—2by—2cz‘=0.
...(5)
The spheres (4) and (5) will cut orthogonally if
(-fl)(-a-A/a)+(-h)(-.h-.A/6)4-(-c)(-c~A/c)-A+0
or
(Q3+ft8+c>)+3A = A or A--(oH^Hc’*)/2.
Ex. 14. Prove that a sphere iS=»0 which cuts the two spheres
5j=0 and S'acaO at right angles will also cut the sphere AiiSi+A25a'=0
at right angles.
(Allahabad 1975; Garhwal 82)
Sol. Let the equations of the spheres be given by
fex®
+2*+2mx+2vy-}-2h-2+rf™ 0,
a X*
+2«iX -f-2viy+2wi2+di=0,
and
Sa a X*+y8+2^+2waX+2vay+2Wa2+</a=0.
The sphere 5=0 cuts the sphere 5i=0 orthogonally, so we
have
2mmi+2vvi-|-2wiv2c=^
...(1)
The sphere 5«=*0 cuts the sphere 5a=»0 orthogonally. so we
have
2««a+2vva+2H'tVa'=-'i/+d3.
The equation of the sphere Ai5i-[ Ag^a=0 is given by
^1^1 *i' Agi/a
Aj+Aa
*H/+2*+2|
...(2)
AiVi-f-AaVa
A1+A2
),+2(
Ai+Aa )-0. ...(3)
The sphere 5»0 will cut the sphere Ai5i+Aa58=0 (i.e., the
sphere (3)J orthogonally if
'? -
Ai^i+AaVa
AiM'i + AaWa
2a Ai«i -{■ AaUa
( Ai + Aa )+2v( Ai+Aa )+2 h.( Ai + Aa )
'
^sAa
Ai i Aa )
297
The Sphere
or
7,u (Aitti-{-AaMjj)-l“2v (AiViH“A2Va)-|-2H> (AiVVi-|-AaiVa7
=d (A,4*Aa)+(rfiAi+i/aA*)
or
Ai (2ttili4-2vVi+2wH'i)-|'A2 (2«Ma+2vVa+2»vWa)
=Ai (rf+<^i)+Aa (rf+^a)
or
■—Ai (^+</i)4*A2 (d^di)
\
^ sing (1) and (2)]
which is true for all values of A| and Aa. Hence the result.
Aj
Ex. 15.
and
Pfove that the tangent planes to the spheres
4.^2+29+2uiX 4 2vi>> 4 2^12 -|0
4 ;>*+2*+2«a*+2 V2>'4-2 WaZ+i/a «= 0
at any common point are at right angles if
2ttiM242viVa4-2M'iWa—i/i4</a-
(Allahabad 1975,78)
Sol. By § 12, we know that at a common point the angle
between the tangent planes to the two spheres is defined as the
angle of intersection of the two spheres. As the angle between the
tangent planes is given to be a right angle, the given spheres will
cut at right angles (re. orthogonally). Hence this is the same
problem as § 12.
Ex. 16. Find the equation of the sphere which cuts orthogonally
each of the four spheres
X*
+z” - a*+6*+c*, X* 4- y* 4- z® 4- 2ax=o*,
Jc*4-)'“+zH26;'-- 6®, x^-\-y’^-\-z^^2cz=-c^.
Sol.
Let the equation oi the required spbeie be
x*+.V®4-z®4-2ux42v>-+'i»vz4i/=^0. .
..●(1)
The equations ot the giveu spheres are
...(2)
x^ 4;^“ 4 z* 4 2flx—a^ =■ 0,
4zH 26;;—i>“« 0,
and
**4)'*4z*42c2-- c2<=0.
...(3)
...(4)
...(5)
If the sphere (i) cuts the sphere (2) orthogonally then apply
ing the condition ‘2«i«2-|-2viV842wiH;8'=^i/i+i/8’, we get
2M.042v.042iv.p=rf-(a94-i>a+c2)
or
...(6)Similarly it the sphere (1) cuts ihe sphue-. (i), (4) and (5)
orthogonally, we respectively hav«
.. \
...0)
...(8)
2h'c ■ a
<,9.
298
Analytical Geometry 3-D
Putting the value of d from (6) in (7),(8) and (9), we get
2tt=(6*+c2)/a. 2v=(c*+flW, 2w=(fl«+
Substituting the values of 2u, 2v,2w and d in (1), the equa
tion of the required sphere is given by
JC+{(c>+flW}
z
+(fl*+b*-j-c*)=?0.
Ex. 17. The variable sphere x*+y*+z*+2MX+2vy+2B>2+l=0
always cuts the sphere 3x*-i-3;^“+3z®—6x+10y+z—8=0 orthogo
nally, Show that the point(u, v, w) moves on afixed plane.
Sol. The equations of the given spheres are
...(1)
jc2
284.2mx+2v;>+2k>z-1-1=0
and
*‘+>Hz>-2*+^;-+5 *-'3=0.
...(2)
The spheres (1) and (2) will intersect Orthogonally if
2m.(-1)+2v.(5/3)42m;.(1/6)=1-8/3
or
—6tt+10v+»v= —5 or 6m—lOv—w-t-5=0.
The locus of the point (u, v, w) is
6x-10;»-2+5=0
which is a fixed plane.
Ex. 18. Show that the spheres .
jc24/4z^=25, xH>'®+z*~18^-24y-402+225«0
touch and show that their point of contact is (9/5, 12/5, 4).
(Berahampnr 1981S)
Sol. The equation of the first sphere is
...(1)
its centre Ci is (0, 0,0) and radius ri=5..
The equation of the second sphere is
jt9^y^28-18x-24y-4024-225=0.
...(2)
Its centre Ca is (9, 12, 20) and radius
ra= v'(9*4'12*4202-225)«20.
The distance between the centres^ CiC%
=V{(9-^0)*+(12r-0)a+(20-0)2=v'(625)='25.
Also the sum of the radii=ri+ra=5+20=25.
Thus the distance between the centres of the spheres(1) and
(2)is equal to the sum of their radii. Hence the given spheres
touch externally. (See § 13).
To find the point of contact. Let (Xj, yu Zi) be the point of
contact. The equations of the tangent planes at (xi, yi, Zi) so the
spheres(1) and (2) respectively are
The Sphere
299
...(3),
xxx^yy^’\r2^l^7.5
and xxi+yyi-^zzi—9(x+Xi)—12(3^+;'i)—20(2+zi)+225«
or x(Xx-9)4^:f(J'i-12)+2(zi-20)
...(4)
-(9xi+l2;^i+20zi-225)«0.
Since the spheres (1) and (2) touch each other, therefore the
tangent planes (3) and (4) must be the same.
Comparing their coeflBcients, we have
Xx-9_yi-l2_Zi-20 9x,+12j’i+Mzi-225_,.
,23
Zl
Xi
yi
«.. .
or
●
xj—9=«fcxij^i—12*=/fj'i.Zi—20=*A:Zi
Xi=9/(l~fc). >'i“12)/(l~fc). 2i«=-20/(1-A:).
».(6)
Also from the last two relations of(5), we have
9xi+12;»i+20zi-225=25A:
81
or
,I l^Lv^-225=»25fe [putting the values from (6)]
or 62^-225(l-fe)«25 (1-A:) fc,or k^^^k+16^0
or (/c44)>«0 or fc=>—4.
Putting the value of fe in (6), we get
Xi=9/5,>»i=»l2/5.zi*=4.
The co-ordinates of the required point of contact are
(9/5,2/5.4).
§14. The length of the tangent.
Tofind the length of the tangentfrom a point A (Xi, yu Zi) to a
sphere.
Let the equation of the sphere be
...(1)
S^x^+y*+z^+2ux+2vy+2wz+d>=0.
Its centre is
—v,—K»)
/O,
and radius=V(«*+v®+H»3-^)Let AP be the tangent line
from the point ^ to the sphere
C
(1) and P be the point of con- ^
tact. The length of the tangent
is defined as the distance between
the points A and P.
Clearly the radius CP is perpendicular to i4P. Hence
AP^=AC^-CP^
=l(Xx+«)"-l-(>'i-l-v)>-l-(Zi+»v)*}-(«!+v*-fw*-</)
...(2)
=XiH yi® 4-2a®+2mXi+2v;;,-h 2wzi-l-d.
The expression on tLw right hand side of(2) is usually deno
ted by Si l.e.
300
Analytical(leometry 3-D
AP^^S^-^Xi^-\ryx*+Zx^-^2uxy+2yyx-\-2wzi-{-d.
This represents the square of the length of the tangent AP, or
the power of the point A. [See § 6 (B)].
Working Rale First make the coefficients of x*,y'^andz^
each unity in the equation of the sphere; then replace x,y and z by
the co-ordinates of the pointfrom which we are to determine the
length of the tangent; the value so obtained represents the square of
the length of the tangent,
'x
j
§ IS (A). The radical plane.
Definition 1. The radical plane of the two given spheres is the
locus ofa pointfrom where the squares ofthe lengths of the, tangents ‘
to the two given spheres are equal.
Definition 2.
The radical plane of the two given spheres is the
locus of a point whose powers with respect to the two given spheres
are equal.
Clearly the two definitions of the radical plane given above
are equivalent.
To find the equation of the radical plane.
. Let the equations of the two given spheres be
St^x^+y^-i-z^^2u,x-h2viy+2wiz+di‘=0,
and
Si+y‘+2‘+2u2X-h2vgy-i-2waZ-l-d2-^^0.
Let (Xi, yj, zj) be the co-ordinates of any point A. By defini
tion the locus of the point A is the radicfil plane if the powers of
A with respect to the two given spheres be equal. So we have, the
power of A w.r.t. the spheie
«the power of A w.r.t. the sphere(5'2=>0)
V-I-2w,Ai-j+2WiZi
i.e.
or
2 (iVi-Ma) Xi l 1 (V|- Vj) Vi+2(Wi~Wa) Zi-^-diThe locus of A (a*i, ,vIt Zi) i.e. the radical plane of the
two given spheres is
or
2(Ur u-i) X i 2(Vi- ra).F-i- 2(Wi-u’a) 2+rfj—</a«0
Si—*Sa=0.
§ 15(B). The properties of the radical plane.
Property I. The radical plane rf two spheres is perpendicular
to the lint joining their centres.
The d.r.’b ot the uomial lo the radical plane (1) arc «i—Wa,
Vi—Va, Wj -w..
Ihe co-ordinates of this etntres of the spheres 6\=0 and 5a=0
are (—«i, —-V|, — Wi) and (-u^, -->'2, —Wi) respectively.
The Sphere
301
Hence the d.r.’s of the line joining the centres* are «i—«a.
t>i—Vg, M>i- H'g which are same as that of the normal to the radical
plane.
Property II. The radical plane of two spheres passes through
their points of intersection.
Let the equations of the two given spheres be given by Si*=>0
and Si*=»0i The equation ofthe radical plane Is given by S'!—5#=0.
Clearly the points which satisfy both
and Si=>0 also satisfy
Si-~Sa=*0i
Hence the radical plane S'!—5*a«=0 passes through the common
points of intersection of the spheres 5'i=0 and 5’a'=0;
(Avadh 1982)
§ IS (C). The radical line (or radieal axis).
Definition. The radical planes of three spheres taken two-at
a time pass through a line which is said to be the radical line (or
radical axis) of the three spheres.
Let the equations of the three spheres in their standard form
be given by 5i==0,52=0 and 58=0.
Taking two spheres at a time, the radical plane of
5i=0 and 5a=0 Is 5i —5a=0.
Similarly the radical plane of 5a==0 and 5a»0 is 5a—58=0
and the radical plane of 5s=0 and 5i=0 is 53—5i=0.
These three radical planes clearly pass through the line
5i= 52=58
which is the equation of the radical line (or radical axis) of the
three given spheres.
§ 15 (D). Radical Centre.
Defloltion. The radical lines of thefour spheres taken three at
a time meet in a point which is said to be the radical.centre of the
four spheres.
Let the equations of the four spheres in their standard form
be given by 5i=0,58=0,53=0, 54=0.
Taking three spheres out of these four spheres *Ca i.e. 4
radical lines will exist. The equations of these four radical lines
are given by ' ’
Si^Sa^Sa't Sa—Sa—S^i 58=54=5ij 54*=>5i=58.
These lour radical lines clearly intersect in a point, called the
radical centre, given by 5i=5a=58=54.r
Clearly the radical centre will also lie on all the radical planes
of these four spheres taken two at a time.
§16. Coaxial system of spheres.
Definition. A system (orfamily) of spheres is called a co-axlal
302
Analytical Geometry 3-D
system ofspheres Iffor all spheres any two of them have the same
radical plane.
Remark. We have proved in property I of § 15(B)that the
line joining the centres of any two spheres is perpendicular' to
their radical plane. In the case of co-axial system of spheres any
two spheres have the same radical plane, hence the centres of all
the spheres of a co-axial system always lie on a straight line.
Note. Some authors write co-axal for co-axial.
(A) To find the general equation of the system of co-axial
spheres.
Let Si=«0 and'5s=0 be the equations of two spheres. The
radical plane of these spheres is given by Si—St=^0.
Now
...(1)
{Si—Sa)=0
represents a system of spheres for different values of ft and any
two members have the same radical plane. Let in and ft2 be the
two values of ft and so two members of the system (1) are given
by
iSi-}-fti (S'!—S’a)=0 and
— 5ia)=0.
Subtracting, the radical plane of these two spheres is given
by
('S'l—5^a)®“0 dr Si — Si^O [V- fti^^fta].
Hence any two members of the system of spheres given by (1)
have the same radical plane. Hence (1) represents a co-axial
system of spheres.
Again (1) may be re-written as
(1-h/t) 5i-/t5a«0 or 5i-{ft/(l +ii)} 5a=0.
This is of the tbrm
5i-f-A5a=0
...(2)
The equation (2) is the general equation of a co-axial system
of adheres.
Similarly if
be a sphere and P=0 be a plane then
S+XP^O
..(3)
represents a co-axial system of spheres.
(B) To find the simplified form of the equation of co-axial
spheres.
(Punjab 1980; Madras 76)
Let us take the centres of all the spheres of the system to lie
on the x-axis so that the y and z co-ordinates of the centres of all
the spheres of the system are zero whereas the x-co-ordinate is
different for different spheres.
The equation of such a system of spheres is given by
:e>+/+2*+2MX-i-rf=iO.
...(1)
303.
The Sphere
Let two members of the system (1) be
The radical plane of these two members (spheres) is
2(tfi—*/a)
t/a“0.
...(2)
The eqaatioQof the radical plane (2) depends upon the values
of d and hence will be in general different for different pairs of
spheres belpnging to the system (I).' If any two spheres belonging
to the system (1) should have the same radical plane, then d must
be an absolute constant. Then the common radical plane of any
two spheres belonging to the system (1) is the plane X'?>0.
Hence the equation (1) is the equation of a co-axial system
of spheres with the fqmmon radical plane as the plane x«=»0,
where u is a variable and d is constant.
(C) Limiting points of a co-axial system of spheres.
Definition. ~The centres of the spheres of a co-axial system which
have zero radius are called the limiting points of the co-axial system*
To find the limiting points. Let the equation of the co-axial
system of spheres be given by:
(1)
x*+>*+2*+2ttX-l-rf=0,
where u is a parameter and d Is an absolute constant.
When written in the centre-radius form, the equation (1)
becomes
...(2)
(x-f-tt)2+(;'’-0)»-Hz-0)*«(«*-d).
The radius of the sphere(2) is ^{u^—d). If the radius is
zero, then
d=0 or u~:^\^d. The co-ordinates of the centre
of a sphere of the system (1) are (—m,0,0). Putting the value
of n<=3±<\/^, the centres of the spheres of zero radius (f.c., the
limiting points of the system) are(±V^» 0, 0).
Thus the limiting points of the co-axial system of spheres(p
areC^?'^, 0, 0),(0, 0).
‘ Role. To find the limiting poirits of a co-axial system^ equate
to zero the radius of the system and then the corresponding co-ordlnates of the centres give the limiting points.
SOLVED EXAMPLES(E)
Ex. 1. Prove that the members of the co axial system intersect
one another^ touch one another, or do not intersect one another
according as d <, ●=> or > 0.
Sol. Let the co-axial system of spheres be given by
.(1)
x*^y^^z^-\-2ux-\-d=ti
where d is an absolute constant and n is a parameter.
304
Analytical Geometry l-D
The common radical plane of the system (1) is
The radical plane xc>0 intersects the sphere (1) in a circle
given by
or given by
rf, X=»0.
...(2)
The equations (2) represent a circle of radius y/{-d) in the
7^-pIane.
Nqw the following three cases arise:
Case I. When d < 0 i.e. d is —ve.
In this case (—rf) is +ve and the radius VC—d) is real t.e.
the circle of intersection is real, and hence the two spheres ofthe
system Intersect.
Case 11. When d(=0; The radius y/{—d) of the circle
reduces to zero I.e. the circle reduces to a point circle and hence
the two spheres of the system touch.
Case 111. When d > 0 i.e. d Is +ve. In this case the radius
y/{—d) is imaginary i.e. the circle of intersection is imaginary,
and hence the two spheres of the system do not intersect.
Ex. 2. Prove that every sphere that passes through the limiting
points ofa co-axial system cuts every sphere of that system ortho
gonally.
Sol. Let the system of co-axial spheres be given by
2mx-h
0.
..(1)
The limiting points of the system (1) are
0,0) and
0, 0).
Let the equation of any sphere be
+2*+2tfiX-h 2 +2 -b =0,
...(2)
If(2) passes through the limiting points of the system (1),
then we have
and
or
”1”0”1“0"*2tti
0*1"0■}“
0
di-luyy/d-^dx—Q and </—
rfj=.0.
Solving these, we have
Hence the equation (2) of-the sphere becomes
+zH 2 V,
2Wi2 0.
...(3)
If the sphere (3) cuts the system of spheres (I) orthogonally,
then using *2uiUx-\-2viV^-{-2wiWi^di-\‘di\ we have
2.«.0+2.0.v/-|-2.0.»vi=rf-rfor 0=0
which is true for all values of Vx and Wy.
The Sphere
.305
Hence (3) intersects the system of co-axial spheres ortho¬
gonally.
Ex. 3. Prove that the equation
where p and v are parameters, represents a system ofspheres passing
through the limiting points of the system xa-H/+z*+2M.v+</=-0 and
cutting every, member of that System orthogonally,
Sol. This is another way of stating the Ex. 2 above. Simply
replace Vi by p and Wi by v here.
»
Ex. 4. Show that the spheres whi^'h cut two given spheres along
[Kaupur i9t8]
a great circle alt pass through two fixed points.
Sol. Taking the line of centres as the x-axis and the equa
tion of the common radical plane as x=0 (/.c. ^2-plahe), the
equations of the two given spheres are given by
...(1)
X®-f
z*
^=0,
...(2)
and
X®+ -I-z®+2i/2X-|-<i=0.
Let the equation of anv other sphere be given by
...(3)
x®-f v®-rz®-f 2ux-i-2v>H-2>vz-f-A:=0.
The sphere (3) will cut the sphere (1) along a great circle, if
their radical plane namely
2x (U-Ui)+2vy+2wz+k -d=>0
passes through the centre (—«i,0, 0) of the sphere (1), and there
fore, we have
...(4)
-2mi {u-Ui)-\-k-d=0.
Similarly the sphere (3) will cut the sphere (2) alongtia great
circle if
..(5)
—2u2(u-Ui)-\-k—d=0.
Subtracting (5)from (4), we get u=fUi-^Uf.
putting the value of w in (4), we get fc=2i#i«|-ftf:
Since the spheres (1) and (2) are given, the quantities Ui, Ws
and d are constants and hence u and k are also constants.
Now it remains to proVe that the sphere (3) passes through
two fixed points.
Thi abscissae of thfe points of intersection of the sphere (3)
with the x-axis i.e. with y=0=z are given by
x®-l-2t/.v-f/:=0.
.
Both the roots of this equation depend upon u and k, which
have already been proved to be constaiits.
Hence every sphere given by the equation (3), [different
spheres exist for different values of v and w] passes through two
fixed p >ints on the x-axis.
306
Analytical Geometry 3-D
Ex. 5. Find the limiting points of the co axial system of
spheres determined by the spheres
3x-3>^+6=0, xM-6z+6=0.
Sol. The equations of the given spheres are
.-(0
'3>;+6=0,
and
...(2)
5"=x*+7*+z3-6v—z+6=0.
The radical plane of the spheres (1) and (T) is given by
S~S'=0 i.e. f=3x+3;;+6z=0 or P=x+y+2z=0,
d
The equation of the co-axial sysicm of spheres d:t:rrr
by the given spheres fl) and (2) is given by S-fAP=0
or
(x4.v-f2?)“=0
or
..-(3)
(3-f-A)-y (3 ~A)-l-‘2Az+6=0.
■ nd iis radius
^ -3-A’ 32
-A
/J(3+A)2 . (A -3)8
...(4)
HA)2 -6[=-| V(6A“ - 0)
■ Vl 4
' 4
We ktiow that the limiting points are the centres of the
spheres when radius=0. Hence equating the radius given by (4)
to zero, we get
6A2 -6=0 or A*=l or A=±l.
Putting the values of A in the co-ordinates of the centre of
the sphere (3), the limiting points are f - 2,1, - 1) and (- ● 1, 2, 1).
Ex. 6. Prove that the locus of points whose powers with respect
to two given spheres are in a constant ratio is a sphere co-axial with
the two spheres.
Sol. Let the equations of the two given spheres be
● (1)
+2^+2w,X 4-2v,.v+2WiZ+1/,=0,
and
.
..(2)
y2-f-zH2t/2X-|-2v.>>M 2WaZ-t-t/o= 0.
Let P (xj, yj, zj) be the variable point.
By the given condition.
the power of P (x,.
z,) w.r.t. the sphere
~ k,(constant)
the power of P (Xj, >’i, Zi) w.r:t. the sphere (2)
-yiH.Vi^4 zi^ 4- 2t<iXi-f 2viyi-f2u>iZi
,
or
V4-Zi*4-2Ma:Vi+2va,Vi-j-2»V2Zi4Cross-multiplying and simplifying, the locus of P (xi, yi, Zi)
is given by
2 {ku. «0 „. 2(A:v2 -V,)
^
:r24-y*4-z*4- ~k-\
y+-2{kw2—Wi)
k-\ '"
k -1
The centre of(3) is
+(kd2~dx)
k~l
...(3)
The Sphere
307.
The equation (3) clearly represents a sphere.
Now subtracting (3)from (1), the radical plane of the spheres
(I) and (3) is given by
2{Ui- Ui) JCH-2(vx-Vs) y -f2(Wi-Wg) z-\-{di...(4)
Similarly the radical plane of the spheres (2) and (3) is
2(«i - Ma) x-\-2 (vi Va) y^-2(Wi-Wj) z+(</i-t/,)=6
...f5)
The equations (4) and (5) are same and hence the sphere (3)
is co-axial with the given spheres (1) and (2).
Ex. 7. Show that the locus of a point from which equal tangents may be drawn to the spheres
j^j^y2^z2^2x
2z 1 =0, x^-{-y^’hz^-x+4y-6z~2‘-=0
. is the straight line (x -1)!2--{y -2)/5=(z— 1 )/3.
Sol. Let P(x, y, z) be the moving point. Then the lengths
of the tangents from the point P to the given spheres are
\Z(^*+.v®+z* -1), V'(^*+:v*+zH2x~2>-4-2r-l)
and
V(^*+,v*4-2®—
4v—6z—2).
By the given condition, these lengths are equal. Hence on
squaring and equating, we get
1 =xH.vH^®+2x—2;>-h2z-1
...(!)
—x'^+y'^^■z^—x-^Ay—6z—2,
From the first two relations, we get
..:(2)
2x-2;;+2z=0.
From the first and third, we get
...(3)
-x+Ay—6z—\—0.
The symmetrical form of (2) and (3) is
...(4)
(x-l)/2«(j> -2)/5=(z-l)/3.
Remark. By definition the line (4) is the radical line (t.e.
radical axis) of the three, given spheres.
Ex. 8. Find the equation of the radical axis in the symrnetrical
form of the spheres
S'iSx*+J'*+z*+2x+2;;-1-2z4-2=0,
5‘a=x*+:V®+z*+4x-|-4z-l-4=0,
- 4z-2=0.
and
(Garhwal 1978)
Sol. The radical plane of the spheres 5'x=0 and 5a==0 is
...(1)
5i-5^2=0 t.e. X .>'-^-z^-l—0.
Again the radical plane of the spheres S'i=0 and 58=0 is
..●(2)
X—4;;+6zH-4=0.
S^^S^—0 i.e.
The equations (1) and (2) are the non-symmetrical form of the
radical axis. The symmetrical form of the radical axis is
x/2-=(^-l)/.^=z/3.
308
Analytical Geometry 3-D
JEx. 9. Show that the locus of the centre of a circle of a radius
<2, which always intersects the co-ordinates axes (rectangular) is
X V(a^~ y^~z»)+y ^/(a^-z^ x^)+z ^(a^ -x^-y^-)=a^
Sol. , l^t the centre of the circle be (a, p, y) and let it cut
intercepts p, q, r on the axes so that (a, /5, y) lies bn the plane
X
K
c.
.
y
z
q
r
B
y
.
...0)
=1.
...(2)
^
Since radius of the circle= a.
/. Distance of(a, P, y) from (p, 0,0)=a
(_a+p)-*+j5=*+f=fl2
=». - p=y/(a^y*)-f a
Similarly q^y/{a^~y^~oy)-\-p
r=v'(«*-a^-y?®)+r
1
1
a-V(a^
●●
P “ «+
g—
a2-(fl3-j88~..8)
P^—y^)
Similarly
1 y-y/(,i^^ofi -/?2)
r ~(ofi+P^-\-f)-a^
Put these values in (2)
4-y {y—V(a®—a**
»\ Locus of(a, (i, *●) is
xi/(a^—y^—z^)-j-y \/(a^- z^-x)+z y/(a^- x^ -y^)=a^
Exercises
1. Find the equation of the sphere which passes through the
origin and makes equal intercepts of unit length on the
axes.
Ans, j^+y-+2^—x - y—z=0.
2. A point moves so that the ratio of its distances from two
fixed points is constant. Show that its locus is a sphere.
3. Find the centre and radius of the circle
^“+P*+2®+A'-f-p+2:—4=0, x-l-y-i-z=0.
Ans. Centre is the point (0, 0, 0) and radius is 2.
4. Find the equation of the sphere through the circle
x-t-3y-F4z=5
The Sphere
309
and the point (1, 2, 3).
(Agra 1982; Gauhati 78; Gorakhpur 81)
Ans. 3(x^+y^+z^)~2x-3y-4z~22-0.
5 Find the equations of the spheres which pass through the
circle
x2+/+2a=4, 2JC^-4J;^-2=9
and touch the plane 2=0.
Aus. x^-^y^+i^—2x—4y—z{-5=i)
and
5 (x*+/+z*)-8x—16;;-42-}-i6=0.
6. . Show that the two circles
x^'\-y^-\-z^-y-\-2z=i), x-y-\-z=l;
x^-{-y^+z^-\-x-3y-i z-5=0, 2x—y+4z=\
lie on the same sphere. Find its equation.
(Lucknow 1979, Avadh 79)
Ans. ^®+)'2+22+3jf-4>+52—6=0.
7. Find the equation of the sphere whose centre is the point
(1,2,3) and which touches the plane 3x+2y-\-z+4=0.
Show also that the radius of the circle in which the sphere is
cut by the plane x-\-y+z—0 is
Ans. x:--\-y^-\-z-2x~4y-6z=0.
8. Show that the spheres
x^+f-¥z^-4x-2y+lz 3=0
and
x2+>;2+2*-8a:-8v-10z+41=0
touch externally.
9. Find the equation of a sphere which cuts four given spheres
orthogonally.
10. Prove that all the spheres, that can be drawn through the
origin and each set of points where the planes parallel to the
plane xfa-\ yjb+z!c=.{) cut the co-ordinate axes, form a
system of spheres which are cut orthogonally by the sphere
x^-\-y'^-\‘Z^-\-2ux-\-2vy-\-2wzt=^Q^
if
au-\-bv-\‘Cw—Q.
10
The Cylinder
§ 1. Cylinder.
(A) Definition.
(Kanpur 1979; Lucknow 77)
A cylinder is a surface generated by a moving straight line which
moves parallel to afixed straight line and intersects.a given curve or
touches a given surface.
The curve is called the.guiding carve and the fixed straight
line is called the axis,of the cylinder.
Any line on the surface of a cylinder is called its geueraior.
(B) Tofind the equation ofa cylinder whose generators are
parallel to the line xfl=yfm=zln and intersect the conic
ax^’¥2hxy+by^-{-2gx-\-2fy+c=0, ^=0.
(Agra 1981; Garhwal 78S ; Kanpur 77, 81; Rohilkhand 81)
The generators of the cylinder art given parallel, to the line
xll=^ylm=z/n.
Hence the axis of the cylinder is given by the equations (1);
Let P (xiy y,; zi) be.a point on the cylinder, then the straight line
through the point P (xi, yi, Zj) and parallel to the axis (1) is a
generator and so its equations are
y~.Vi z -Zi
/
“
m
■ (2)
The equations of the conic are given to be
o<2+2^xy-l-fiy3+2gjf+2/yfc=0,z=0.
The generator(2) meets the plane z=0 in the point given by
x-Xj y—yt
/
“
m
Or-Zi .
t.e
Iz
●(**-«*>●>’*
fflZj
.0).
Since the generator (2) meets the conic (3), hence the point
(^1 /zi/rt, yi—wzj/n, 0) will satisfy the equations of the conic
given by (3), and so we have
311
The Cylinder
+2g(x.-^)+2/(
mz
n
0
or a (h.Xi—/zi)^+2A («a:i—/zi)
(nvi- wzi)®
+2g« («a:i--/z,)+2//i (nj»i-OTZi)+n*c=0.
The locus of P(Xi, yi, z,) /.e. the required equation of the
cylinder is given by
a{nx-hf-\-2h{nx lz)\ny-mz)-irb(ny-mzy
+lgn{nx~h)-\-lfn{ny—mz)-{-cn^=Q. ...(4)
Coro-lary 1. If the axis of the cylinder be taken as z-axis
whose d.c.’s are 0, 0, 1, then putting /= 0, w=0, n=l in the equa
tion (4) above, the equation of the cylinder becomes
(5)
ax^+2hxy-hby^-{-2gx-h2fyn-c=0.
Remark. The readers should recall that the equation{5) repre
sents a conic in two dimensional geometry whereas in three dimen
sional geometry the equation \5) represents a cylinder with its
generators parallel to the z-axis (the co-ordinate which is absent in
the equation).
Corollary 2. In view of corollary 1 above, the equation of
the cylinder whose axis is the z-axis and whose generators inter
section the circle
z-0 is given by
x^-\-y^==-a^.
(C) The equation of theform /(jf, y)=0 represents a cylinder
(Rr'asthan 1?75)
with its generators parallel to the z-axis:
Consider a curve in the x;/-plane whose equation m two
dimensional geometry is /(x, y)=0. If P (a, p) be any point on
it, then we have /1«,/?>=0.
Now the co-ordinates of P in three dimensions are (a, 0).
Draw a line ● through P(a, j5, 0) parallel to the z-axis. Take
0(ft, 3, z) any point on this line. The co-ordinates of the point
Q clearly satisfy the equation /(x, y)««0. Therefore the co-ordinates of every point on the line PQ satisfy the equation/(x,7)=0
and hence the whole line PQ lies on this locus.
Thus the assemblage of lines (parallel to the z-axis drawn
through the points on the curve in the x:;;-plane) is the required
locus which is, therefore, a cylindrical surface having generators
parallel to the z-axis.‘
Similarly, the equations f(y, z)c=*0 and /(z, x)«0 represent
cylinders with their generators parallel to x-axis and v-axis res
pectively.
312
Analytical Geometry 3-Z)
Corollary. The equation of the cylinder which intersects the
curve/(jc, z)—0, 4>(Xy y, r)~0 .and whose generators are para
llel to z-axis will be obtained by eliminating the co-ordinate z
between the equations/(jf, y, z)=:0 and
y, z)=0.
Similarly the equation of the cylinder intersecting the curve
f{Xy y, z)=0, ^{Xy y, z)=0 with the gener%tO|fS parallel to .v-axis
(or >>-axis) will be obtained by eliminating,the co-ordinate x (ot y)
between the. above equations.
§ 2. Right circolar cylinder.
(A) Definition. If the generators are always at a constant,
distance from the fixed straight line, the cylinder so generated is
called a fight circular cylinder. The fixed straight line is called
the axis and the constant distance is called the radius of the right
circular cylinder.
(B) Tofind general equation ofa right circular cylinder.
(Jodhpur 1978 ; Gorakhpur 79 ;Kanpur 83 ; Rajasthan 76;
Meerut 86 P)
.Let the radius of the cylinder be rand the equations of its
axis be
(x~a)/l=(y ■ 0j/m=(z ~y)/n.
...(1)
Clearly the axis (1) of the cylinder passes through the point
(a, /i, y), say the point
Consider a current point
P(X,.V, z) on the cylinder, /
PQ perpendicular from the U
point P to the axis (1). Then
V
i'G=radius=r.
'
r=r:
Pi
a
Now AQs=i\\Q projection of the join of A{a, j3, y) and (x, y, z)
on the axis (1) whose d.r.’s are /, my n
(X -tt)./■{‘{y—Pi.m-\-(z -y).n
i4P«the distance between the points ^ and P
=
-/?)H(z-y)2).
From the figure, we clearly have
AP^-AQ^=PQ»
or
{(x^u)^-\-(y—ijf^(z—f}—iLi^Z!!l±!Piy-‘^)-^n(z- ■7>y
or
{(x~«)2+(^^-^)a+(2_y)2}
-{/ (x-aj-hm {y-fi)-fn (z-y)}a=r2
tp)
The Cylinder
313
or {n (y-p)-ni (z- y)>*+{/(z ~ y)—n (x - ■ a)}^
...(2)
+im (A'-a)-/ {y -j3)}2==r*
The equation (2) is the required locus of P and hence this is
the required general equation of the cylinder.
Corollary. If the axis of the right circular cylinder be z axis
then putting a=fi=y=*0, /=wi=0, and n= I in the equation (2),
the equation of the right circular cylinder in its simplest form is
given by
§ 3. Tangent plane to a cylinder.
Tofind the equation, of a tangent plane to the cylinder whose
equation is
<7x*4-2/2xy+6j^2+2g.x:+2/v+c=0
at tho point P{Xt, yi, Zi). Also to prove that this tangent plane touches
(Gorakhpnr 1982)
the cylinder along a generator.
The equation of the given cylinder is given by
...(1)
ax^+2hxy+by^+2gx-[-2/jj;+c=0.
Since P(xi, yu Zi) a point lying on the cylinder (1), we have
...(2)
ax^^-^2hxi}\+ bye+2gxx+2fy^+
0.
Let the equations of any line through the point P (Xi, y^ Zj)
be
...(3)
(x-xi)//=(v-y0/m=(z -Zi)fn=r (say).
The co-ordinates of any point on the line (3) are
(/r-f.Vj,
«r+z,).
Therefore the points of intersection of the line (3) with the
cylinder (1) are given by
a (/rH-Xi)«+2/i (/r-{-Xi)(mr+yi)+b (mr+yiY
+2g (/r-|-.Yi)+2/(mr-|-;;i)-|-c=0
or r^ (aP+2hlm-\-bm-)-\-2r {/ (axiA-hyi-^g)-^ m
4-(flXi®+2/iXiyi+byi^4-2gxi4-2/;»i4c)=0
or r* (fl/24-2/i/m4^/M^)+2r (/ iaXi-\-hyi+g)+m (Ajci4-A7i4*./)>=0
...(4)
[using (2)]
The equation (4) being a quadratic in r, gives two values of r,
Clearly one value of r is zero. Hence, if the line (3) is a tangent
line to the cylinder (1), then the other value of r must also be
zero and the condition for the same is that the coefficient of r
must zero
Le,
...(5)
/ {aXi-\-hyi-{‘g)+m {hxi~\-byi~\-f)^0.
The tangent plane to the cylinder (1) at the point/>(xi, >vzi)
is the locus ol' the tangent line (3) for all values of /,
n and is obtained by eliminating the variables /, m, n between the equa-
314
Analytical Geometry 3-Z)
tions f3) and (5). Hence the equation of the tangent plane to the
cylinder (1) at the point P (Xi,
z,) is given by
(x-xi) iaxi+hyi-\-g)-h(y~ yi)(Axi+hyi+/)=0
or
(axi-^hyi-\-g)-{-y {hXiA-byiA-f)Haxi^i-W+^hXiyi
+gxi+fyi)=0
or a:(.aXi-\-hyi-^g)-\-y ihxi+byx+f)-{-(gxi+/i^i+c)=0[using (2)]
or axxi-hh ixyi-\rX,y)-^byyi-\-g {x+Xi)-\ fiy-{-}\)+c=0. ...(6)
Second part. The equation (1) of the cylinder i> of the,form
./T^» 3')=0 and hence, in view of § I (C) above, the generators of
the cylinder (1) are parallel to the z^axis. Again P (jc^, v„ z,) is a
point on the cylinder (1) and, therefore, the equations of a gene
rator through the point P Uj, y„ Zj) are
{X - Xj)/0=(y y^)f0={z -z,)/l
(say),
...(?)
The co-ordinates of any point on this generator are
(Xuyur'-^Zi).
Therefore, using the formula (6), the equation of the tangent
plane to the cylinder (1) at the point (xi, yi, r'-j-Zj) is given by
A {xyx-\-x^)A-byy^+g {x-\-Xx)-\-f{y-^yi)+c=0, l..(8)
Clearly v/e see that the equations (6) and (8) are identical.
The equation (3) of the tangent plane remains the same for all
values of r, it being free from r and, therefore,there is same tan
gent plane at every point of the generator (7). Hence the tangent
plane to the cylinder (1) at the point P (jCi,
Zj) touches the
cylinder (1) along a generator through the same point
^ixuy^.zi).
§ 4. £nvelopiog Cylinder.
(A) Definition. The enveloping cylinder is the locus of the
tangents to a surface {sphere or conkoid) which are parallel to a
given line or in other words the enveloping cylinder is the cylinder
whose generators touch a given surface and are directed in a given
direction.
(B) Tofind the equation of the enveloping cylinder of the sphere
JC®+3^“-|-z2=a2 whose generators are parallel to the line
x(l=yjm—zln.
tBohilknand 1982 ; Kanpur 76 ; Meerut 84)
Consider a point jP (a, /i, y) on the enveloping cylinder.
Since the generators of the cylinder are all parallel to the line
xjl—yfm ~zlny therefore the equations of the generator of the
cylinder through the point P are
315
the Cylinder
(x-a)//=(.v ~ p)/m=(ir--y)/n=r (say).
The co-ordinates of any point on the generator (1) arc
(/r+oc,
nr+y).
The equation of the given sphere is
...(1)
...(2)
The points of intersection of the generator (1) with the sphere
(2) are given by
(/r4-a)*+(mr-b p)2+(«r
or r* (/*-f»j2-i-n2)+2r (/«+w^+«7)+(«’^+P*+y’*—
’- O)
The equation (3) being a quadratic in r gives two values of r
and hence two points of intersection of the generator (1) with the
sphere (2). If the generator (1) touches the sphere (2) then the
roots of the equation (3) must be equal and the condition for the
same is
(/a+mp+ny)2=(/*+ m2+n")(aHAH7* - «*)
[using B‘==4AC].
The locus of P(a, % y) z.e. the equation of the envelop
ing cylinder of the sphere (2) is given by
fl*){l^+m^‘hn^)°{lx-hmy+nz)\
...(4)
SOLVED EXAMPLES
Ex. 1 (a). Find the equation of a cylinder whose generators are
parallel to the x=;'/2= -z and whose guiding curve is
3x2--|-2y2=l, z=0.
Or
Find the equation of a cylinder whose generators are parallel to
the line x~vll=—z aid passing through the curve
3x*+2>-2=l, z=0.
Sol. The equations of the given guiding curve ^are
...(I)
3xH2>'*=1,z=0.
The equations of the given line are
....(2)
x/l=y/2=z/(—1).
Consider a current point P (Xi,
Zj) on the cylinder. The
equations of the generator through the point P (xi,.Vi, Zi) which
is a line parallel to the given line (2) are
. (3)
(X - xi)/1=(y--;^i)/2=(z Zi)/(-1).
The generator (3) meets the plane z=0 in the point given by
i.e. (xi-i-zi, n-|-2zi. 0).
316
Analytical Geometry 3-D
Since the generator (3) meets the conic (1), hence.the point
(■^1+^1, yi+2zi, 0) will satisfy the equations of the conic given by
(I), and so we have
3{.'Ci+zi)H2(.ri+22,)2=l
{yi^+4y,Zi-\-4zi^)=l
^Xi^-h2yi^-\-11 Zi^-\-8y^Zi-\- Gz^Xi -1=0.
The locus of/* (.rj. V,, Zj) i.e. the required equation of
the cylinder is given by
3jc2+2^2+ 1 lz2-j-8;;2-|- 6zx -1 = 0.
Ex. 1 (b). Find the equation of the cylinder whose generators
are parallel to the line,x—y(( 2)=zf3 and passing through the
curve jf24-2;;2= 1, z ^ o.
(Agra 1982; Gorakhpur 74;
Meerut 83 S, 84 S; Kanpur 78; Lucknow 79)
or
or
3
Sol. Proceeding exactly as in Ex. 1 (a)
equation of the cylinder is given by
3 (,Y® +'y^+7^) - Izx -f 8 vz
Ex. 2
Find the equation to the cylinder
parallel to the line x =yj{ - 2) = z/3, and the
ellipse
;c2 4.2v2=1, z=3.
above the required
3 0.
whose generators are
guiding curve is the
(Avadh 1982; Allahabad 80; Agra 78; Garhwal 79, 81;
Kurukshetra 76; Meerut 77. 89; Punjab 77; Rajasthan 77)
Sol. The cquatfons of the guiding, curve are
x2+2j»8=!, z=3.
●●●(1)
The equations of the given line are
^/1=M ■2)=z/3.
(2)
Consider a point P (.Xi, )>i, z{) on the cylinder. The equations
of the generator through the point/> (.y^,
z^) which is a line
parallel to the given line (2) are
y-yi z-zi
^
-2 ● 3 ‘
...(3)
The generator (3) meets the plane z=3 in the point given by
^ -^1 y~yj 3~ :
1
- 2 “ 3 I.e. (xi - K+l,;’i+fzi-2, 3).
Since the generator (3; meets the conic (I), hence the point
(xi-izi-M, Vi-f fri-2, 3) will satisfy the equations of the conic
given by (1), and so we have
(^i-i^i+l)H2 (>»,-|.fz,-2)2=1
(3xr-2i+3)2-|-2 (3>^i+2z,-6)2=9.
,. The locus of P (x^ yi, zj) i.e, the required equation of
the cylinder is given by
or
The Cylinder
or
or
or
317
(3x-z+3)'^+2(3y+2z-6)^=9
Qx*-rZ*+9—6zx+I8x—6z
+2 {9y^+Az^-\-36+ \2yz-36y-24z)=9
9jc2+ 1 8jfH9zH24j;z-6za:+ 18x—72>'—48z+72-0
3jc®+6;;®f3z*+8;;2 2zx+6x -24;k -16z+24=0.
Ex. 3. Find the equation of the circular cylinder^ whose generating lines have the direction cosines^ /, m, n and which pass through
(Agra1980)
thefixed circle .\:*4-z“=fi*, in ZOX plane.
Sol. The guiding curve (circle) is given to be in the ZOX
plane i.e. y—0 plane and hence its equation are
x^+z'^^--a\ y=0.
Consider a point P(xu yi^ Zi) on the cylinder. The equations
of the generator through the point P {Xi^ y^i Zi) and with d.c.’s
fm^n are
...(2)
(x—Xi)//=(>’—Vi)/m=(z -Zi)/«.
The generator (2) meets the plane j;=0 in the point given by
x~xi O-yi z—zj.
i.e. (Xi ■ /Vi/w, 0, Zi - nyjm).
n
m
I
Since the generator (2) meets the curve (1), hence the point
(xi-lyilm,0, zi—nyjni) will satisfy the equations of the curve
given by (1), and so we have
. (xi-lyi/mY-\-iz^-nyJmy=^a^
or
.% The locus of P (Xi, y^ zO i.e. the required equation of
the cylinder is given by
{mx—lyf+(mz—ny)^^m^d^'
Ex. 4. Find the equation of the surface generated by a straight
line which is parallel to the line y=-mx, z=nx and intersects the
(Avadh 1981 ; M.U.90)
ellipse x-la--]-y‘’fb‘=\, z=0.
Sol. The equations of the guiding curve (ellipse) are
.. (»)
The equations of the given line may be written as
.. (2)
x/l=y/m=z/«.
Consider a point P (xi, y^ Zj) on the cylinder. The equations
of the generator through the point P (Xi, j'l, Zi) which is a line
parallel to the given line (2) are
...(3)
(x - Xi)l\ =(y -yi)!m -(z - z{)!n
The generator (3) meets the plane z=0 in the point given by
mzx
y -yi 0-Zi .
i.c.
yi— n
n
■ 1 ~ m
318
Analytical Geometry 2~D
Since the generator (3) meets the co :ic (1), hence the point
ixi—ziln, yi—mzjny 0) will satisfy the equation of the conic given
by U), and.so we have
(1;V)
(l//>^) 0’x-mzi//i)2=-1
b-(nxi -Ziy*-\-a\{nyi mz^y~~d'h‘^n^.
The locus
P f.vi, Vi,.z0 / e. the required equation of
the cylinder is given by
A- (nx—z)--\-ir (ny--mz)-~a'br'n\
or
Ex. 5 (a>. Find the equation of the cylinder which intersects the
curve ax--\-by^-y-cz-=\, lx-{ wy~rnz=p and whose generators are
parallel to the axis of x.
(Meerut 1978, 82 ; Lucknow 82)
Sol. The equations of the guiding curve are
ox~-i-hf-y-cz^=\
U)
and
lx~\-my-\-nz—p.
(2)
Now the equation of the cylinder whose generators are para
llel to x-axis will not coniain the terms of x. [See § 1 (C)]. Hence
the required equation of the cylinder is obtained by eliminating
X between the equations(I) and (2), and so is, given by
“f
/
V
or
or
or
From(2),
/
/a(p ● my - nzf-\-bl'y'-\-cF.
a(pHmlv--! ;j-2= -lpmy—2pnz-\- 2mnyz)+bFf+cl'Z~=^F
{anr4-bl‘).vM-(nn’+c/") z'-f2flm«vz—2nw/7V
~2anpz+{ap- -/')=0. Ans.
Ex. 5(b). Find the equation of the quadric cylinder which
intersects the curve a:c-yby--^cz' 1, lx-\my-rnz^p and whose
generators are parallel to z-axis.
(Kanpur 1982. 83 ; Buodelkhand 79 ; Meerut 85 P)
Th: equations of the guiding curve are'
ax^+by--\-cz^=\
..(I)
and
/x4mv+7zz==p.
.. (2)
Since the generators of the cylinder are parallel to the z-axis,
therefore the required equation of the cylinder is obtained by
eliminating the r co-ordinate between the equations (1) and (2),
and so is given by
Sol.
or
ax--~hy^^c {{p-~lx~my)jnY=\
f:i-y^-2/,lx+2lmxy-2pmy)=n=
The Cylinder
or
(c/2+fl«2)
319
y^^-2clmxy - Icplx- lcpmy
+(c/>?- /I*)—0. Ans.
Ex. 5 (c). Find the equation of the cylinder .with generators
parallel to the x-axis and passing through the curve
ax^-\‘by^=lcZy lx+my-)raz=^p.
Sol. The equations of the guiding curve are
..(1)
ax^-\-hy^~-7cz~0
.. (2)
and
lx-^my-\-nz=p.
Since the generators are parallel to the x-axis, therefore
eliminating .v between the equations (1) and (2), the equation of
the required cylinder is given by
a {{p my-nz)llf'\-b}’^—lcz~Q
or
a
m'y^^nH^ ~lpmy-\-2mnyz-^2pnz)-\-bFy^~2cFz=^
or
bF) 1*2+an^z^+2amnyz- 2apmy-2 {apn-f cF)z=0.
Ans.
■Ex. 5(d). Find the equation of the cylinder with generators
parallel to z~axis and passing through the curve
ax^+by^—2czt lx+my-y nz:=p.
(Lucknow 1977; Kanpur 79)
Sol. Proceeding as in Ex. 5 (6) above, the equation of the
required cylinder is
anx^-\-bny^-\-2c{lx4my)—2pc=-0.
Ex. 6. Find the equ ition of the right circulur cylinder of radius
2 whose axis passes through (1, 2, .3) and has direction cosines proportional to 2, -3, 6.
( Vgra 1976, 7R; Lucknow 83; Garhwal 78, 82;
Meerut 74, 77, 83, 84 P, 85; Kanpur 80)
Sol. The axis of the cylinder passes through (1, 2, 3) and has
d r.’s 2, —3, 6 and hence its equations gre
.. (1)
(x-l)/2=(>» 2)/(-3)-(z -3)/6.
Consider a point p[x,y,z)in\ the cylinder. The length of
●the perpendicular from the point F (x, v, z) to the given axis (1)
is equal to the radius of the cylinder i.e. 2. Hence the equation
of the required cylinder is given by (See § 2 (B) equation (2). Put
3; and r—2 in the equation 2)
/=2, m-= -3, rt =-6; a-- i, ^=2,
of the cylinder'.
(6 {y~2)~{ - 3) (r-3)}M {2 (z-3)-6 (x-l,]‘^
-hU'-3) (X -1)^2 (y-2)r^nt^ {{2)^-^-(^-3)M~i6)-}
or
(6v-f- 3z -21)“-! (2z - 6x)--:- (-3x -2y-J-7)'^=4 (49)
or 45x'*f-40,v--i- 13zM 367z-^-24zx-h 12xy 42x—280.v
-l26z^-294=-0. Ans.
320
Analytical Geometry 3*/)
Ex. 7(a) Find the equation of the right circular cylinder of
radius-1 and having as axis the line
. (.'C--l)/2-(v-2)=(z -3)/2.
(Avadh 1978: Kanpur 79; Punjab 8t: Meerut 8SS)
Sol. The equations of the axis of the cylinder are
..(1)
{X 1)/2=(;; -.2)/I =(2- 3)/2.
Consider a point P(x, y^ z) on the cylinder. The length of
the perpendicular from the point P (x, y, z} to the given axis(1)
is equal to the radius of the cylinder / e. 2. Hence the required
equation of the cylinder is given by [See § 2(B), equation (2)]
{2(j>^-2)-1.(2-3)}H{2(z-3) - 2(x-l)}-+{!(x~l)-2(p-2)}2
=(2)*.{(2)'-f(l)H(2)n
or (2y-z~\f-^(2z-2x-4y-4-{x-2y+3f==36
or 5x’"+V+5z'-4;;z 8zx-4x’v+22.r-16y-14z-10=0. Ans.
Ex. 7(b). Find the equation of the right circular cylinder of
radius 2 who^e axis is the line (x—1)/2=;;/3=(2 -3)/l.
(Berahainpur 1976(S); Rohilkband 82; Punjab 78)
Sol. Proceeding exactly as in Ex. 7(a) above, the equation
of required cylinder is given by
IOxH5y-+ l3z*-12xy-6yz-4zA--8x+30;^-74r+59=0.
Ex. 8. Find the equation of the right cylinder whose axis is
xl2^y(3=zi'6 and radius 5.
(Gorakhpur 1978, 81)
Sol. The equations of the axis of the cylinder are
(x-0)'2=(y -0)/3=(z-0)/6.
...(1)
Consider a point P(x,,v, z) on the cylinder. The length of
the perpendicular from the point P (jc, v, z) to the given axis (1)
is equal to the radius of the cylinder i.e. y. Hence the required
equation of the cylinder is (See § 2(B). Note that here
a—j5=y=0;/~2,/m-3, n=('.\ and r=5]
(6y -3z)M-(2z-6x)’f(3x-2j0“=52 {(2)*+(3)2+(6)-^}
or
Ans.
45x2+40/+13z^-36;;r-24rx- 12xy-1225=0.
Ex. 9. Find the equation of the right circular cylinder whose
axis is x—2—Zy j;=0 and which passes through the point (3, 0, Oj.
(Agra 1981)
Sol. The equations of «he axis of the cylinder may be
written as
(x-2)/l=(y-0)/0«(z-0)/l.
..(1)
First we shall iind the raJius r of the cylinder.
We know that
r=the length of the perpendicular from a point (3, 0, 0) on the
cylinder to the axis (1)
The Cylinder
321
1
+{0.(3-2)-1.0}') ,
=I/V2.
Consider a point P (jc, y^ z) on the cylinder. The length of
the perpendicular from the point P(xyyt z) to the given axis (1)
is equal to the radius of the cylinder. Hence the required equation
of the cylinder is [See § 2 ^B)]
Q.z>2+{l.^-l.(^-2)?+(0.(;r-2)-l.>'}“-(l/V2)“(l+O+l)
or
x+2)*+;;®=l
Ans.
or 3C*+2y24-22—2z;c—4x+4z+3=0.
Ex. 10 (a). Find the equation ofa right circular cylinder des
cribed on the circle through the points A (q, 0, 0), B (0, a,0)and
C {0, Ot a)as the guiding curve,
^ —S6I. We are given three points .d (n, 0,~0), B(0,fl» 0)ahd
C(0,0, a). Let the fourth point be taken as O (0, 0, 0). The
equation of the sphere OilBC is [See Ex.6(a), set(A) of the
chapter on sphere]
...(1)
x'i^yi-\.z^—ax—ay—az=0.
The equation of the plane ABC is x!a-\-yfa-\-zfa= \
or
...(2)
x-\-y-\-z=a.
The guiding circle is given by the equations (1) and (2)
together.
Since the cylinder is a right circular cylinder, hence the axis
of the cylinder will be perpendicular to the plane (2) of the circle
and thus the d.r.’s of the axis of the cylinder are 1, 1, 1.
Let P (jci, yx^ Zi) be any point on the cylinder. The equations
of the generator through P(xj, y^, Zi) having d.r.’s 1, 1,1 are
...(3)
(x-xj)/l«=(:p-;»i)/l=(z-zi)/l=r (say).
Any point on the generator (3) is(r+x^, r+j'i, r+Zx).
Since the generator (3) meets the guiding circle, hence the.
point (r+x,,
r+Zi) will satisfy the equations(1) and (2)
of the circle, and so we have
('●+Xi)*+(r+>^i)H(r+Zi)2-a (r+JCi)—a
(r+Zi)-0
and
(r+Xi)-i-(r+:Fx)+(r+Zi)=«
or 3r*+r (2xi f 2yx+2z,-3fl)+(Xi*+;?i*+Zi*—axi—
azj)=0
...(4)
and
-. 15)
r=i(a-Xx-;^x-Z|).
322
Analytical Geometry 3-i)
Eliminating r between.f4) and (5), we get
34 {a—Xi -yi—Zi)^-\-k (a -xi -yi-Zi){2xi+2yi+2zi—3a)
A-W-]ryi^-hzi^—aXi-ayi-aZx)=0
or (a^+Xi^+yj^+zi^--2axi—2ayi—2azi+2xiyi+2xiZi+2yiZi)
+\2axi+2ayi+2azi-3a^—Xi^-2xiyi—2xiZi+3axi-2xjyi
~W—2yiZ^3ayi—2xiZi~2y^zi-2zj^+3azi)
+3{xi^+yj^+zi^—axi—ayi—azi)=0
or 2
-2(jFiZi+Zi*i+Jf,:);i)--2a2=0
or Xi^r¥yi*-\-Zi^~yi2i~ZiXi-Xiy^=a\
.% The locus of P f.e. the required equation of the cylinder
is given by
z*—yz—zx^xy-a®.
Ex.10 (b). Find the equation of the right circular cylinder
described on the circle through the points ^4 (I, 0, 0), B (0, 1, 0)and
C(0, 0, 1)as the guiding curve.
(Gorakhpur 1975;Burdwan 75; Meerut 87P)
Sol. Proceeding exactly as in Ex. 10 (a) above or putting
0=1 there, the required equation of the cylinder is
x^-{-y^+z^—yz-zx-^xy=\.
Ex. 11. Find the elation of the right circular cylinder which
passes through the circle x®+y’+z®=9, x—;^+z=3.
(Avadh 1978; Meerut 79, 83(S), 89(S); Kanpur 80, 81;
Rohilkhand 79, 80; Jodhpur 73; Gorakhpur 77)
Sol. The equations of the guiding circle are
and
x^+y^+z^=-9
x-^+z=3.
●■●■(I)
...(2)
Since the cylinder is a right circular cylinder, hence the axis
of the cylinder will be perpendicular to the plane (2) of the circle
and so the d.r.*s of the axis of the cylinder are I, - 1, 1.
Let P(xi, yi, Zj) be a point on the cylinder. The equations
of the generator through P(xi, j'l, Zj). having d.r.’s 1, —1, 1 are
...(3)
(X—xi)/i=(7
l)=(z zi)/l =r (say).
Any point on the generator (3) i,s (r+xi, —r+^i, r+Zi).
Since the generator (3) meets the guiding circle, hence the
point(r+Xj, —r+>?i, r+Zi) will satisfy the equations (1) and (2),
and so we have
and
. (r+Xi)»+(-r+>^i)H(r+z0“*=9
(^+.Xi)-(-r+j^i)+(r+Zi)=3
323
The Cylinder
or
and
3r»+2r{Xi4-Vi+Zi)4-W-t-:Ki*+^i*-9)=0
(3-X1+/1—
.(4)
.v(5)
Eliminating r between (4) and (5), we get
i(3-Xi+;Ki-zOHf (3-Xi+;>|-z,)(jfi+.Vi+Zi)
+W+>i*+Zi*-9)=0
or
The locus of P (xj, yi,
or the required equation of the
Ans.
x^-\-y^+z*
+yz—
zx:\rxy~9.
cylinder is
Ex. 12. Find the equation of the enveloping cylinder of the
sphere
z*—2x+ -1=0 having its generators parallel to
the line x—y=z.
(Avadh 1981; Rohilkhand 83; Kanpur 83; Meerut 83,88,89S)
Sol. [This questionjs based on § 4}.
The equation of the sphere is
●..(1)
x®4-3^4-2*—2x+4y~l=iC.
The generators of the enveloping cylinder are parallel to the
●●(2)
line
x^y=z.
Consider a point P (a, |5, y) on the enveloping cylinder. The
equations of the generator of the cylinder through the point
P (a, % y) and parallel to the given line (2) are
...(3)
(X -a)/l
=(z-y)/'l =r (say).
Any point on the generator (3) is (r+a, r+p, r+y). The
points of intersection of the generator (3) with the sphere (1) are
given by
(r+a)>+(r+;9)*+(/'+y)*-2 (r+a)+4 (r+)?) -1=0
or
3ra+2r (a -|y 4-1)+(a2+^^+y«-2a+4fi -1)=0. ,..(4)
The equation (4) being a quadratic in r gives two values of r.
Since the generator (3) touches the sphere (I), the two values of r
obtained from (4) must be equal and the condition for the same is
{2 (a4-j5+y+l)}* - 4.3.(a«+)52+y"-2a4-4^-l)-0
[using B*‘—4AC=0]
or
(a-j./?4-y+l)2-3 (aa+^s*H-y* -2a+4)?--l)=0
or
oc'--\-fi^+y^-0y—yet—afi—4a+5^—y—2=0.
The locus of/* (a,
y)/.c. the required equation of the
enveloping cylinder of the sphere (I) is given by
Aos.
x^+y'^+z^—yz - zx -xy—4x-{‘5y—z-2=0.
Ex. 13. Show that the enveloping cylinder of the conicoid
ax^:^by^-\-cz*=\ with generators perpendicular to the z-axis meets
the plane z=0 in parabolas.
(Meerut 1984P)
324
Analytical Geometry 3-Z)
Sol. The equation of the cotiicoid is
...(1)
The direction cosines of the z-axis are 0, 0, 1. Hence the
direction ratios of a fine, perpendicular to the z*axis can be taken
as /, m,0 because O./H-O.m-f-1.0=0.
Consider a point P(a,
y) on the cylinder.
of the generator of the cylinder through P (a,
d.r.*s /, m,0 are
The equations
y) and having
(x—a)//=(;>—3)/m=(2—y)/0=i*(say).
...(2)
Any point on the generator (2) is (/r+a, mr-f-fi, y). The
points of intersection of the generator (2) wHh the conicoid (1)
are given by
a {Ir+OLf+b (wr+)?)Hcy®= I
,r%r
(3)
6m“)+2r (aLaL^^B)4-(aa?4-b^^-^cy^-1)=0.
The equation (3) being a quadratic in r gives two values of r.
Since the generator (’2) touches the conicoid (1), the two values of
r obtained from (3) must be equal the condition for which is
{2 {aloL-\ bmfi)Y-A.{aP->rhm^){aa?-yhP^’^cy^-\)^Q
[using B^—AAC=^]
'or
{alaA-bmP)^—[aPA-bm^)
1)=0.
The locus of P fa, ^,y) i.e. the required equation of the
enveloping cylinder of the conicoid (1) is given by
(alx+bmy)^—(aP+bm‘^){ax^+by^+cz^—\)=0. ...(4) Ans.
The sections of(4) with the plane z=0 is
—
(^x+bmyy—(aP*^bni^)(ax^-\-by^ ~l)=»0, z=0
or
a*Px^-\^2jablmxy-\-b^m^y^—a^Px^—abm^X“ abPy^
—6®w®y®+(fl/®-H>m®)=0, z=0
or
2ablmxy—abm^x^—dbPy^-\-(aP-{6m®)=0,z=0
or ●
ab (m*x*—2lmxy+Py^)—(aP-T bm^), z=0
or
ab Xmx—ly)^=^
5m®), z«=0.
...(5)
In the first of the equations (5), the second degree terms form
a perfect square and hence (5) represents a parabola in the z-0
plane.
Ex. 14. Find the equation of a right circular cylinder which
envelopes a sphere ofcentre {a, 5, c) and radius r, and has its gene
rators parallel to the direction /, m, w.
Sol. Proceed exactly as in Ex. 12 above. The required
equation of the enveloping cylinder is given by
{I {x—a)+m iy—b)-\-n {z—c)Y
=(/®+/»® \-n-) {(x—a)*+(jy--6)*-!-(z“C)®-r®}.
11
The Cone
§ 1. Cone. Definition. A cone is a surface gerierated by a
moving straight line which passes through a fixed point and intersects
a given curve or touches a given surface.
The fixed point is called the vertex nnd the given curve (or
surface) is called the guiding carve (or guiding surface) of the cone.
The moving straight line is called a generator of the cone.
Quadric cone. A cone which is cu* by a straight line other than
the generators only at.two points is called a quadric cone or in other
words *a cone whose equation is of second degree is called a quadric
cone\
§ 2. The cone with the vertex at the origin.
To prove that the equation of a quadric cone with its vertex at
the origin is a homogeneous equation of second degree in x, y^ z.
(Lucknow 1981; Kanpur 78)
Let the cone with itsr vertex at the origin O be represented
by the general equation of second degree in x, y and z i.e. by
ax24-^>'^4-cz‘*+2/yz+2gzx+2/rxy+2t/x+2vy+2wz4<f«=0. ...(I)
Consider a point P (Xj, yi, Zi) on the cone. The equations
of the generator OP are
X--0 y-0 2—0
or X
Xi -0 ~yi~0~2i - 0
xi
y
=-^^r (say).
yx 2,
...(2)
Any point Q on the generator (2) is (rxj, ryi, rzi). Since the
line OP is a generator of the cone (1), therefore every point on it
like Q must lie on the cone (1) whatever the value ofr maybe.
Thus, we have
(aXi2+W+C2i2+2/yi2i4 2g2iXi+2Axiyi)
+2r (MXi+vyi+w2i)+rf=:0.
...(3)
Since (3) holds for all values of r, therefore it is an identity
in r and so the coefficients of r*, r and constant term must be
separately zero/.c., we have
326
Analytical Geometry 3-D
...(4)
ax,*+W +cziH 2y>iZi+2^ZiXi-f 2Axi^i=0,
and
d=0.
...(6)
(5).
uxi+vyi+wzt=0
The relation (5)shows that the point P (Xi,
Zj) lies on the
plane mx+vj?4-wz=0 if «, v and iv are not all zero but this con
tradicts our assumption that the surface is a con.e. Hence
u=iv=w=0.
'
The relation (6) is clearly true since the cone passes through
the origin.
Substituting the values of «, v, w and J in (1), the equa
tion of the cone vyith its vertex at the origin is given by
...(7)
ax^+by^-\-cz^+2fyz-{-2gzx-\-2hxy=0
which is a homogeneous equation of the second degree.
Converse. To prove that every homogeneous equation ofsecond
degree in x, y and z represents a cone whose vertex is at the origin.
(Lucknow 1981; Kanpur 78)
The most general homogeneous equation of second degree in
X, y, z is given by
...(1)
ax^-{-by*A-cz*+yyz'jr2gzx-\-2hxy=0.
If equation (1) is satisfied by any point P(Xi, yu ^i). then for
all values of r, the point (rxi, ry^ rzi) clearly satisfies the equation
(1). Since (rxi, ryi^-rzi) are the general co-ordinates of a point on
the line through the origin O (0, 0,0) and the point P(Xi, yu Zi),
therefore every point on the line OP lies on the equation (1) and
so the line OP lies wholly on the surface (1).
Therefore, the surface (1) is generated by the straight lines
through the origin and hence it is a cone with its vertex at the
origin.
Note. Incase abc-\-2fgh af^~-bg^—ch^=0 the equation (1)
will represent a pair Of planes. See the chapter on The Plane.
Working method. If the two equations representing the guiding
curve are such that the one equation is of first degree then the requi
red cone with the vertex at the origin is obtained by making the other
equation homogeneous with the help of thefirst equation.
[Note that by § 2 the equation of the cone with the vertex at
the origin is homogeneous.]
If both the equations representing the guiding curve are not
of first degree, then we introduce a new variable
say. We make
each of the two equations homogeneous (in x, y, z and t) by
multiplying the terms with appropriate powers of t. Now the
327
The Cone
required equation of the cone with the vertex at the origin is
obtained by eliminating the new variable t between these two
homogeneous equations.
SOLVED EXAMPLES(A)
Ex. 1. Find the equation of the cone with vertex at (0, 0, 0)
and passing through the circle given by
x2-{-3;2+z2+x-2>'+3r-4-0, x-y+z=2.
(Bimdelkhand 1978)
Sol. The equations of the given circle are
...(1)
4=0,
...(2)
and
x-y+z=1ot
Making (1) homogeneous,with the help of(2), the equation of
the required cone with the vertex at the origin is given by
(JC-2;;+3z)(x-y:]-z)-4.i{x-y+z)^=^0
-3x;;+4zx+.2y*-5>^z+3z2)
or 2
-2(x*4-/+2*-2x;'+2zx—2>^z)=0
or
x*+2>>>+3z*+x;'—;^z=0.
Ex. 2. Find the equation of the cone whose vertex is (0, 0,0)
and which passes through the curve ofintersection^ of the plane
lx-irmy’\-nz=p
(Meerut 1983)
and the surface ax*+6/+cz*= L
Sol. The equations of the given curve are
ax^-^by^-^cz^—U
...(2)
and
(/x+m.v+»z)/p=l.
Making (1) homogeneous with help of(2), the equation of
the required cone with the vertex at the origin is
(flx*+by^+cz*)^{(/x+./w^+”^)/py
Ahs.
or
p* {ax^-\-bf^->rCz^)={lx-\-my’\-nzf.
Ex. 3. Find the equation of tne cone whose vertex is the origin
and base curve is given by
(Gorakhpur 1976)
ax^+by^=2z» lx-\-my-\-nz=p.
Sol. The equations of the given curve ire
(1)
ax*-{-by^=2Zy
(2)
{ix’i-my+nz)lp=U
and
Making (1) homogeneous with the h;lp of(2), the equation of
the required cone with vertex at the origin is
ax?+by^^2z.{lx+my^nz)lp
Ans.
or
p {ax^-\-by^)=2z(/x+mp+nz).
32g
Analytical Geometry 3-Z)
Ex. 4. Frove that the equation of the cone whose vertex is ihe
origin and base the curve
0 is/{xkizy ykjz)=0.
(Kanpar 1979, 81; Meerat 87,89 S)
Sol. Let xll=ylm=zln
...(1)
be a generator of the cone.
Since it meets the base curve 2=ife,/(x, >>)=0, therefore
y k
I
m
k.
— n ,so that
I m
n ky y~ n
Putting these values of x, y inf{x,:j^)=0, we get
...(2)
Bliminoting /, m, n between (1) and (2), the locus of
xll=^ylm=zln is the cone
^(t
k.^k )=0.
Alternative SolntioD. We know that the equation of a cone
whose vertex is at the origin is homogeneous in x, y and z. From
z==ky we have kjz=\. Therefore making/(x,>>)=0 homogeneous
with the help of z=*A:, the equation of the required cone IS
fixklzyyklz)=^0.
Ex. 5. Find the equation of the cone whose vertex is the origin
and base the circle x=^Oy y^-\-z^=>b^ and show that ike section of the
cone by a plane parallel to the plane XOY is a hyperbola.
(Meerat 1973,84)
Sol. The equations of the circle are
y^+z*^b'^
-.(1),
and
xla=l.
..(2)
Making.(1) homogeneous with the help of(2), the equation
of the required cone with the vertex at the origin is
(x/fl)2 or (y^^-z^)==b^x\
...(3) Ans.
The section of the cone (3) by a plane parallel to the plane
XOY i.e. by the plane z=c is the conic given by
a^{y^4-c^)^b^x\z=c
or
which is clearly the equation of a hyperbola.
Ex. 6. The plane x/fl+j>/*+2/c=l meets the co-ordinate axes
in Ay By C. Prove that the equation of the cone generated by the
lines drawn,from O to meet the circle AB^is
^?
yz {b(c-{-clb)+zx (c/a-{-a/c)+xy {afhy\-bla)=>0.
(Agra 1979; Bondelbhaiid 78; Ponjab 82; Nagpur 78;
Robilkhand 77)
S29
The Cone
...(1)
Sol. The given plane is xja-{-ylb+zlc==l.
The points B, C are (a, 0, 0),(0, 6,0) and (0,0, c).
The equation of the sphere Oi4;BC is
...(2)
x^-\-y^-\-z^-ax~by—cz=i\
[Refer Ex. 6(a), page 246, in the chapter on sphere]
The plane (1) meets the sphere (2) in the circle ABC. Thus
the equations (1) and (2) together give the circle ABC.
Making (2) homogeneous with the help of(1), the equation
of the cone generated by the lines drawn through the origin O to
meet the circle ABC is given by
:^+y^-\-z^—{ax+by-{-cz)(x/a4-J'/^4-x/c)=0
or
yz {blc+clb)+zx (cfa-\-alc)A-xy{afb-{-bla)=0.
Ex. 7. Find the equation to the cone with the vertex ut the
origin and which passes through the curve
x^-^y^+z^+x—2y+z-4=0, +y^+
2x— "iy^r4z—5=0.
Sol. The equations of the guiding curve are
...(1)
x*+y*+^*+Ar-2y+3z-4=0,
and
x^+y^+z^+2x-3y+4z-5=0.
...(2)
Subtracting (2) from (1), we get
—x+y—z+l=.0 or X—y+z=l.
...(3)
Making (1) homogeneous with the help of(3), the required
equation of the cone with the vertex at the origin is
x^+yi^z^+{x-2y-\-3z)(x-y+z)-4{x - y+z)’^==0
or
2x^4-y^—Sxy—3yz-{-4zx~0.
...(4)
Ans.
Note. We can also make the equation (2) homogeneous with
the help of(3) and then also we shall get the equation (4) as the
equation of the required cone.
Ex. 8. Find the equatjoii of the cone with the vertex at the
origin and which pa.^ses through the curve
x*l(^+yW-^2^lc^=\, x^la^+yVfi^=2z. (Meerut 1971)
Sol. The equations of the guiding curve are
x^la^4-yW+zVc^=i, and x^jaL^-TyW=2z.
Making these equations honiogeneous with the help of a new
variable t [See working method on page 326], we get
...(1)
and
..(2;
Eliminating -V between (1) and (2), we get
2z
330
or
Analytical Geometry 3-D
4z2
\2
Vfl» ■*'62
j“\a2 ■*’y92
This is the required equation of the cone, with the vertex at
the origin.
Ex. 9. Planes through-OX, OY include an angle a. Show that
their line of intersection lies on the cone
2^ {x^-{-y^-^z^)=x^y'^ tan^ a,
(Agra 1978; Meerut 76, 87, 89; Lucknow 76, 80; Gurunaoakdev 76;
Nagpur 76)
Sol. The equation of any plane through OX i.e., the line
y=0, 2=0 is
or
...(1)
O.x^ 1 ●7+Ai^=0»
Similarly the equation of any plane through OY i.e., the line
x=0, z=i=0 is
...(2)
x+AaZ=0 or l.x+0.j'+AaZ=0.
The angle between the planes (1) and (2) is given to be .‘a’,
hence we get
0.1 + 1.0+A,.A,
AjAa
cos a
V(l+Ai'‘)V(l+Vj" V{(I+V)(H-A/J}'
or
or
or
Squaring, we get Ai^Aa^ sec^ a=(l+Ai2) (l+A#*)
Ai^Aa* sec* a=-l+Ai*+A2*+AiV
Ai*AB*(sec*a-1)=1+A,*+A2*,
AiV tan* a=H-Ai*M-Aa*.
.. (3)
Eliminating Ai and Ag between the equations (1), (2) and (3),
the locus of the line of intersection of the planes (1) and (2) is
(-y/z)* (●- x/z)* tan* a=.H-(- y/z)*+(-x/z)*
or
x*y* tan* a=z* (x*+y*+z*).
This is a homogeneous equation of 4th degree and hence
represents a cone with its vertex at the origin.
§ 3. The line xll=yim=zjn is a generator of the cone whose
equation [^homogeneous) is given by
/(x, y, z)=ax*+hy*+cz*+2/yz+2gzx+2/«y=0,
if and only if its direction ratios /, m, n satisfy the equation of the
cone i.e., if and only iff(l, m, n)=0.
The equation of the given cone is
/ (x, y, z)=ax*+hy*+cz*+2»z+2gzx-l-2Axy=0.
...(1)
The equations of the given line are
x/l—ylm=z{n=r(^s2iy).. .
;..(2)
The co-ordinates of any point P on the line (2) are (/r, mr, nr).
$31
Ttw Com
The line (2) is a generator of the cone (1) if and only if each
point on the line (2) lies on (I) i.e., if and only if the point
<lr,mr,nr) satisfies (1) for all values,of r i.e., if and only if
(at^+bm^+cn^-\-2fnw-\-2gnl+:hIm) r^=0 for all values of r i.e., n
and only if al^ ^bm^^cn^A-2fmn+2gnl+2hlm^0 i.e., if and only
if/(/, w, «)=0.
jj
§ 4. Tofind the general equation of a cone of second degree
which passes through the co-ordinate axes, the axes being rectangular.
(Avadh 1980; Agra 76, 78; Garhwal 78, 82;
Kurukshetra 76 ; Meerut 87P)
Since the cone passes through the co-ordinate axes, its vertex
will be at the origin and hence its equation will be homogeneous
of second degree. Let the equation of the required cone be given
ax^-{-by^-\-cz^+2fyz-\-2gzx+2hxy==0.
...(0
by
The d.c.’sof the x-axis are 1,0,0. The x-axis will be a
generator of the cone (1), if the d.c.'s 1, 0,0 of the x-axis satisfy
(1)[See r 3 above] and hence we get a=0.
The d.c.’s of the ;;-axis. and z-axis are 0, I, 0 and 0, 0, 1
respectively. Similarly the and z axes will be'generators of the
cone (1) if 6=0,c=0.
Substituting the values of
c in (1), the required equation
of the cone which passes through the co-ordinate axes i.e., the
cone which has co-ordinate axes as generators is given by
/>’z-l-gzx-!-6x;^=0.
SOLVED EXAMPLES(B)
Ex. 1. Show that a cone can be found to contain any two sets
of three mutually perpendicular concurrent lines as generators.
Or
Show that a cone of the > second degree can be found to pass
through any two sets of rectangular axes through the same origin.
(Rohilkhand 1979)
Sol. Let one set of three mutually perpendicular lines be
chosen as the co-ordinate axes and the other set of three mutually
perpendicular lines through the same origin have their, direction
cosines /i, m,, «i ; /a, m^, «a ^**d /a, m3, W3.
The general equation of the cone passing through the co
ordinates .axes (i.e., one set of rectangular axes) is
...0)
fyz-\-gzx-\-hxy=^0
[See § 4 above]
332
Am^ical Geometry i-D
If the cone (I) passes through the first two lines of the second
set, then we have
and
●
./>W2«2+«'«2/2+/j/2W8=0.
...(2)
...(3)
Adding (2) and (3), we get
/.(%«,
(«i/i+«2/2)+A (/i«h+/2«a)=0
f
(—fnans)+g(—nil^)+h (—/3mj)=0
or
[V
/Mi/ii+W3«2+m3«3=:0, etc., the lines being mutually
perpendicular]
or
fmzfh^-gnak+hl^ms^O.
...(4)
●L ?!
<■*> «●><’»'* ‘hat the cone (1) also passes through
the third line of the second set.
'
Hence a cone can be found to pass through any two sets of
rectangular axes through the same origin.
Ex. 2. Show that the equation of the cone which contuins the
three co-ordinate
the lines through the origin having direc
tion cosines
w,,
and I,,
n, is S l^l, (m^na-m^n,) yz^O.
Solution. First we shall show that a cone of second degree
can be found to pass through five concurrent lines.
Take the point of concurrency of the lines as origin. We
know that the equation of a cone with the vertex at the origin is
ox^+by^+cz^-\-2fyz-i-2gzx+2>ixy=0.
Let
Dividing throughout by V, we get
x^-h(bla)
y^+(cla)
z^-{-2 { f/a) yz-\-2 {gja) zx-\-2 {hfa) xy=0
or
.x2-{-h>3+<?'zH2/>+2g'zx+2/t'x;;=0,
...(2)
where
h'=h/fl, c'—cja etc.
The equation (2) of the cone contains fi ve arbitrary independent constants na ely A', c',/', g' and A'and as such a cone cun
be found u, sou-fy five independent- conditions. Therefore, a cone
can oe made to pass through fi ve concurrent lines,
i;he general equation of the cone through the three co-ordinate axes is [See § 4]
fyz-{^gzx-i-hxy=0.
I
through the lines whose d.c.’s are
‘> ●‘=●’5 will satisfy the equation
(I) of the cone iSee § 3J and hence we have,
+g//i/i 4 hlimi=0,
●●(2)
and
.Onant+gn.Ja-\-hlami={).
...P)
Solving (2) and (?), we get
333
The Cone
h
g
Wim,(Wj/i - ni/a) ^
{l%mx-.~ h^t)
h
f
g
or
kk
rngWi) “wJima(Wi/s—WaM
«i«8 Kh^~k*^x)
Putting these proportionate values of/,g, h in (1), the requi
red equation of the cone is given by
Proved.
H {\xk (Wi/t2-m,«i) yz)=Q.
f
kk
Wi/ia)
Ex. 3^ (a). Show that the lines drawn through the point(a, fiy y)
whose direction cosines satisfy al^+bm^-\-cn' ~0 generate the cone
a (x—a)^-{-b (y~fi)^+c(z—y)*=0. (Madras 1977; Berahmipar 76)
Sol. The equations of any line through the point (a, y?, y)
are
ix—a)ll==(y-fl)lm=(z- y)/n.
Its direction cosines /, m, n satisfy the relation
fl/2-4 bm-^ cn^—0.
...(1)
.. (2)
Eliminating/, w,/I between th.^ equations(1) and (2), the
locus of the liiie (1) i.e. the equation cf the required cone is given
. by
a {x- oi^^+b iy—^f+a (z- y)2=0.
Ex. 3(b) Find the equation of the cone generated by the
straight lines drawn through the point (1, 2, 3) whose direction ratios
satisfy the relation 2/2-}-3m=—4«2=0.
Sol. The equations of any line through the point (1, 2,3)
are
...(1)
(;c-l)//=(y - 2)/m=(z -S)//?.
Its direction ratios /, m, n satisfy the relation
.. (2)
2/2+3m2-4/j2=0.
Eliminating /, m, n between the equations (i) and (2), the
locus of the line (1) i.e., the equation of the required cone is given
by
2(x-l)H3(;^-2)*-4(z -3)*=0
or
2x*-|-3/ -4z2-4x-12j;-j-24z - 22=?0.
Ex. 4. Find the equation of the cone with vertex at the origin
and direction cosines of its generators satisfy the relation
/*4-2m2-3«2=o.
Sol. The equations of any line through the origin are
xll-ylm=‘zln.
Its direction cosines satisfy the relation
/H2ma-3«8=0.
...(1)
...(2)
-334
Analytical Geometry 3-D
Eliminating /, w, n. between (1) and (2), the equation of the
required cone is given by
x^-\-ly^—Zz^—0,
Ex. 5.. Find the equation of the cone through the co-ordinate
axes and the lines in which the plane IX'^my+nz—Q cuts the cone
ax^-\-hy'^-\-cz^-{-2fyz-\‘lgzx-\-Vixy=0.
Sol. The equation of the given cone is
S=ax^-\-by^^cz'^ ■'\-2fyz-\-2gzx-\-2hxy=^ 0
and the equation of the given plane is
P=/x+Tw;?+wz *=a0.
...(1)
...(2)
The equation of any surface through the intersection of (1)
and (2) is
S-t-AP=0,
.●(3)
where A is an arbitrary function of jc, y^ z.
Since the required cone passes through the co-ordinate axes,
its vertex is at the origin. Now if (3) represents a cone with the
vertex at the origin, then the equation (3) must be homogeneous
and hence we choose A to be a linear function of jc, y and z, say
h=ux-\rvy-\‘WZ.
Putting this value of A in the equation (3)., the equation of,the
cone is given by
ax*Arby'^-\-cz^-\-2fyz+ 2gzx-\-2Jhxy
+(MX+vy+wz) {lx-\-my^nz)=Q.
...(4)
The d.c.’s of the x, y and z axes are 1, 0, 0; 0, 1,0 and 0, 0, 1
respectively. If the cone ^4) passes through the co-ordinate axes
then the d.c.’s of the co-ordinates axes will satisfy the equation (4)
and so we have
fl -l-M/“0, 64-vm=0, and c-f-w«=0.
These give u= —a/l, v= - bjm, w—
cin.
Putting these values of m, v, w in (4), the equation of the
required cone is given by
ax*-^hy^-^cz^-\-2fyz+2gzx-\-2hxy
iaxll+byfm+czin) (lx-\-my-\-nz)=0
or
yz
or
/ (6/1®4-c/w® - 2fmn) yz+m (cl^ -j-a/i® - 2gnl) zx
0
(2/
-f/j (fl/«2 i-bF^2hlni) xy=0.
Ex. 6.
Ans.
OP and OQ are two lines which remain perpendicular.
335
The Cone
and move so that the plane OPQ posses through OZ. If OP descri-.
bes the conefiyix, z/;c)=0, prove that OQ describes the cone
^ \x’\
Z
ZX li
Sol. Let the ^uations of the line OP be
...(1)
xlh=ylmi=zlnx
..(2)
and that of OQ be x//2=3>/ma=z/na«
It is given that the lines OP and OQ are perpendiculat
therefore
...(3)
/i/,+Wim,+«in2=0.
Now it is given that the plane OPQii.e. the plane containing
the lines OP and 00 passes through OZ (i.c. x=0=;k)- The
equation of any plane through OZ is
..(4)
x+Ay=0.
Let the equation , of the plane OPQ be given by (4). Since
the plane OPQ given by (4) contains the lines OP and OQ,there
fore the lines OP and OQ both will be perpendicular to the nor
mal of the plane (4), and hence we get
/i.l+mi A-l-ni.0=0. and /j.l +m2.A+w« 0=0
●.(5)
/a/ma.
or
Again it is given that the line OP given by (1) describes the
cone/(y/x, z/x)=0. Therefore the d.c.’s /j, mi, «i of the line OP
will satisfy the equation of this cone, so that we have
...(6)
/(mi//i, ni//i)=0.
Dividing (3) by /i, we have
/a + (wi//,) ma+(rti//i)
or
(Wa//a) ma i (ni/M »a=0
[*.* from (5), mi//i- ma//|J
...(7)
or
ni!h= - /a/«a—WaVCW
Putting the value of m^jli from (5) and n^jli from (7) in (6),
we get
ma*
/a
/
Wa
l/a’l"
Hence the line (2) i.e. the line OQ whose d.c.’s are /„ m,, Wt
generates the cone
.r-*
proved.
336
Analytical Geometry 3-Z)
§ 5. The equation of the cone with a given vertex and a given conic
as base. Tofind the equation of a cone whose vertex is the
point fa, yff, y) and base the conic
ax^-\-2hxy+by^+2gx-i‘2fy-\-c=0, z=0.
(Agra 1980; Delhi 75; Indore 76; Lucknow 78; Kanpur 76)
The equations of the base of the cone are given to be
.. (1)
ax*-^2hxy+by^f2gx+2/y+c=0,z=0.
The equations of any line through the vertex (a,
(x -a)//=(y-p)/w=(z-y)//i.
y) are
..(2)
The line (2) meets the plane z=0 at the point given by
X—« y^fi 0-y
/
m
n i.e. at the point
,P"-^* ojIf this point lies on the given conic (1), then we get
●8
fc=0.
● ● (3)
The.relation (3) is the condition that the line (2) intersects
the conic (1), and hence the locus of the line (2)/.le. the required
equation of the cone is obtained by eliminating /, m, n between (2)
and (3).
From (2), we have
I X—a
m y-P
n z~y , and n
z-y
Putting these values in (3), the required equation of the cone
is given by
+2h\u.
+2/j^-(f:J)y(+c-0.
Multiplying throughout by (z - y)® and simplifying, we get
a (az-yx)2-f 2/7 {u.z -yx) {flz -yy)+b {fiz-yy)^
+2g (az-yx) {z - y)+2f(Pz -yy) (z-y)+c (z -7)*=0.
§ 6. To find the condition.for the general equation of the second
degree to represent a cone and to find the co-ordinates of its Vertex.
(Avadh 1982; Allahabad 77, 80; Jodhpur 78; Rajasthan 75)
●
The Cone
337
The iDOSt general equation of the second degree is
2/j'z-l:2gzx-ir2hxy -I- 2«x+2v;/+2wz+d=0. ...(1)
Let (1) represent a cone with the vertex at (a, /3, y).
Shifting the origin to (a, /5, y), the equation (1) becomes
o (x+a)«+h(y+P)*+c(z+y)>+2/(y+j8)(z+y)
+2g(z+y)(x+a)+2A (x+a)fy+i8)+2tt(x+a)
+2v (y+jS)+2H*(z+y)+d-0
or flx®+6);*+cz®+2/i)>z+2gzx+2Axy
+2x.(fla+/ii8+gy+M)+2y (/ia+6^+/y+v)
.+2z(ga+//5+cy+w)+(na*+bp*+cy®+2/i0y
■..(2)
+2gya+2Aaj8 +2«a+2vj8+2>vy+d)-0.
Now the equation (2) represents a cone with the vertex at the
origin and hence it must be homogeneous. Therefore, the coeflSdents of X, y, z and the absolute term must be zero separately.
So we must have
...( 3)
aa+Ai3+gy+tt=»0,
...( 4)
Aot+h]8+/y+v=a 0,
■●●( 5)
get +/j8+cy +
0,
and cra*+h/5*+ cy*+2/jBy+2gya+2Aocj3+2tta+2v^+2»y+d=0
i.e., a (fla+/rj8+gy+w)+j5 (Aa+h/5+/y+v)
+y (g«+/J8+cy 4-w)+(«a + vj3+wy+d) = 0.
Using (3), (4) and (5) this last condition gives
na+v^ + M-’y + d-O.
The required condition that the equation (1) represents a
cone is obtained by eliminating ot, p\ y between the equations (3),
(4), (5) and (6) and hence is given by
(7)
a
h
g
u
=0.
I
h
b
/
V
S
f
c
w
d I.
w
u
V
If the condition (7) is satisfied; then the equation (1) represents
a cone. The co-ordinates («, p, y) of its vertex are obtained by
solving the equations (3), (4) and (5) and are given by
X
i
y
-p
a n u
h g u
a g
a h g
b
f
V
few
h f V
h
g
8 f w
c
w
b
V
''!
h
b f
8
f c
338
Analytical Geometry 3-D
Working rule for nomerical problems. For solving numerical
examples, an easier method can however be adopted. We intro
duce a new variable t. Make the given equation (J). homogeneous
by multiplying the different terms In it by proper powers of t.
Let the homogeneous expression so obtained be denoted by
^(.X, y^ z, /) and thus the homogeneous equation is given by
P(Xty,2, t)max*-\-by*-^cz^-{-2fyz-\-2gzx-\-2hxy
+2uxt-{-2vyt-\-2wzt+dt^=0.
Now find BF BF BF . BF ■ ^
By* Tz
^
observe that the equations ^^=0/^=0,^=0 and |^=0 at
r-1 clearly represent the equations (3).(4),(5) and (6) respecti
vely. Now solve any three equations for x,y,i and if the
remaining fourth equation is satisfied by. these values of x, y and
zthen the given equation (1) will represent a cone with the vertex
at the point(x, y, z) found by solving the three equations.
SOLVED EXAMPLES(C)
Ex. 1. Prove that a line which passes through (a, y) and
intersects the parabola z*=4ax, y=>Q lies on the cone
(pz^yyy-4a (p -y){fix-^oty)=0.
(Meerut 1983)
Sol. Clearly in the present problem, we are to find the equa
tion of a cone with the vertex at (a,|3, y) and base curve as
z^~4ax,y=^0.
-.(I)
The equations of any line through (a, j8, y) are
(x-a)//=(>;-^j8j/m=(2-y)//i.
. (2)
The line(2) meets the plane y=0 at the point given by
x~a_0-j8 z-v .
y~
-r=-^“—'c.at the point ^ «~ m* 0,
’^
ml
This point will He on tbexurve (1), if
iy^rii^jm)^=»4a ia—lBjm).
...(3)
Now from (2), we have
. / X—«
n Z—y and
m y--p
Putting these values of n/m and tjm in (3) i.e., eliminating
/, m,n between the equations (2) and (3), the required equation
of the cone is given by
339
The Cone
or
or
(y;;- J8z)2—4« (ay-fix)(y-fi)
Proved.
(fiz-yy)^^4a (fi-y)(fix-ay).
Ex. 2(a). Find the equation of a cone whose vertex is the point
(a, fit y) and whose generating lines pass through the conic
x^/a^+y^lb^ -- 1, z«= 0. (Gorakhpor 1975; 82; Gauhati 77)
Sol. The equations of the given conic are
The equations of any line through the vertex (a, fi , y) are
...(2)
(x—a)ll^(y-fi)lm=^{z-y)ln.
The line (2) meets the plane 2«=0 at the point given by.
x-a _y—fi,_0 - y
i.e. at the point
n
7
m
If this point lies on the conic (1), then
...(3)
Eliminating /,
n between the equations (2) and (3), the
required equation of the conic is given by
a > SSB 1
or
ft* (az—yxl*4-fl* (]8z—yy)*=a*ft* (z—y)*
...(4)
Ads.
Ex. 2(b). The section of a cone whose vertex is P and the base
curve the ellipse x*la^+y^fb^=>l, z=0 by the plane x=0 is a rect
angular hyperbola. Show that the locus of P Is
+(y?+z*)/6*=1
(Meerut 1978, 84; Pupjab^l (S); Gorakhpur 75)
Sol. Let (a, fit y) be the co-ordinates of the vertex P of the
cone. The equation of the cone whose vertex is the point (a, fit y)
'■ and base curve the ellipse x^la^+yflb*=>l^z^O is.
ft* (az-yx)*+fl* ()3z—yy)*=>fl*d* (z—y)C
...(1)
[See eqn. (4) of Ex. 2 (a) above. Derive it here].
The section of the cone (I) by the plane,
is obtained by
putting x—0 in the equation (1) and is given by
ftVz*+fl* (j3z-yy)*=>a*ft* (z—y)*, x=-0.
...(2)
If the equations (2) repre.<ent a rectangular hyperbola in the
X-0 plane, then in the fi rst of the equations (2) we should hove
the coefficient of y®+the coefficient of z*«0
340
i.e.
or
Analytical Geometry 3*Z)
(aV)+(**«*
-a*6*)=0
The locus of the. point P (a, j8, y) is
6V+fl* iyHz^)=a^b^ ot xya^-\-{y^+z^)/b^>=\.
Proved.
Ex. 3. Find the equation to the cone whose vertex is the point
P (fl, b, c) and whose generating lines intersect the conic
px*-\-qy^=\, z=0.
(Agra 1982; Avadh 81; Calcutta 77; Meerut 85)
Sol. The equations of the base curve are
...(1)
px^ \-qy^—\t z=Q.
The equations of any line through the vertex P (a, b, c) are
...(2)
(X-a)//=(y-b)jm=(z-c)/n.
The line (2) meets the plane z=0 at the point given by
X—a y—b 0—
/
m
n - l e. at the point
This point will lie on the conic (1), if
p (a—Icfn)^+9(6 —mcjn)==» I.
...(3)
Eliminating/, m. R between the equations (2) and (3), the
equation ot the required cone is giv^en by
or
or
p faz—cx)®+9(bz-cy)^-(z—c)*
c* (/?x2 +^y*)+(pa*+ 6*9-1)z*—2c(apzx+69yz—z)= c*. Ans.
Ex. 4, Find the equation of the cone whose vertex w (1, 2, 3)
and guiding curve is the circle x*+y*+z*=j4, x-fy' +z—l.
(Lucknow 1982)
Sol. The equations of the guiding circle are given by
:«®+y*+z*=4
...(1)
andx+y4-z=l.
...(2)
The equations of any line (generator) through (I, 2, 3) are
X—l_y-~2_z—3_x+y+z--6
m
n
...(3)
i+m~tn
[Note]
The line(3) meets the plane x+y+z=l at the point given by
5
x-l-_y-2„z-3_ 1-6
n
/
m
/+m+n /+/»+« .
or
5m
SI
2—
—
3
/+m-t-«*
y+w+n’
i.e, at the point ^ 1 —
/m+w—4/ 2/+2a—3m 3/+3m-2w
\/+m+n ’ /+m+n~* /+m-|-n )●
5n
/+w+«)
The Cone
341
This point lies on the base circle given by the equations (I)
and (2), hence this point will satisfy the equation (1) and so we
have
(m+n -4/)H(2/+2n-3OT)*+(3/+3m-2«)»«4 {i+m+n)* ...(4)
Eliminating/, M, » between (3) and (4), we -get the locus of
the line (3) f.e,. the equation of the required cone, We observe
that the equation (4) is homogeneous in /, m, n, so to eliminate /,
iw, n, we can put even the proportionate values of /, m, n from (3)
in.(4). Then the required equation of the cone is given by
{(_^__2)+(2-3)-4(x-l)}H{2(x-D-1-2(z-3)-3
4{3(Jc-l)+3(:y-2)~2(z-3)}»=4 {(x- l)+(y-2)+(z-3)}*
or
2-4x-1)*+(2x+2z-3j;-2)*+(3x+3;^~2z- 3;V
=4(x+>»+z-6)2
or 5x*+3j>®+z®—6;;2--4zx—2xj>'+6x+8>>H-10z—26=0.
Ex. 5. Find the equation of the cone with vertex (5. 4, 3) and
(Indore 1979)
with 3x*+2>*=6,;;-hz=0 as base.
Sol. The equations of the base curve are
...(1)
3x»+2);*=6, and)^+z=0.
The equations of any line (generator) through (5, 4, 3) are
x—5 V—4 z—3 ^4-z-c7
n
...(2)
I
m
m+n
(Note]
Ihe line (2) meets the plane y+z=0 at the point given by
X—5 y~4 z—3 0-7 ^ —7
n ~^m+n^m+n
/ “w
7«
7m
7/
.3i.e., at the
m+n
m+n)
point I 5 — m+n *
5m4-5n—7/ 4n — 3m 3m—4w
or
m+n
m+n * m+n )●
If this point lies on the base curve given by (1)» then this
point will satisfy 3x»+2/--=6 and hence we have
● ●(3)
3 (5m+5n-7/)*+2(4n-3m)*-6 (m+n)«.
Putting proportionate values of /» m, n from (2) in (3) and
thus eliminating I,
n between the equations (2) and (3), the
required equation of the cone is given by
3 {5 (y-4)+5 (z-3)-7 (x-5)}«+2 {4 (z-3)-3 0>-4)}*
=6{0»-4)+(z-3)}*
or 3 (5)>+5z-7x)»+2 (42-3)/)*«6 (;^+z-7)»
or 147xH87y*+l01z2+90;'z-210zx-210x)>+84;y+84z-294= 0.
342
Analytical Geometry 3-i)
Ex 6.
Prove that the equation'
4x*—y*-\-2z^-lr2xy-3yz-\-i2x- i l;; F6z-t-4c=0.
represents a cone. Find the co-ordinates of its vertex.
(Rohilkhand 1982)
Sol. Let
F{x,y, z)=4x'^-y^+2z^-{-2xy~3yz+nx ~l\y-\-6z+4>=‘0 .. (I)
Now we introduce a new variable t and making (1) homo
geneous with the help of /[See working rule, § 6j, we get
F(x,}\ 2, t)^Ax^~y*-\-2z^-\^2xy—3yzA-\2xt- llj/-f-62r+4r^ ...(2)
Differentiating (2) partially with respect to x, r and f
respectively, we get
dFldx^Sx+2y+l2t, 8Fldy--r ~2y+2x-3z-lU
...(A)
dFldz‘=4z—3y -{-bt,'dF;dt~~ I2x--1 l.v+f’Z+8r
Putting
in each of the relations in (A) and ihcu equating
them to zero, we get
2x--2v-32-11=^0
...(4)
(3).
8x+ 2i4-.l==0
...(6)
●~3j'-f‘●:2^'6=-0
12x-I1:>;+62+8--.-0
Now we shall fi nd x, y, z by solving the equations j(3), (4;
and (5i.
Eliminating x between (3) and (4), w.c get
Solving (5) and (7), we get
J'C= -~2, 2
3.
Pulling the value of y in (3), we get X--1.
Substituting these values i.c. x-— I, y^-- —2, 2~—3 in (6),
we have
12 (-1) -It ( - 2^f6 (- 3)i-8--.0 or0«0
I . «ne cq.<ation (6) is satisfied. Hence the given equation (I)
i ep:e>t‘iii.s a cone with the vertex at ( —1, —— 3).
Ex.7.
Prove (hat the equation
●i.Y ~ f by^+cz^ ■f-2ux+2vy-\- 2wz-\-d-^0
reptesents o ; one if u^ja+v^(b
d.'
(Avadb 1980; Garhwal 79, 81; Kanpur 8?; Meerut 80, 82, 86, 90;
Rnjastiian 75, 77; Delhi 76)
Sol. Let
F {X, >, z]e^ax^A-by^
ezH 2r/x-f 2vj -f IwzA-</=^0.
.. (1)
Now introducing a iiew' variable / and making (1; hemo■:cneous with the help of /.● [See working rule, § 6J, we gc.t ’
i,i^^ax'--}-by^-\-cz^-\-2uxt-\-2Yyi-i-2wzt-]-dtK
.. {2)
Differentiating (2) partially w.r.t. .v, y, z und t lespeciivdy.
ve have
343
The Cone
dFldx^2ax+2ut,oFldy=2by-{-2vt
..,(A)
8Fldz^2cz+ 2wt, dFfdl-2ux-\-2vy-h2wz+2dt \
Putting t= 1 in each of the relations in (A)and then equating
them to zero, we have
2ax-^2u=0,2by+2v^0, 2cz+2iv=0, 2«jt+;2v>H-2wz+2rf=0.
. From the first three equations, we have
x^-ula,y=» — vlbtZ<=>—w/c.
Putting these values of x, z in the fourth equation namely
2wx+2v;;+2M'z+2flf=0, we have
2m(-m/aH-2v (—v/6)4-2tv(-iv/c)+.2</=0
u*/a-\’V^/b+w^lC‘=d.
or
...(3)
Hence (3) is the required condition that the given equation
(1) represents a cone.
We also note that the vertex of the cone (1). is the point
i-u/a, —vjb, —wfc).
Ex. 8.
curves
Two cones with a common vertex pass through the
2* —4ax, y=0. and z*.=4hj», x=0.
The pla, j z 0 meets them in two conics which intersect in four
con-cyclic points Show that the vertex lies on the surface
z* («/fl+;'/^)==4(x^+y).
Sol. Let (a, /3, y) be ihe common vertex. Proceeding as
Ex. 1,above, the equations of. two cones are respectively
given by
iyy~Pz)^-^4a (ay-jSx)(y-fi)
(zx- azj*- 4b {fix-ay)(x —a)
...(1)
..*(2)
The plane 2=0 meets the cones (1) and (2) in two conics
whose equations in xj^-plane are respectively given by
...(3)
Si=/y-4c(ay-jSx)(y-j8), z=0
Sz--Y^x^—4b (j8x-ay)(x-a), z=0
. .(4)
Any curve through the intersection of(3) and (4) is given by
5i+A5a=0,2=.0
i.e.
yY-4a{ay~^x)(y-^).
+A [y*x*—4h (i3x—ay).(x—a)l=0,2=0.
If the last equation represents a circle, then we must have :
i.e.
and
The coeffi. of X"=lhe coefli. of y* and the coeffi. of xy*=0.
A (y*—4h/5;«=y>—.4ca ●
_
...(3)
...(4)
4flj8+4haA«=30 i.e. A=—(c)3)/(ha),
344
Analytical Geometry 3*2)
Eliminating A between (3) and f4), we have
y* (^»«+flj8)=4fl^(a«+^«)
The locus of(a, p, y) is 2® {xla-^ylb)=4 (Jc®+y*).
Ex. 9. Find the equation of the cone with vertex at {2a, b, cj
and passing through the curve x® ^-y*=4a® and 2=0. Find b and c
if the cone also passes through the curve >®=4a(2+a), x =>0. Also
show that the cone Is cut by the plane y=0 in two straight lines
and the angle 6 between them is given by tan 0—2.
Sol. The equations of any generator through (2a, b, c) are
...(1)
{x—2a)ll—(y—b)/m=‘{z-c)fn
The line (1) meets the plane z=0 at the point given by
X—2a y—b —c. ,
● ,/ «
t
n\
—z—
=
/ e. at the point
|
2a
.b
,01
l
m
n
\
n
" /
This point will liei on the conic x*+j;*=4a*, 2=0 if
...(2)
(2a—Icjn)® -I-(6—»ic/n)®=4o®
Eliminating/, m, n between (2) and (1), the equation of the
cone with the vertex at(2a, b,c) is given by
or
...(3)
(2a2—cx)®+(62--cy)®=4fl® (z—c)®
If the cone passes through the curve
...(4)
y*=4a (2+fl). *=0
the curve of intersection of (3) will be same as (4). Now put
x=0 in (3), we get
(2a2)*4-(d2—cj>)®«:»4a* (2—c)®
or
6®2*+c®y*=26c>>2—8ca®2+4o*c*
...(5)
Compafing (4) and (5), we have
● a » c* — 8ca® 4a*c*
*=’0*y=-4-- 4a® ^ 6=0,c*=»—2a»
Now putting 6*»0, c=->2a in (3), the equation of the cone
becomes(on simplification).
x*+j>®4-22X—4a2—4a®=0.
...(6)
The plane >>=0, meets the cone (6) in
x*+22X—4a2—4a®*=>0',>>=0
it clearly represents two straight lines and the angle 0
between them is given by
Tan 0
1+0
«2.
345
The Cone
§7. The tangent line and the tsDgent plane to a cone.
Tofind the condition that a given line through the point (a, P, y)
on the cone
ax^-^by*-\-cz*+2fyz-{^2gzx+2hxy^0
be a tangent line to the cone and to find the equation of the tangent
plane to the above cone at the point (a, y) on it.
the equation of the given cone is
F(x, y, z]^ax^-\-by^+cz^+2fyz+2gzx-\-2hxy^-0. ...(1)
Let P (a, /3» y) be a point on the cone (1), at which the
tangent plane is to be obtained, Also let /, m, n be the actual
direction cosines of the given line, Therefore, the equations of
the given line through (ot, y) are
...(2)
(jc-a)//=(y-i5)/m=(2-y)/n“r (say).
Since the point P(aj j3, y) lies on the cone (1), we have
F(a. 13, y)saa?-\‘b^*->rcy'^ fV)3y+2gy«+ 2h»p^0. ...(3)
The co-ordinates of any point on the line (2) are (a+/f,
y+«r). If the line (2) intersects the cone (1), then for some values
of r this point will lie on the cone (1) and, therefore, the points
of intersection of the line (2) and the cone (1) are given by
a (a+/r)*+b (]8+mr)«+c(y+«r)*+2/(^+mr)(y+«r)
+2g fy+«r)(a+/r)+2A (a+/r) O+mr)«=0
or r* (fl/*+bm*+c/i®+2/wn+2gn/+2A/w)
+2r {(oa + /ii8+gy) /4-(/ta+ftl3+/y) m+(ga+/j8+cy)fi)
+(oa*+bl3*+cy® -b 2/]^y-l-2gya+2Aa^)=0.
Putting the value from (3) and using F(/, m. n)for the coeffi
cient of f*, the above equation becomes
r*F{l,m,n)’\-2r{{a<t+hp-\rgy)l+{h(z+bp’^fy)nt
+(g«+/l3+cy)
...(4)
The equation (4) being a quadratic equation in r shows that
the line(2|cuts the cone (1)in two points and, therefore, if the
line (2) be a tangent line to the cone(I), the two values ot r should
be coincident. Clearly one value of r is zero and hence the condi
tion that the line (2) be a tangent line to the cone (1) is that the
other value of r should also be zero. Hence the required condition
is obtained by putting the coefficient of r in (4) equal to zero and
is
(fla-b A]S-1-gy) /-}-(^«+^^+/y) ^+(g«+/iP+cy) n=0. ...(5)
The tangent plane. The tangent plane at a point P (ot, p,y) to
the conn is the locus of the tangent lines through P (a,‘ y) drawn
in all directions.
Hence the tangent plane to the cone (1) at the point P(a, ^,y)
346
Analytical Geometry 3-D
is the locus of the tangent line (2) under the condition (5) and is
obtained by eliminating /,
n between the equations (2) and (5).
Thus the equation ot the tangent plane at P(a, y) is given by
(aot-\-hP i gy) {x~a)-\-{h<K-\-bp-\-fy'i(y—^)
-|-(5a4//34-Cy) (z-y)c=0
or .V iaa+hp+gy)+y (/ta +/>l3+yy)+z(ga+//3+cy)
-(aofi+ bjS*+cy*+2/j3y+2gya+2hdp>=0
or
.V {ax { h^+gy) yy (hoL+h^+fy)+z {gOL+/p t cy)=0 ...(6)
[using (3)]
Note. The equations (5) and (6) can be more conveniently
remembered and written with the help of differential calculus as
follows :
Differentiating (3) partially w.r.t. a,/3. y respectively, we
have
dFlda =2(ax +hp f-gy). dF/d^^2(hx+bp^fy)
dFjdy-2[gx \-ffi +ey).
Thus the condition (5) that the line (2; is the tangent line to
the cone (1) may be written as
, 8F^
8F ' dF
'T«+"'a^+":0^=o
...(5')
The equation (6) of the tangent plane at the point P(a, p, y)
to the cone (1) may be written as
0.
(6')
Corollary. To prove that the tangent plane at any point
^(«.P. y) t-i the tangent plane at every point of the generator through
F(a, p, y). In other words, the tangent plane at P touches the cone
along t!h> generator through the point P.
The vertex of the given.cone (1) is D (0, 0, Oj, and the tan
gent plane at P(a, p, y), i.e.(6) clearly passes through the vertex
0(0,0. 0).
The equations of the generator OP ate cleat ly given by
X
x-0^y-0_z-0
£ =^k (say).
l.e.
a~0”iS~0 ^0
§ y
Any point on this generator’is (/ca, kp, ky). The equation of
the tangent plane at {kx, kp, ky) to the cone (1) is
‘ x \akx-\-hkP-\-gky)^y (^a^:h^i3-H/^yHz {gkx-\-fkp-\-cky)^[)
or
*(aa+/r^+gyj+y {//a-f/>jS+/y)+z (ga+/j8+cyj-0
which is the same as the tangent plane at P [a, p, y) given by (o).
347
The Cone
Hence the tangent plane at every point of the',generator OP
is the same as the tangent plane at the point P. This proves our
desired statement.
We also conclude that the tangent plane at any point P
touches the cone along the generator OP.
This generator OP is called the generator of contact.
The vertex of the cone is called a singular point since all the
tangent planes of the cone pass through the vertex of the cone.
§8. The condition of tangency.
To determine the condition that the plane
+
may
touch the code
ax*+by-\- cz^. f2fyz+2gzx +■ 2!i\y - 0.
(Avadh
Agra 77; Kanpur 77)
The equation of the given cone is
p (X. y. z)sax*+M-*T
The equation of the given plane is
2gzx+2hxy==0.
...(1)
ux ^vy-^wz~0.
...(2)
Let the given plane (2) bt the tangent plane to. the given
cone (1) at the point (a, j8, y). The equation of the tangent plane
to the ccae (1) ai (a.
y ) is
...(3)
X iaix-\-h^-Pgy)i-y {hci-\ bji fy)+zigx-\-J^+cy)=0.
The equations (2) and (3) represent the same plane. Hence
comparing the coefficients of x, y and z, we get
1
w
u
-= - (say).
£ia~j-//ja+gy /ia + /)^+/y g<r.-pffi-]rcy p.
...(4)
aa4-/ijtJ+gy—/uM=»0,
and
hx-{-bfi+fy-~pv=>0t
...(5)
««+y/i+cy—/4tv«0.
...(6)
Also (a, j3, y) lies on the plane (2), so we have
ua+vj3+W’y«=*0.
Eliminating a. /3, y and
...(7)
between the equations (4), (5),
(6) and (7), the required condition is given by.
u
a
h
g
h
;
b
f
V
g
J
c
u-
U
V
w
0 i
«0.
.. (8)
348
Analytical Geometry 3-D
The equation (8) [on expanding the determinant] may be
written as
Au*-\-Bv^-\-C\^+2Fvw+2Gwu+2Huv=0,
.,.(9)
where A^B,C,F,GyH are respectively the co«factors of the
corresponding small letters a,b,c,f,g,h in the determinant A
given by
A= a
h
g
A^abc+2fgh—af^—bg*'-~ch^.
h
b
f
g
f
c
Thus
C=ai-A«=^
■
...(10)
Gt=-hf-bg-~i ^,H==fg-ck=i^
§ 9. The Reciprocal cone.
Definition. The reciprocal cone of the given <one is the locus
of the lines through the vertex and at right angles to the tangent
planes of the given cone, Ue. the reciprocal cone af the given cone is
the locus of the normals through the vertex to the tangent planes of
the given cone.
(A) To find the equation of the reciprocal cone of a given
cone.
Let the equation of the given cone be
ax*^by*+cz*+2fyz+2gzx+2hxy=k0
and let
ux-j-vy-^wziaO
...(2)
be the equation of a tangent plane to the cone (i). The condi
tion that the plane (2) touches the cone(1) is given by [see equa
tion (9), § 8]
Au*
-i-Cw*-I-2Fv w+2Gwtt+2fl«v=0.
...(3)
The d.r.’s of the normal to the plane (2) are u, v, w. Hence
the equations of the normal to the plane (2) passing through the
vertex (0,0, 0) of the given cone (1) are
x/u=>ylvi=>z/w.
..,(4)
Eliminating«. V, w between (3) and (4), the locus of the
normal(4) is given by
Ax*+By*+Cz*^2Fyz-^2Gzx+2IJxy^0.
..(5)
The equation (5) being a homogeneous equation of second
349.
The Cone
degree represents a cone with the vertex at the origin. Thus the
cone(5) is the reciprocal cone of the given cone (1). ●
Working role. In order to find the reciprocal cone of the
given cone flx*+^J'*+cz*+2y>'z+2gzx+2Axj^*=*0, we have to sinaply replace the small letters o, b, Cffyg, h by their corresponding
capital letters A, B, C, F, G, H, where the capital letters are the
cO'factors of the corresponding small letters in the determinant
A“ a
h
g
h
b
f .
g
f
c
(B) To find the reciprocal cone of the cone
...(6)
Ax^+By*+Cz^+2Fyz+2Gzx-{-2Hxy=^0.
To find the reciprocal cone of the cone (6) we should only
replace A, B, C, F. G, H by their co-factors in the determinant
H
G
A
H
B
F
G
F
C
Let A\ B\ C', F', G\ H' be the co-factors of A, B, C, F, G, H
in the above determinant. We have
A'r=BC--F»‘^.ica-g*)(ab-h^)-(glt -af)\
[See equations(10) of § 8]
-r a^bc4- 2afgh -a^p-abg*—ach^
csa{abc-\-2fgh af^—bg^'~ch^)=ohSimilarly F'=bA. C’===<’A»
and F*=‘GH^AF^f^, G'^^HF-BG =^gA. H'^FG-CH^hLNow the reciprocal cone of the cone (6) is given by [See
working rule of § 9(A)above]
+C'22+2Fyz-¥2G^zX’\-2H‘xy^0
or
A(flJc*+/>/+cz»+2/>^z-l-2gzx+2Axy)«=0
..(7)
or
axH
-b 2/yz-b 2gzx+2Axy=0
[as A 9^0 because (i) represents a cone].
But in § 9(A) we have proved that the reciprocal cone of(7)
is given by (6). Hence we see that the cones (6) and(7) are such
that each is the locus of the normals through the vertex (the origin)
330
Analytical Geometry 3-D
to the tangent planes to the other and on account of this property
the cones (6) and (7) are the reciprocal of each other.
Important Note. We have defined above that the reciprocal
cone is the locus of the normals through the vertex of the tangent
planes to the given cone. Hence in order to show that a given
plane /»c=0(ky) is a tangent plane to a given cone F (jc, j;, z)=0
(say), we should show that the normal to the plane P=>0 through
the vertex of the cone F(x, y, z)<=0 is a generator of the recipro
cal cone.
SOLVED EXAMPLES (D)
Ex. 1. Find the equation of the cone reciprocal to the cone
fyz-}-gzx-\-hxy<==Q,
Sol. The equation of the given cone is
/yz+gzx+/ixy=0.
Let the reciprocal cone of(1) be
Ax^-+By^+Cz*+2Fyzi-2Gzx-{-2Hxy=>0.
...(I)
...(2)
Comparing the equation (i) with the equation
ax^-v
cz^-^2fyz-\-2gzx-\-'2hxy—0, we have
a=»0r 6 -=(), c=--u. f -~hf.""ig and h=>\h,
F-^^*gh-ar^\g.\h~0=:\gln G=Wi
Putting these values in (2), the equation of the cone reciprocal
to (I) is given by
—}/‘Lv^ - JgVf2.ighyz-{- 2.ihfzx y 2.Jfgxy =0
(3)
/*A* i
^
—2ghyz -2hfzx—2fgxy^0.
Ex. 2. Prove that the general equation to a cone which touches
the co-ordinate plane is
or
a^x^+6-y*+cV—2hcy2-l.cazx 2abxy=0.
(Agra 1975 80; Gorakhpur 82; MeeruV TO, 79(S); Knrukshetra 76)
So!. The co ordinate axes are the normals to the co-ordinate
planes and hence the cone which touches the three co-ordinate
planes is the reciprocal cone to the cone which contains the three
o ordinate axes.
The equation of the cone which contains the three co ordinate
axes is given by
fy2-ygzx-\-hxy^Q.
[See § 41
...(I)
●No.v
lewritlnv the steps of Ex. 1. above, the cone reciprocal
to (1) is givci n’
351
Xhe Cone
Px-^-g^y^+
which is of the form
-2ghyz— Ihfzx—2fgxy=^O',
...(2)
+Py^+c^z*—2bcyz—2cazx—2abxy*=»0.
Ex. 3. Show that the tangent planes to the cone
fyz+gzx-\-hxy=0 are perpendicular to the generators of the cone
[Meerat 84 (Re)]
Px^+gY+h^z^-2ghyz-2hfzx- 2fgxy=^0.
Sol. In view of the definition of the reciprocal cone, to
prove the given problem it is sufficient to prove that the two
cones are reciprocal to each other.
Rewriting the steps of Ex. 1 above, the cone reciprocal to
fyz+gzx+hxy=0 is
/^x^'+gY+hh^ -2ghyz- 2hfzx-2fgxy=>0.
Ex. 4. Find the equation of the corre reciprocal to the cone
ax^+by^+cz^---0,
(Rohilkhand 1981; Kanpur 74; Meerot 85)
Sol. The equation of the given cone is
ax^-\-by^+cz^^O.
Let the cone reciprocal to the c >nc { 1) be
Ax^+
+Cz^+2Fyz+2Gzx+2
...(1)
0.
...(2)
Comparing (1) with the equation
ax^+by^-{-cz^+2fyz-\-2gzx-\-2hxy== 0,
we have
a=a, b^b, c=r,f=0,g=0, A =0.
A=**bc- /*’*t=Ac—0=»Ac.
Similarly B=ca and C—ab.
0.0-fl.0«0.
Also F=*'gh—af
Similarly (?=*0 and
Putting these values of/f, B, C, F,(7, H in the equation (2),
the required equation of the reciprocal cone is given by
bcx^+cay^-\-abz^=>0 or x*/«+;’*/*+z*/C“0.
Ex. 5. Prove that the perpendiculars drawn from the origin to
the tangent planes to the cone flX*+oy*H-cz*«=0 lie. on the cone
x^(a+y^lb+z^lc= 0.
(Kanpur 1978)
Sol. By the definition of reciprocal cone (§ 9), we know that
the locus of fhe normals (i.e perpendiculars) drawn from the
vertex (/.c. the origin in the present case) to the tangent plt^nes of
a given cone is called the reciprocal cone. Therefore, in the
present case, we are requtred to find the cone reciprocaf to the
cone ax*+ Ay2+cz‘‘*=>0. Hence it is the same problem as Ex. 4
above.
352
Analytical Geometry 3-JJ
Ex. 6. Find the condition that the plane ux-\-vy+wz—0 may
(Meernt 1984 S)
touch the ccne ax*-\-by^-^cz^=t^.
Sol. The equation of the cooc is
...(1)
The equation of the plane is
0.
...(2)
We know that the condition that the plane mx+v;^ +>vz=»0 be
a tangent plane to the cone ax^-\-by^+cz'^-\^2fy-\-2gzX’\-2hxy=^0
is (See § 8)
Au*-{-
+Cw®+2Fv M»+2Gwu+2Huv =» 0.
Here
a- a^ b^b,c=c,/~0,
A=**bc-f^^*=>bc-Q^-bc.
Similarly B=ca, C=ab.
Also F^^'*gh--af*\=^0.
Similarly G== 0 and H=0.
..(3)
/t=0.
Putting these values in (3), the required condition of tahgency
is given by
or
Actt*-t-cflV**+a6w*4-04-0+0 0
u^fa-{-v*/b^w^fC'=- 0.
Alternative method. In view of the 'Important note* to § 9,
the plane ux-\-vy-\-wz-=-l^ will be a tangent plane to the cone
ax*-\-by^Arcz^~ 0 if the normal to this plane through the vertex
of the given cone /.»?., through the origin in this case is a genera
tor of the cone reciprocal to the given cone namely
ajc2+t»j'*+cz*=0.
Now by Ex. 4 above the equation of the cone reciprocal to
the cone ax'^-\-by^-{-cz'^^D is
--0.
...(1)
The equations of the normal to the plane «x tv>'H-W2—0
passing through the vertex (0, 0. 0) are x/M=>/vt-z/w.
.. (2)
.
(2)is to be a generator of the cone (1), then the direction
ratios u, v, w of the line (2) should satisfy the equation (1) and so
the required condition of tangency is given by
u*/^4-v*/64-m'2/c«=0.
Ex. 7. Prove that the equation VCAli V'(«>')± V(Az)-^0
represents a cone that touches the co-ordinate planes.
(Bundelkhand 1979; Lucknow 74» 81; Kurokshelra 76;
Rajasthan 74)
353
The Cone
Show also that the equation of the reciprocal cone is
fyz+gzx-^ hxyc=i{).
(Agra 1973, 77; Rohilkband 80; Kaopar 75;80;
Meerut 83S, 84P,85P,89S)
Sol.' The given equation is
...(1)
Vif^)±V(gy)±y/(hz)^Q
or
Vifx)±Vigy)--T{h2).
Squaring, we have
fx■\■gy±2^/ifgxyj^hz
or
{fx-\-gy-hz)^T2Vifgxy).
Again sqmnng^px*-{‘^y^+h*z*+2fgxyr~2fhzx--2ghyz=4fgxy
or
..(2)
/*x*+gV+A«2»—2/gx;»—2/Azx—2gA;>r«=0
which is a homogeneous equation of degree two, hence the equa
tion (2) Ue, the given equation (I) represents a quadric cone with
the vertex at the origin.
To'fiud the reciprocal cone of (2) [he. of (1)]. In the equation
of the cone (2), we have
a =./*,. b=»g\ c^h\ /= -gA, g« -/A,|A» -/g.
if=»*‘Ac--/***=g*A*—g*A*s=0. Similarly RaaO, C«0.
Also F=“gA-fl/”=(-/A) (-/g)-/» (-gA)-2/*gA.
SimilarIy.C?=»2!g»A/and
Hence the equation of the cone^reciprocal to (2) \i,e. to (1)] is
given by
Ax*-{‘By*-^-Cz*-\‘2Fyz-\-2Gzx^‘2Hxy^H
or 0+0-|-0+4/*gAv2-t-4g»A/zjc+4A*/gJC3'«*0.
Dividing throughout by 4/gA, we have
*»
Proved. '
fxy+gzx’\‘hxy^i).
Ex. 8. Prove that the perpendiculars drawn from the origin to
the tangent planes to the cone
3x*-^4y*-\^5z^-lr2yz+4zx+6xy^0 lie on the cone
(Kanpur 1979)
19x*+lij^8+32»+6j^2-102x-*26*)^«-0.
or Prove that the tangent planes to the cone
3x*+4;>*H-5z*+2yz+4zx+ bxy^tO
are perpendicular to the generators of the cone
19x*+1 1>»*+32*+6j'2— lOzx —26x;>=0.
Sol. In accordance with the definition of the reciprocal cone,
in the present problem it is required to prove that the two given
cones are reciprocal to each other.
The equation of the first cone is given by
3x*-f 4>'>+5z*+2j>z+4zx+6xj«0.
..(1)
354
Analytical Geometry 3-Z)
Let the cone reciprocal to (1) be
Ax»+By^-i-Cz^-\-2Fyz-\-2Gzx+2Hxy=0.
...(2)
For the cone (1), we have
a^3,b==A,c°5,/=L^=2, A=3.
/. ^=^»c-/*=4.5-(l)fc=i9.J=cfl_ga=i5-4=lI.
C=^ab--h^^l2^9=>3. F=gA-fl/=6-3=.3.
G=hf--bg=3-Z
5, jy=/i?-cA=2-15=-13.
Putting these values in the equation (2), the cone reciprocal
to (1) is given by
19^*+llyH3z*+6yz-10z;c-26xy=0
which is the equation of the second given cone. Hence proved.
§10. The angle between the lines in which a plane cuts a cone.
Tofind the angle between the lines In which the plane
ux-\-vy+wz=0 cuts the cone
axHby^+czH2fyz+2gzx-}-2hxy=0,
Let the plane
ttJC+vy+wz=0
cut the cone
(Rajasthan 1976)
...(1)
Fix, y, z)&ax^+by*+cz*-\-2fyz+2gzx-h2hxya^0
...(2)
In a line given by xll=>ylm=>zln
where I, m, n are the actual direction cosines of the line (3). The
line of intersection (3) passes through the origin since the.vertex
of the cone (2)is (0, 0,0) and the plane (1) also passes through
the origin.
Since the line (3) lies in the plane (1), therefore we have
...(4)
Again the line (3) is also a generator of the cone (2) and
hence its d.c.’s namely I, m, n will satisfy the equation (2) and so
we have
fl/H
+c/|8+2/m/H-2gw/+2W/n=0.
.-(5)
From (4), we have «=—(tt/+v/n)/w. Putting this value of n
in the equation (5), we get
{-(tt/+vm)/»v}*+2/m {-(M/+vm)/w)
+2g/{-(m/+vm)/w)+2/r/w=0
or . l* (aw*-\-cu*^2guw)+2lm(cuv.--fuw—gvw-\-hw^)
+m*(hw*+cv2--2/vw)<=i0
or (aw^‘\-cu*~2guw)(//m)*+2 icuv-fuw-gvw+hw^)(llm)
■iibw^+cv^—2/v w)=0.
...(6),
The equation (6) is a quadratic equation in Ijm and hence it
355
The Cone
shows that the plane(1) cuts the cone (2)in two lines, Let the
direction cosines of these two lines be /i, mi, ni and /a, ma, n%.
Thus the roots of the equation (6)are lilnti and /a/m*. By theory
of equations, using the formulae for the sum and product of the
roots of the equation (6), we have
li_
h_
-2(euv—fuw—gvw-\-hw*)
cu*-[-aw*^2guw
mima
/ima+/a^i
c»A (say)
— 2{cuV’-fuw—gvw+hw^)'^cu*-\-aw^—2guw
...(7)
k bw*+cv*—2fvw
mi mt~'cu*-{-aw^-2guw
mi“^ma
or
and
A/a
m,ma
«i«a
^ .x
6w*+cv“4-2/yw“«<*+flH'*—2lg«m>“av*+6si*—2Avif “ *
...(8)
keeping in view (7) and writing the third fraction by symtnetry.
(/iWa—/ami)*=(/im8+/aWi)*—4/i/a rwima
=4A‘(cttv—/«w—gvw+
—4A*'(dw*+cv*—2/vw)
2gttw)
=4A*h>® {—(Au*-\-Bv*+CV*+2Fvw+2(jwtf+2^uv^}
=4A«vy*D*
where the expression within the brackets is denoted by /)* ^nd is
given by the determinant
/)*●= a
h
g
u
h
b
f
V
8
f
^
w
u
V
w
0
Also the capital letters At B, C, F, G, H are the cqfactors of
the eorresponding small letters n, b, c,ft g, A In the determinant
a
h
8
h
b
f
8
f
c
Hence
/ima—/ami=>2AivD.
Similarly minz—m^m^T^uD
} ..(9)
356
Analytical Geometry 3-D
Let 0 be the aogle between the two lines of intersection, then
cos B^liU+mimt-^nina
“A (6wHcv*-2/v»r)-|.A («*+flW»-2g«wi+A (avHbu*-2huv)
«A [(d+c) «*+(c+a) v*+(fl4.^) „,8_ 2fvw-2guw-2huv]
f*A [(a+6+c)(u*+vH w*)-(am*+fcv®+cw*+2/vw
+2g'wu+2Ativ)l
[on adding ffnd subtracting
[(a+ft+c)
«.)] [In view of(2)]
,'=4A* (tt*+v*+w*>
[Substituting from (9)]
or sin ff=>2Wy/(u*-\-v*+w»),
sin 6
2D “x/ftt^+v^+w*)
tan 0
costf”(a+6fc)(tt>+v»+w>)-/!’(tt, V, w) ...(10)
Dedoction I. Tofind the condition that the plane (1)cuts the
cone(2)In two perpendicular lines (l.e. generators).
(Delhi 1976)
fn the case the plane (1) cuts the cone (2) in two perpendi
cular generators (i.e, lines), the'angle 0s=>^7r i.e, tan d=tan Jtc=00.
Hence the required condition from the formula (10) is given by
(fl+b+r)(uH v*+ w*)^F(u, V, w).
...(II)
Dedoction 2. Tofind the condition that the plane
«Af+vy+wz=0 cuts the cone ox*4-6y*+cz®=0 in two perpendicular
generators (lines).
In this case the direction cosines of the two perpendicular
lines are given by (he relations
M/+vm+iv«=0 and fl/*+6m*+c/i*=0.
Also F(Xf y, z)c=ax*-^by^+cz^.
.*. F(u, Vt w)=^au*+bv*+cW*.
From (I I) the required condition is given by
(a-l-6+c)(tt*+v*+w®)=au*-|-6v“+cw*
.. Of (6+c)i#*+(c+a) v*+(a+6) w*=0.
...(12)
Dedoetion 3. Tofind the condition that the plane (1)cuts the
cone (2) in two coincident generators. In other words to find the
condition that the plane (1) be a tangent plane to the cone (2).
In this case tf=>0 i.e. tan 0=tan (?=0. Hence the required
condition from (10) is given by
2Dx/(m*+v*+w*)=0 i.e. Z)=0 as «*+v*-{-w*9&0
l.e. D*=0
J.e. iiu^^iIr*+Ctv'+2Fvw+2Gwu+2^«v«Q
,...(13)
The Cone
357
Note that (he condition (13; is the same as is the condition
(9) obtained in § 8.
SOLVED EXAMPLES(E)
Ex. 1.(a). Find the angle between the lines of section of the
plane 3x+y-^5z=>0 and the cone 6yz- 2zx-{-5xy=0.
Sol. The equation of the given plane is
4*+3'+5z=0.
●●(1)
The equations of the given cone.is 6yz—2zX’i-5xy^O ...(2)
Let the equations of a line of section of the cone (2) by the
plane (l^be given by ji://=>'//w=3r/fl
...(3)
Then 3/-frti+5/1=0 ..(4) and 6/M»~2/i/-i-5/m—0 .i.(5)
Eliminating m between (4) and (5), we get
6(-3/-5/I) n-2nl+5l(^3/-5/i)=0 or 30«*+45/i/+15/*=0
or 2/i«+3/i/+7»=0 or (2/i+/)(n+/)=0.
.*. 2/i+/=0 or /i+/=0.
When 2/i+/=0 i.e. /=.—2/i then from (4), /n=n.
//(-2)=/w/l=/i/I.
Again when /i+/=0 /.e. l==-n then from (4), m=i -2/i.
//l=/w/2=/i/(-l).
Hence the equations of the lines of section are '
r2=r=T
1
—1
r=—
Let 6 be the angle between these lines of section.
(-2).1-K1).(2)4-(1).(-1)
Then cos 6 =
1
the acute angle $ between the lines of section is giv^n by
0=co8-' (1/6).
Ex. 1 (b). Find the equations to the lines in which the plane
2x-\‘y—z=0 cuts the cone 4x*—y*+3z*=0.
Find also the angle between the lines of section,
(Avadb 1981; Agra 82; Robiikhand 77; Madras 78)
Sol. Let the equations of a line of section of the cone
4x^—y*+3z‘=0 by the plane 2x+y—z=0 be given by
xll=ylm=zln,
...(1)
Then
2/+/W—/i=0
...(2)
and
4/*—iw*+3«*=0.
.,.(3)
/i=2/+/w.
From (2),
...(4)
Putting the value of n from (4) in (3)» we get
4/*-/n*+3 (2/y/n;*=0 or 8P+6//ii+/n*=0
358
or
Analytical Geometry 3-D
(4/+w)(2/+/w)*=»0.
4/+w=0,2/+w=0.
When 4/+ot=0 Le. m=—Al then from (4), n=»—2/.
//lc=m/(-4)«n/(-2).
Again when 2/Thm«0 Le, m= —21 then from (4), n=0.
//1=ot/(-2)=/i/0.
Hence the equations of the lines of section are
X
y
z
^ X
y z
1 --4“-2
1 “-2=0 ■
Let 0 be the angle between these lines of section.
Then cos
1.1+(-4).(-2)+(-2).0
vu<r+(-v+(^2j»).vui)*+(--i)*+(o«
9
V(2I)V^(5)
fl=.cos-V{27/35}.
Ex. 2. Find the angle between the lines whose d.c*s are given
«/+vm+H»«t=iO
by the equations
and
fl/2+bm*+<?»*+2/mn+2gn/+2A/m=0.
Sol. Clearly it is required to find the angle between the lines
of section of the cone ax*+by^+cz^+2fyz+2gzx+2hxy*=»0 by the
plane wx+vy+wz=0. Hence see § 10 above.
Ex. 3. Show that the condition that the plane ux4-vy+wz=>0
may cut the cone ax*-\-by^+cz*=0 in perpendicular generators is
(6+c) tt*+(c+fl) v*+(fl+b) w*«=0.
Sol. This problem has already been solved in deduction 2 of
§10*. But here we are giving its complete solution.
Let the equations of a line of section of the cone ax*-^by*
+cx*a»0 by the plane ux+v;;+tvz=0 be given by
...(1)
x/l=>y/m^zln,
Then
...(2)
o/*+bm*+c/i*<=0
and
...(3)
ii/+vm+w«=0.
M}
Eliminating n between (2) and (3), we get
aF-^bm*+e {—ii/-l-vm)/w}*=>0
or
(aw*-fCM*) /*+2cMv/m+(bivf+cv*) m*=0
or
...(4)
(aw*+cM*)(//iw)*+2c«v (//m)-i-(bw*+cv*)«=0
This is a quadratic equation in Urn and hence it shows that
the plane Mx+vy+w2«=0 cuts the given cone in two lines (or
generators). Let lu mu Ri and la, ma, n« be the d.c.*s of these lines.
Then /i/iri and la/ma are the roots of the equation (4). The pro-
The Cone
duct of the roots of the equation (4)is given by
/j_
/a
mi * ma ow^+ cu*
«iWa
Ilk ^ miffla
ft»V*+CV* CU*+flW* flV*+6u**
writing the third fraction by symmetry.
Now the generators (l.e, the lines) will be perpendicular if
/i/a+mifWa 4-iiina =T=0
i.e. if
(bw* -|-cv*)4-(cu*4-flw*)+(flv*4“^M*)=*0
Proved.
i.e. if
(6-l-c) u*+(cj^a) v*+(a4-ft) w>=*0.
Ex. 4. Show that tfte plane ax+by+cz^O cuts the cone
yz+zx+xy^O in perpendicular lines If l/o4-l/fr4-l/tf=*G.
(Garhwal 1982; Indore 78. 79; Kanpur 80,82;
Meerut 81. 84S, 85S. 86. 89; Lucknow 77. 80;
Rohilkhand 78.82; Rajasthan 76. 78)
Sol. Let the equations of a line of section of the cone
yz+rx+xy=0 by the plane ux4-ftj^4-cz=0 be given by
(1)
x//=ym=z/«.
(2)
Then
mn4-n/4‘^'»=G
● (3)
fl/4-6m+cn«=0.
and
Eliminating n between the equations(2) and (3). we get
m {-(fl/4.6m)/c}4-{-(u/4-hm)/c)/4-/w=0
or
al^-\-(a+b-c) /m+ftm*=0,.
...(4)
or
a (//m)*+(tt+6—c)(//m)4-^=0*
This is a quadratic equation in Ifm and hence it shows that
the given plane cuts the given cone in two generators,(f.e. lines).
Let /i. mu and /a, Wa, wg be the d c.*s of these two lines so that
the product of the roots of the equation (4)is given bv
l\ k
mi * ma a
ilk Wima ni«3
(by symmetry).
1/a^ 1/6 “ 1/c
These lines of section will be perpendicular if
lik+mimi-^nint^O ie. if ]/a+\/b-\-}lc^0.
If
Ex. 5(a). Prove that the angle between the lines given by
x4.y4-z=0, ayz-{-bzx-^cxy=0 is|fr
u+64-c=0 hut ^TT//l/a4-l/64-l/c=0
(Agra 1981;Meerut 75; Pnnjab 79. 82)
360
Analytical Geometry
Sol. Let the equations of a line of section of the cone
ay2-^bzx-^cxy<=^(ihy the plane :c+;;+z=0 be given by
xlltB»ylm>=tzfn.
...(1)
Then
amn+bnl+clm=0
...(2)
and
/+rn+»=50.
...(3)
or
or
Eliminating n between the equations(2) and (3), we get
ami^l^m)fbl(-m^l)-^clm^O
W*+(a+6—c)/m+am^=^0
^
c)(//m)+fl=0.
...(4)
This is a quadratic equation in//w and hence it shows that
the given plane cuts the given cone in two fines {i,e. generators).
Let lu ntu «i and U, m^, be the d.c.’s of these two lines, so that
we have from (4), the product of the roots»^ .
mi ma b
*
a
b
~(by s”.nimetry)
A (say).
Case I. The angle between the lines of section is
In this case, we have
or
..(5)
/i/*+miW2+»i»8*“0 or aA+6A+cA=>0, using (5)
a+b+c«0, [V A96O].
Proved.
Case II. The angle between the lines of section is
The sum of the roots of the equation (4) is given by
d+b—fi
oil
c
. .
O 0'
/iWa+/tWi
q—a—b
OTll»o
.
^^*=A [from (5)].
...(6)
(/im8-/2Wi)*«(/ima+/gm,)a-4/i4«iOTg
«A* [(c-fl-b)*-4.ab], using (6) and (5)
«A* [fl*+b«+c*-2bc-2cfl~2flb].
By symmetry, we have
●.
imina-mant)»={nila^hali)*^\i [fl*+b»+c^-26c-2cfl-2flbl
Now
tan* dca £ (lima—Umi)*
(/i/a+miWa+ffina)* *
Since the angle between the lines is jr/3, therefore we have
tan* |«r< 3A* (a*+b*+c*~2bc-2co—2flb)
^ ~~ A* (o+b+c;»
“
●»»»■
or
3 (fl+b+c)*«3 (fl*+b>+c*- 2bc- 2ca-2ab)
361
The Cone
or (fl®
or
c*+26c+2ca+7ab)=a*+6*+c*—26c—2cfl—2a6
4(6c+cfl+a6)=»0 or l/a+I/6+ l/coO.
Proved.
Ex. 5(b)|. Show that the angle between the lines given by
zx
xy
ond JH.+
\
0 Is |tt.
q-r rr-p p-q
Sol. Putting fl=>l/{g—r),6=1/ r-p),
g),
the equation of the given cone becomes
flpz4*6rx+ cxp=0.
Here we see that 1/a-f l/6+l/c=g—r+r-p+p—9=*0.
Hence the angle between the lines of section of the cone
pz/(^—r)-fzx/(r—p)+jcp/(p—g)=Q by the plane x|-p+z=»0i8 fw.
[See Ex. 5(a) above].
Ex.6i Find the angle between the lines of section of the plane
6*—p—2z=0 and the cone 108x*-7p*—20z*=»0.
Sol. Let the equations of^ line of section {Le. generator)
of the cone 108x*—
202*«0 by the plane 6x—p—2z=0 be
given by
...(1)
x/li^ylm^zln.
Then
108/»-7m2-20n«=0
...(2)
and
6/—ffi—2n=s»0.
...(3)
Eliminating m between (2)and (3), we get
108/*-7(6/~2«)2-20n>=0 or 144/»-168/n+48n*=0
or
6F-Vn+2n^=0 or (3if-2«)(2/-n)«0
or
3/=2», 2/=n.
When 3/=>2n, from (3) we havem=>3/.
...(4)
3/t=m«=2n . or l/2=»wi/6«n/3.
When 2/=in, from (3), we have m=2/.
...(5)
2/=w«=« or //l=w/2=in/2.
The direction ratios of the two lines of section are given by
the relations (4) and (5). If 6 be the angle between these two
lines of section,.then we have
cos Qi_,
/>/8+WlW»+«tW8
V t^1*+wii*-b
V(V+^**
_
2.1+6.2+3.2.
_ 20 _ 20
VU2j“+lb)*+(3/)VUl)‘H-l2)*+(2)»} ”7.3”21*
^=cos"* (20/21).
Ex. 7. If the plane 2x—y+czt=0 cuts the cone yz+zx+xp=0
in perpendicular Unes^find the value of c.
362
Analytical Geometry 3-2)
Sol. Let the equations of a line of section of the cone
yz-{-zx+xy=0 by the plane 2x—y+cz=a0 be given by
xfl=yfm=zln.
...(1)
Then
/Mfi+n/-f-/m=0
...(2)
and
2/—/w+crt=>0.
...(3)
Eliminating m between the equations (2) and (3), we get
(2/+ cn) «+«/+/(2/-f-c/i)^6 or 2/*+(3+c) n/-i-c/i*=0
or
2(///i)a-f-(3+c)(//n)+c-0.
...(4)
This is a quadratic equation in //n and hence it shows that
the given plane cuts the given cone in two lines. Let A, mi, n»
and72, m2, »2 be the d.c *s of these lines, so that we have from (4)
the product of the roots
/i
.
«1
/
»2
=-y, so that
/i/8/c=ni/i2/2.
or
.(5)
Now eliminating / between the equations (2) and (3), we get
mn f« {(m~c/i/2}+{(m—c/i)/2} m=0
m*-|-(3-c) mn-c/|2=0 or (m/n)»+(3~c)(m/«)—ccaO.
the product of the roots=— . —
Wa
m,m2/(~2c)=>Mi/i8/2.
c, so that
...(6)
From the relations (5) and (6), we get
l\lz miOJi rti/ia
...(7)
According to the given problem, if the lines of section are
perpendicular, then we have
/i^+wtima+niWa^O
or
c—2c+2=*0 or c<=>2.
Ans.
Ex. 8. Show that the plane^ ax-\-by-{-cz=Q cuts the cone
yz-\-zx+xy=^0 in two lines inclined at an angle
1
(aHb^+c^~2bc-2ca-2abWn
bc+ca^ab
J*
and by considering the value of this expression when o+A+c«=0,
show that the cone is of revolution arid that its axis is x=>y=z and
vertical angle tan-^ {—2'^2).
Sol. Let the equations of a line of section of the cone
yz+zxr{-xy^0 by the plane ax-^by^czt:^0 be given by
xjl=»ylm=zln.
(1)
Then
mnr|»n/+/m«=iO
- (2)
363
The Cone
or
or
...(3)
fl/+frm+c/i=0.
and
Eliminating n between the equations (2) and (3>, we get
(n/+6w)/c} /+//i«0
m
aP+{a+b-c)lm-^bm*=>0,
a
-c)(//m)+6=0.
Let the roots of this equation be A/ma and A/mt, then we get
II. !l=*.so that^^*=^=^’‘=A (say)(by symmetry),
m, IBs a'
I/a
l/» '/«
...(4)
a
/Wi
or
or
/ima-l-/aWi_c-fl-^
a6
bifiitn^
/jWla+ /a^i
using(4).
ic-a-b)l{ab) 1/6
Now (Ama-/aWi)2=(/ima+/aWi)*-4/i/aWiWa
(c—a—b)* 4 I
((c—a—6)*—4fl6}
a6
0*6*
=A.{
r
A*fe*
^2(fl*+6*+c*-26c-2ca-2fl6}«^-S5T»
...(5)
where fc*=so*4-6*4'e*'”26c—2cO““2o6.
2(l,ma-/ai«i)>=A%“{l/(a>6*)+l/(b’a*)+(l/('‘’)}
A»fc* (a’+**+a*)
and
W.+m.ms+n,»,=A(l/a+l/i+l/c)=.A(6e+aa+aft)/(a6c).
lines of section (I'.ea generators)
Let ^he angle between the two
be 0, then we have
tan 0«=»
or
or
/i/a+Wi^Wa+^i^a
A (6c+Cfl+fl6)
({a*+6*+c*—26c—2ca—2ab)(o*-l-6*+c*)}^^*
...(6)
tan 0*=*
'
(6c+co-1-o6T
[Using (5)for fc].
Again if a+i+c=0,then (a+6+c)»=0
a —2(6c+co+fl6).
a®4-6*+c
Putting this value of a»+i>‘+c‘ in (6). we have
K-4 ta.-j.hfl4-fltn.i-2 (hc+ca+oh)})*'^.
tan 0*=*
jjc^ca-{-qb
...(7)
= V'8='±2\/^*
Since fl+64-c«=0, therefore the line x/l=»>'/l=z/l lies on the
Ipane ax+by-\-cz=^.
....
364
Analytical Geometry 3-2)
The equation (7) therefore gives the angle between the generotors of the cone cut by a plane passing through the line
It is required to prove that this line is the axis of
the given cone. It will be so if we prove that the angle between
this line namely x/l=y/l=z/i and any generator of the cone is
half of the angle I? given by (7) t.e, vertical angle of the cone
contained between the generators in which a plane through the
axis cuts the cone.
Now if xll~y/m=z}n is a generator of the cone
j'2r+2X-f-x;'=0, then mn-{-nl-{-lm—0.
.(8)
Let a be the angle between this generator, and the line
xl\^yj\t=szj\ then we have
cos a=
/♦1+in.l-l-yi.l ■
V{(/*+«*-!-«*)
_
I'^m+n
V3 .V{(/+w-i-/i)a-.2(/wn+fl/-f./w)}
1
since mn-|-n/-|-//n=aO from (8).
/.
tan a=\/2.
Also tan ?ff— ^
* —2^2
I —tan»a 1—2
From (7) and (9), tan tf=tan 2a ^ -2a/2.
e^2a. or
2V2.
...(9)
...(10)
Hence the line x=y^z Is the axis of the cone and the cone is
of revolution.
Ex. 9, Prove that the equation to the planes through the origin
perpendicular to the lines of section of the plane
lx-\-my-\-nz=0 and the cone ax*-\-by*-\-cz*=^Q is
x» ibn^+cm*)+y* (cl*+an*)+z^ {orn^+bl*)^2amnyz
—2bnlzx^2clmxy=Q.
Sol. Let the equations of a line of section of the cone
ax -Fhy>-|-C2>«0 by the plane lx+my^nz^=h0 be given by
xlh=yfp=zfv.
● (0
Then aA*+h/t>+cv»=.0 and
or .
Eliminating v between
these relations..we get
„
aA*-|-V+c {-(/A-|-w/x)/n)«=o
●●(2)
This is a quadratic equation in V/* and hence it shows that
the given plane cuts the given cone in two
lilies. Let X„p„ v,
365
The Cone
and Ag, /«g, vg be the d.r.*s of these two lines so that* from (2) we
have,
♦
the product of rootsa^.
AiAg
^
'*jyz
bn*+cm* cl*’\-an* am*+bl^
(by symmetry)a.Ar(say).
...(3)
C3
Also the sura of the roots=s —
/*! /**
.
lime
an*+c/’**
Ai/ttg+Ag/<t _ fjtxiij
=a:[from (3)]
e/“+a«*
—2/me
jMiVg+AtgVi
V|Ag+VgA|
—2mna ~ —Inlb (by symmetry) ...(4)
Now the equations of the two lines of section are
xlhi^y/fii^zhi and x Xi^ylin^zht^
Therefore, the equations of the planes through the origin
perpendicular to these lines of section are
AiX+/*!;;+viz=0 and Agx+/*ty+vgz=0.
Hence the combined equation of the planes through the origin
perpendicular to the lines of section is
or
or
(AiJC+/tiiy+viz)(Ag*+/*gy+vgz)--^0
AiAgAC*+/ti,4tgy*+ViVgZ*+(/iivgH-/i8Vi);»z
+(AiV8+Agv,) zx+{Xtiii+\iiii) xy^O
(bn*+cm^) x*+(c/*+flo*) y*+(am*+-6/») z*
—2amnyz^2bnlzx—2cimxy—0, - using (3) and (4).
Proved.
Ex. 10. Prove that the conditions that the lines of sections oj
the plane lx-^my+nz<=0 and the cones
ax*+by^+C2^es»0, fyz-^gzx+hxy>=0 may be coincident are
6n*+cm* c/>+fl«* arn^+bP
him
(Meerut 1977)
fmn
Sol. Let the equations of a line of section of the cone
flx*+by*+cz*«=>0 by the plane /x+my+wzcaO be given by
jf/A=y//a=3z/v. Then, we have
aA*+6/Lt*4-cv*=0,/A+Wft+uvssaO.
Proceeding as in Ex.9 above, we get
zss
ViVji'
6/1*-t* cm* e/*+o//*
om*+6/“
AiAg
^
...(1)
366
Analytical Geometry 3-Z)
Again let the equations of a line of section of the cone
fyz+gzx+hxy—O by the plane /jc+m^+nz=0 be given by
xlp=ylq=zlr.
Then, we have
fgr-]rgrp-{‘hpq=0, lp+mq+nr*=^0.
Eliminating r between these relations, we have
fg {-Op+mg)/n)+g {-Up+mg)fn)p+hpg^O
...(2)
or gl(plg)^-{‘ifl-^gm—hm)(plq)+fm=>0.
This is a quadratic equation in pjq and hence it shows that
the given plane cuts the conefyz+gzx4-hxy=0 in two lines. Let
Pi» ^ii ri andpa. ^a. ra be the ;d.r.’s of these two lines, so that
from (2), we have
El
or piPi (///)
gigi^gi
gigi iglm)
.●●(3)
Now if the lines of section of the plane /x+mp+«2=0 and
the cones a**4-^y*+cz*=»0,/y2r4-gzx+AAcy=0 be coincident, then
we should have
AjAa _Mi/*a_ViV2
P^P^
g\gz Tira
^ cl^ -r on* _ am^+bl*
nr
///
' gim
hjn
Multiplying the denominators throughout by /m«, the required
conditions are given by
bn^-\-cm^
am^+bP
him
fmn
” gnt
Es. 11. Show that i\he planes which cut ax*+by*+cz^^Q In
perpendicular generators \ ouch the cone
Sol.
JcV(^+c)+-yV(c+a)^z*/(a+6)=0.
The equation jf the given cone is
ax^-\^by^-{-cz^=‘0.
(M.U. 1990)
..(1)
Let the equation of any plane through the vertex (0, 0, 0) of
the cone (1) be
ux+vy+wz—0.
.. (2)
Suppose the plane. (2) cuts the cone (1) in perpendicular
generators. Then proceeding as in Ex. 3 above, we have
(6+c) wH(c+fl) v* + (fl+i>j V=0.
...(3)
The Cone
367
From (3) we observe that the normal to the plane (2)through
the origin i.e. the line jc/m=;^/v«=»z/w lies on the cone
...(4)
Now the plane (2) is a tangent plane to the cone which is
reciprocal of the cone (4). The equation of the cone reciprocal
to the cone (4) is
...(5)
x*lib+c)-\-y^Jic+a)-\r2*Ka+b)=0
[See Ex. 4 page 347].
Hence if the plane (2) cuts the cone (1) in perpendicular
generators then it touches the cone (5).
Ex. 12. Show that the locus of the line of intersection of tangent
planes to the cone ax^-\-by^-\-cz^^ii which touch along perpendicular
generators is the cone
a*(6+c)
(c+o)
(a-\-b)z^=0.
(Kanpur 1977,81)
Sol. The equation of the given cone is
...(1)
...(2)
Let
xll‘=ylm=zln'
be the line of intersection of the two tangent planes to the cone
(1) which touch the cone along two generators OP and OQ where
P and Q are the points (ai,
yO and (a2, jSg, ya) respectively.
The tangent plane to (1) at (ai, /5i, yj is
fla,x+^j8jy+cyiZ=0.
Since it contains the line (2), therefore
...(3)
flai/
m+cyi/i=0.
Similarly since the tangent plane to (1) at the point (cc2, fiz, ya)
also contains the line (2), theref^ore
...(4)
aaa/+bPitn+cyan «=*0.
From (3) and (4) it is clear that both the points («i,
yi)
and (aa, jSg, ya) lie on the plane
.(5)
a/x +^/ny+cnz«=»0
through the origin. Hence (5) is the equation of the plane OPQ
containing the two generators OP and OQ the tangent planes
along which intersect in theiine (2).
If we put al=Ut brn=Vt cn=w, the equation of the plane (5)
becomes
..,(6)
ux-\-vy+wz<=0.
The plane (6) cuts the cone (1) in perpendicular generators
OP and OQ. Proceeding as in Ex. 3 above, the condition that the
plane (6) cuts the cone (1) in perpendicular generators is given by
368
Analytical Geometry 3-D
(6+c)iia+(c+fl) v«+(fl+6) w*=0
(b+c)(fl/)*+(c+fl)(6m)*+(a+6) (<:»)●= 0
a* (6+c) l*-\-b* (c+fl) m®+c* (a+6) n*=»0.
The locus of the line of intersection x/lssyfntf=»2}n is
<j* (6+c) jc*+6* (c+a)
(a+6) 2*=0,
which is the equation of a cone.
Ex. 13. Prove that the locus of the line of intersection of two
perpendicular tangent planes to ax*-\-by*^cz*=0 is
a (6+c) x*+b (c+a) y*+c (a+b) z*=0. (Rohilkband 1982)
Sol. The equation of the given cone is
>ax*+by*-{-c2*<=^0.
...(1)
Let
xll=>ylm>^z/n
...(2)
or
or
be the equations of the line of intersection of two perpendicular
tangent planes to the cone fl).
Let the equation of any tangent plane to the cone (1) be
«x+vy+wr=0.
...(3)
Th en the normal to the plane (3) through the vertex (0, 0, 0)
of the cone(l) namely the line x/tt=y/v«=x/w will be a generator
of the reciprocal cone x*/®+3'*/^+zV^=“0» so that we have
«*/a+vV&+w*/c“»0 Or bcu*+cav*-\-abw*=0,
...(4)
If the plane (3) contains the line (2), then
»/+vm+w/ica0.
...(5)
The equations (4) and (5) give the d r.'s of the normals to
those tangent planes to (I) which pass through the line (2). Since
(4) is a quadratic in M, V, w, therefore (4) and (5) will give two
pairs of values of u, v, w. Thus two tangent planes to (1) will
pass through the line (2). Let a,, v„ w, and Wg. v,, Wg be the d.r.’s
of the normals to these two tangent planes. Since the tangent
planes are given to be perpendicular, therefore we have
...(6)
or
or
Now eliminating w between (4) and (5), we get
bcu*+cav^-\-ab {—(tt/+vm)/«}»=0
{bcn*-\-abP) u*+2ablmuv-\-(can^-^a$m*) v*t=0
ibcn*-^abP) (ulv)*-\-2ablm (u/v)i-(can* f fl6m*)=b.'
Its product of the roots»^.
Vi
●
●●
Vg
ViVg
ean‘+ abm‘ “
bcn*-\-abP
WiWa *
°
Putting these proportionate values in (6), we get
symmetry).
369
The Cone -
or
{can^+abm^)-t-(abl^+ben^)+● (6cm* 4* ca/*) “0
a (6+c) /*+6 (cffl) m*+c (fl+6) «*=0.
The required locus of the line of inierseciion (2) is given
by
A (6+c) x*+6 (6+fl)>>*+c (fl+6) z*=0
which being a homogeneous equation of second degree represents
a cone with the vertex at the origin.
Ex. 14. A. line OP Is such that the two planes through OP, each
of which cuts the cone ax*+6>'*-1-cz*=0 in perpendicular generators
are perpendicular. Prove that the locus of OP is the cone
2
(2a+6+c) x*4-(a+26+c) >*+(a+6+2c) z 0.
(Jodhpur 1976)
...(1)
Sol. Let
x//=j>/m=z/n
be the equations of the line OP.
The equation of any plane through OP may be given by
,..(2)
wx+vy+wz=0
where
..(3)
ul+vm+wn=0.
Now proceeding a$ in Ex. 3 above, the condition that the
p'ane (2) cuts the cone a.<*+6;^*+cz*=0 in perpendicular genera
tors is given by
(6-j-C) M* + (c-i A) V*-|-(o-f6) w* = 0.
..,(4)
The d.r.’s u, v, w of the normal to the plane (2) are given by
the relations (3) and (4). Also the equation (4) being a quadratic
in M, V, w shows that there will be two planes like «x+vy+ivz=*0.
Let Ml, V,. Wi and u^, v.^ Wa be the d.r.’s of the normals to the two
planes. Since the two planes are given to be perpendicular, there
fore we have
Mi«2 + Vi Va +
(5)
Wa=0.
Now eliminating w between (3) and (4), we have
(6-f-c) M*-l-(c-f a) v*-|*(A-f6) {-(M/-fvm)/n}*“0
or {(6-f c) n*+(A-b6) /*} («/v)*-|-2 (a+6) Im (m/v)
+{(c+A) «® -I- (a -1-6) m*}=0,
which is a quadratic in m/v.
(c-f a) n*.-)-(A-|-6) m*
Its product of roots<=>
Vi ’ Va
(6-|-c)n*-HA-l-6) /* '
ViVa
w^Ma
(c-f A) rt*-f{A+6) m’* “(a f 6) /*-l-(6+c) a*
iVjiVa
(6-fc) m*-l-(c-f A)/*
(by symmetry).
370
Anilyttcal Geometry 3-/>
Putting these proportionate values in (5), we get
{(c+fl) n>+(fl+i») m^}^{{a^-b) /*+(h+c) «*}
+{(/>+c) m*+(c+fl) /*}=0
or
(2a+6+c) /●+(o+2h+c) m2+(o+^+25/
/. The locus of OP l.e. of the line (1) is given by
(2fl+b+c)*»+(a+2b+c)/+(fl+h+2c)2«=0
which being a homogeneous equation of second degree represents
a cone with the vertex at the origin.
§11. Three mutoally perpendicular generators.
7b find the condition for the cone
F (x, y^ z)^ax*-\-by^+«*+2fyz-V2gzx+ 2hxy^0
to have three mutually perpendicular generators.
(Agra 1974; Rohilkhand 80; Madras 76)
The equation of the given cone is
...(I)
F(x, z)^ax*-\-by'^-^cz*-\-2fyz-\-2gzx-\-2hxy^Q.
Let xlu=yfv==zlw be the equations of one of the generators
^rte cone (1) so that Its d.r.’s will satisfy the equation (1) of the
cone and, therefore, we have
.. (2)
F(u, V, w)=0.
The equation of the plane through the vertex (0, 0, 0) of the
cone (1) and perpendicular to the generator xlu^y/v—zlw is
.. (3)
MX + VJ>+»VZ=0.
Now from ^Deduction 1 to § 10* the plane (3) will cut the
cone (1) in two perpendicular lines (t.e. generators) If
(a+6+o) (tt"+v*+w*)=F(M, V, w)
or (a+6+c) (tt*+v*+w*)=0, using (2)
or fl +h+c=0[ V u*+v*-f-w*9fi0 being the sum of squares ofreal
numbers not all zero]
The condition a4-^+c«=0 being independent of it, v, w shows
that the plane through the vertex of the cone and perpendicular
to any generator cots the cone in two motually perpendicular
generators and each one of these two generators is also perpendi
cular to the first generator. Hence we conclude that the cone (1) has
a set of three mutually perpendicular generators if
a^b-\-c=^0
.. (4)
Le, if the coefficient of x* + the coefficient of y^
+ /Ac coefficient of z*«=»0.
We also see that the condition fl+6-|-c=0 is .independent of
«, V, w and hence a cone, in general, has an infinite number of
sets of three, mutually perpendicular generators.
371
The Cone
Remark, if it is required to prove that a plane ux+vy-{-wz=>0
cuts the cone(1)in two perpendicular lines, and if we show that
the cone (I) has a set of three mutually perpendicular generators
i.e,
then the normal to the plane
through the vertex must be a generator of the cone (1) and hence
the d.r.*s u, v, w of the normal must satisfy the equation (1) of the
cone t.e. F(u, v, w) must be equal to zero.
§ 12. Three mutually perpendicular tangent planes.
To find the condition that the cone
Fix, y, z)=flx*4-^>y*+«*4'2/yz+2gzx+2A*y«0
may have three mutually perpendicular tangent planes.
The equation of the cone reciprocal to the given cone is
[See § 9(A)
...(2)
Ax*+By^-^Cz^+2Fyz-\-2Gzx+2Hxy=0.
The given cone will have three mutually perpendicular tangent
planes if its reciprocal cone (1) has three mutually perpendicular
generators and the condition for the same is
^+R+C=0^
or (hc-/*)+(cfl-g*)+(ad—A^)=0
or
SOLVED EXAMPLES(F)
Ex. 1. Show that the plane /x+my+nz=0 atts the cone
ib—c)x*+(c—fl) y*+(u-A^z*+2/yz+2gzx+2Axy=0 in perpendi
cular lines if
(b—c) F-i-(c—a) m^-^(a-b) z*+2fmn+2gnl+2hlm‘=0,
(Meerut 1983S)
Sol. The equation of the given cone is
(b -c) x*+(c—fl) y^+(a~b)z*+2fyz+2gzx-i-2hxy*=^0 ...(1)
Here we see that the sum of the coefficients of
and 2*
In the equation of the given cone <=>(A- c)*f(c— a)+(n—A)a:0.
Hence the given cone (1) has an infinite number of sets of
three mutually perpendicular generators.
[See§ 11]
Now if the plane /x+my+nzaO cuts the given cone (1)in two
perpendicular lines (i.e. generators) then the third mutually per
pendicular generator is the normal to the plane /x+my+nzesO
through the vertex (0, 0,0) of the cone (1) i.e. the line xll=y(m
=‘z!n is the third generator of the cone (1) and hence its d.r*s I,
m,R will satisfy the equation (1) of the cone. Therefore, the
required condition is given by
372
Analytical Geometry 3-Z)
(b^e)P+(c-a)nP+(a-b)f‘+ifmn+2gni+2Mm=0.
E». 2, Find the locus of the point from which three mutually
perpendicular lines can be drawn to intersect the conic
z^O,ax^+b)^..i,
(Rajasthan 1974)
p
determined by
(«» P. Yh Now first we shall find the equation of the cone whose
vertex is the point P(a, p, y) and the base curve is the conic
z=’0, ax*+by*=^l..
The equation of the above cone is
[See Ex. 3 on page 340]
a(<zz~yX)^-]^b(Pz--Yy)^=(z^y)\
...(I)
The cone (I) will have three mutually perpendicular genera>
tors if
the coeflf. ofx>+the coeff of;;«4-the coeflF. of z*=0
oy*+ +(fla*+
— 1)=0
act^+bp*-{-(a-^b)y^=\.
Hence the required locus ofP(«, p, y) is ax^+by*+{a-\.b)
J.
Ex. 3. Prove that the locus'of the points from which three
mutually perpendicular lines ^can be drawn to intersect a given circle
X ■f-y*=a*, z=>0 is a surface of revolution,
Sol. Let the point whose locus is to be determined be
^ C«i P. y). Now first we shall find the equation of the cone whose
vertex is the point P (a, p, y) and the base curve is the circle
Z--0.
...(1)
The equations of any line through P («, j3, y) are
(x - ot)/l=(y - P)fm == {z-y)ln.^
...(2)
or
or
It meets the plane z=0 at the
point
o).
This point will lie on the circle (I) if
' i<z-lylny+{p .-myln)*^a*.
●●●(3)
Eliminating t m, « between (2) and (3), the equation of the
cone with the vertex at P (a, P, y) is
or
{-(S3
(ctZ-^YX)*+(Pz~yy)^ = a* (z-y)*.
...(4)
The cone (4) will have three mutually perpendicular genera¬
tors if
the coeff. of x*+the coeff. of y*+the coeff. of z» 0
y*+yH(a*+^*-n*)=0,, or a>+jS*+2y>=fl=‘.
liii; required locus'of P (a, p. y- ) is x*+y*-^2z* C=fl
The Cone
373
which is the equation of an ellipsoid (a surface of revolution)
obtained by revolving the ellipse
about the z-axis.
Ex. 4. Find the locus of the points from which three mutually
perpendicular tangent planes can be drawn to touch the ellipse
z=0.
Sol.
Let the point whose locus is to be found be P(«, j8, y).
The equation of the cone with the vertex at F(a, $, y) and *the
base curve the ellipse Jc7a>+yV^a=,l,
is [See Ex 2 ta) set
(C), page 339]
’
h*(az- yx)«+fl* ifiz-{-yyf<=a^b*(z—y)*
or h*y*-x’*+flW+(W+a»/3«-a>6»> z^-2a^^yyz-2b* tsyzx
+2a»h
V «0
...(1)
The cone(1) will have three mutually perpendicular tangent
planes if
<4
hc+ca+ah=/»+ga+/i**»
[Sgg § ^2]
aV.(W+<J^i3*-aW)+(h*a*+a2j8*—a*h*).6y+fl*^V
tsa
(-aVy)*+(~b*ay)*+(0)>.
Dividing throughout by y*, we get
aW+a«j8»-a^h>+ W+ay>^p^-a^b*+a»bY--a*B»+
or a*ha (aH/S’^+y^j-a'h*(a*h»)*=0
or a*+i3*+y*=fl“+h8.
Le.
The locus of P(a, y) is x»4-T*+2®=*a*+6*
which is a sphere with centre *t (0,0,0) and radius
Ex. 5. Show that a cone whose vertex is at the origin and which
passes through the curve of intersection of the sphere x2-j-y*+z>=3a*
and any plane at a distance *a* from the origin has three mutually
perpendicular generators.
Sol, The equation of the given sphere is
^®+>®+z**=3a*
Let the equation of a plane at a distance *a’ from the origin
be (normal form of the equation of a plane)
/A+my+nz=a
...(2)
where /, m, n are actual direction cosines.
Now making (1) homogeneous with the help of(2), the equa
tion of the cone with the vertex at the origin and the base curve
given by (1)and (2>, is given by
●x*4->*+z*=3fl2 {(lx-i-my-{-nz)/a)*
or
(lx+my+nz)^>=:0.
...(3)
374
Analyticai Geometry 3-2)
The cone (3) will have three mutually perpendicular genera¬
tors if
the coeflF. of jc*-f-the coeflf. of;>*4-the coefF. of 7^=0.
t.e. (l-3/>)+(l-3m»)+(l-3na)=0
or 3(/*+m*4-/j*)=3 or
which is true. Hence proved.
Ex. 6. Three points P, g, R are taken on the ellipsoid
so that the linesJoining P, Q, R to the origin are mutually perpendi
cular. Prove that the plane PQR touches afixed sphere.
. Sol. The equation of the given ellipsoid is
...(1)
Let the equation of the plane PQR be taken as
Ix+my-^-nz-l.
...(2)
Now making (1) homogeneous with the help of(2), the equa
tion of the cone with the vertex at the origin and the base curve
given by (1) and (2), is given by
xVfl*+y^!b^+z*/c*=(/x -f
-1- nzf
...(3)
If the cone (3) has three mutually perpendicular generators
then we have '
the coeff. of x*-l-the coeff. of;^*+the coefF. of z* --0
or
(l/a*-P)-f(l/h>-w»)-h(l/ca-na)=0
or
P+ m>+«*= 1/flH l/h’’+ l/c»= l/pa (say)
...(4)
where R is constant since a, h, c are given numbers.
Now consider a sphere whose equation is
xa+y«-l-z*=i?».
●●(5)
If the plane (2) touches the sphere (5). then the length of the
perpendicular from the centre (0, 0, 0) of the sphere (5) to the
plane(2) will be equal to the radius of the sphere, /.e. we have
1
which is true by virtue of the relation (4). Hence the plane (2)
touches a fixed sphere (5).
Ex. 7. If x—y^zjl be one of a set of three mutually perpendi
cular generators of the cone 3yz-^2zxT-2xy^Q, find the equations of
the other two generators.
(Kanpur 1980)
Sol. The equation of the given cone is
3^z—2zx—2x>>=0.
● ●●(I)
The Cone
375
In the equation of the cone (I), the sum of the coefficients of
and z* is zero, hence the cone (1) has an infinite set of
three mutually perpendicular generators.
Thus if Jc/i=y/l«=2/2 is one of a set of three mutually
perpendicular generators then the other two mutually perpendi
cular generators will be the lines of intersection of the given
cone (1) by the plane through the vertex (0,0,0) and. perpendi
cular to the given generator namely x/l^y/1^2/2 t.e, by the
plane 1.x-1-1.y-1-2.z=»0. Let x//=y/m=z/n be a line of inter
section, so that we have
● (3)
3m«-2«/-2/m=0
...(2)
and /+OT-f2n=0
Eliminating m between (2) and (3), we get
3(_/_2n)/i-2n/-2/(-/~2n)=0
2/*-/rt-6/i>=0 or (/-2rt) (2/-H3/i)=*0
or
or
/=2rt,/=--f«. .
When /=»2n, from (3) we have m-l-4/i=0 or m«=*-4n.
2/=—m=4rt or //2=/w/(—4)=/i/l.
...(4)
When /=~|fi,from (3) we have m-1-Jn^O or m=—\n.
.(5)
/=33m=—frt or//3«=3iw/l=n/(—2).
Therefore, from (4) and (5), the equations of the other two
generators are
...(6)
x/2 =y/(-4)=2/1 and x/3=yl1 =z/(-2)
Clearly both these generators are perpendicular since
2.3-1-(-4).(1)+1.(-2)=0.
Also each of these two generators is perpendicular to the
given generator x/l=y/1=2/2 since
2.1+(-4).(1)-M.2=0 and 3.1-1-1. l-l-(—2).2s=»0.
Hence the required two generators are given by the equations
(6) above.
Ex. 8. // x/l=y/2=z/3 represents one of a set of three
mutually perpendicular generators of the cone Syz—82X—3xy«=>0,
find the equations of the other two.
(Meernt 1976, 83 P; Rajasthan 77; Kanpur 78)
Sol. The equation of the given cone is
...(1)
5y2-82X—3xy«=»0.
In the equation of the cone (1), the sum of the coefficients of
x*,y* and is zero, hence the cone (1) has an infinite set of
three mutually perpendicular generators.
Thus if x/l=y/2=z/3 is one of a set of three mutually
376
Analytical Geometry 3-i>
perpendicular generators then the other two generators will be
the lines of intersection of the given cone (I) by the plane through
the vertex (0,0, 0) and perpendicular to the given geoerUtor
namely x/l=;;/2=2/3 he. by the plane 1.x+2;;+32=0. Let
xll=ylm—zln be a line of intersection, so that we have
5mn — Snl—3lm~0
...(2)
and
/-|-2/n-}-3/is=0.
...(3)
Eliminating / between (2) and (3), we get
5mn—Bn(—2w—'3/i)—3(—2m—3/i) m=0
or
6m*+30m«-f'24«®?=0 or m*4-5m«+4n*=0
or
(m+4n)(m+n)=0 Or m=—4n, m=-n.
When m=—4n,from (3) we have /—5/r=»0 or /=5n.
4/= —5m=20n or //5=m/(—4)«=w/l,
...(4)
When OT=—/I, from (3) we have l-^n=>0 or l——n.
...(5)
/=w=-n or//l«=m/l=n/(—1).
Therefore, from (4) and.(5), the equations of the other two
generators are
xl5r=yl(-^4)^zfi and xl{=yl\^zj(,— \).
.:.(6)
Clearly both these generators are perpendicular since
5.H-(-4).I+(l).(-l)-0. '
Also each of these two generators is perpendicular to the
given generator x/l=»j>/2c=2/3 since
5.14-(-4).2+1.3-0 and 1.1 +1.2+(-1).3=0
Hence the required two generators are given by the equations
(6)above.
Ex. 9. If x-=*yfl=^z represents'a set of three mutually perpendicular generators of the cone Wyz-\-6zx — \Axy
find the
equations of the other two.
Sol. Proceeding exactly as in Ex. 7 or Ex. 8 above the
equations of the required generators are
x/(-ll)=>/2=z/7 and Jc/2.=;;/(—3)=2/4.
Ex. 10. Prove that the angle between the lines in which the
plane x+y-\-z=»0 cuts the cone ayz-^bzx-^cxy^O will be \-n if
Sol.
The equation of the cone *s
ayz-k-bzx-^cxy>=Q^
...(1)
Since in the equation of the cone (1) the sum of the coeflBcientii of x\ y^ and 2* is zero, therefore the cone (1)has an infinite
set of'three mutually perpendicular generators. If two perpendi*
cular generators are the sections of the cone (1) by the plane
377
The Cone
■*+3'+2=0, then the third mutually perpendicular generator will
be the normal to the plane
f z-=0 passing through the vertex
(0, 0, 0) of the cone /.e., the line x/l--^/l=z/l will be the third
generator, Hence the d.r.’s of the third generator x/l=;>/l=z/l
Hence the required
will satisfy the equation (1) of the cone,
condition is given by
fl .1.1+b.l.l+c.1.1=0, or o+b+c=0.
Note, One way for solving Ex. 10 above has already been
given in Ex. 5 (a) in the Solved Examples set (E) on page 359.
§13. Right circular cone.
Definition. A right circular cone is a surface generated by a
line which moves in such a way that it passes through a' fixed point
(called the vertexi and it. makes a constant angle 9 with a fixed
straight line through the vertex.
The constant angle $ is called the semi-vertical angle of the
cone and the fixed straight line through the vertex is called the axis
of the cone.
(A) The section of a right circular cone by a plane perpendi
cular to its axis is a circle.
Let A be the vertex, AO the axis and 6 the semi-vertical
angle of the right circular cone. Now draw any plane perpendi
cular to the axis AO meeting it at the point O.
Consider any point P on the
section of the cone by this plane
and join PA
Now OP is a line
lying on the plane section of the
cone which is perpendicular to
the axis AO of the cone. Hence
OP is perpendicular to the axis
AO diod thus the triangle POA is
a right angled triangle with
/_PAO=-e.
0/'=^0 tan 0. But the
semi-vertical angle 6 is constant
and the axis AO is-fixed and hence
the length OP is constant for all positions of the point P on the
section, Hence the locus of the point P (i.e. the section of the
right ciicular cone by a plane perpendicular to its axis) is a circle
with the centre O and radius OP-
378
Analytical Geometry 3-/)
On account of this fact the cone is termed as right circular
cone. Thus the right circular cone can also be defined as:
A cone which has a plane circular section whose normal passes
through the vertex is a called a right circular cone.
(B) Tofind the equation of a right circular cone.
(Kanpur 1979; Punjab 81; Lucknow 79)
Let A (a,
the axis AO be
y) be the vertex of the cone and the equations of
(x-a.)jl={y-^)lm==^{z-y)ln,
where /, m, n are the d.r.’s of the axis, Let the co-ordinates of
any point K on the surface of the cone be (x, y, z) so that the
d.r.*s of the line AR are x «»
2—y. If 5 is the semi¬
vertical angle of the cone i.e. the angle between the lines AO and
AR is 6, then it is given by
cos $*=●
/(x--o{)+m (y~B)+n (z-^v)
-f «=*)V{(x-
+(F-7)^’
Squaring and cross-multiplying, the required equation of the
right circular cone is given by
{/(x~a)+m (y-P)A-n (z-y)}^
=
m2-i{(X-a)H-O’-/3)»+(z- y)*} cosM. ...(I)
Particular cases. Case 1. If the vertex is at origin. If the
vertex ^ be taken at (0,0,0) then putting a=j8=y=0 in the
equation (1) above, the equation of the right circular cone with
the vertex at the origin and semi-vertical angle 6 is given by
ilx-\-my+nz)*^ (/*-l-m»+n®) ix^A-y^+z^) cos* d
or
(/x4-wy+«z>*=(/*^m*-|-n*) {x^+y^+z^) (1-sin* 6)
or
(/*-i-w*-l-n*) (x*-f>*+2*)sin* 0«(/*-f.m*-f«*) (x*-l-^*-j-2*)
—{IxArmyA-nz)^
or
(/’*+/wHn*) (x^+y^+z*) sin* d
=(wz—ny)2-i-(nx-/2)a+(/y-»ix)* ...(2)
[By Lagrange’s Identity)
Case II. If the vertex is at the origin, the axis of the cone is
the z-axis and the semi-vertical angle is 6.
(Kanpur 1980; Meerut 79, 82. 83, 86 P, 89 S)
Let OX, OY, OZ be the co-ordinate axes. The vertex of the
cone is at the origin O, the axis of the cone is along the z-axis
OZ and the semi-vertical angle is 6. Let R (x, y, z) be any
current point on the surface of the cone so that OiJ is a generator
of the cone.
379
Tke Cene
The d;c.’s of OZ are 0, 0, 1 and the d.r.’s of OR are x—0,
y—0,z-0
X. y, z. The angle between the lines OR and OZ
Therefore
is equal to the semi-vertical angle 0 of the cone.
Z
<9
y
cos d=
or
or
x.O+r.O+z.I
V(0H0^ 1“)
z* sec* 0«x*+>»*+z* or
z
+2*) V(^*+y*+2“)
x*+;'“=»z* (sec* 0—1)
x24-^*=*z* tan* 0.
...(3)
which is the required equation of the cone.
Case III. If the vertex is at the origin, axis the y-axis and
(Burdwan 1980)
semi-vertical angle Is 0.
The d.c.’s of the j/-axis are 0, 1,0. Hence putting l=n^0
and m=l in the equation (2) above, the required equation of the
cone is given by
siija 0=z*+x*, or (z*+x*)(1 —sin* 0)->»* sin* 0
..(4)
or
2*+x*=y* sin* 0.
Case IV. If the vertex is at the origin, axis the x-axis and
(Agra 1979, 81)
semi-vertical angle is 0.
The d.c.’s of the x-axis are 1, 0, 0. Hence putting m-=w«=0
and /«=>! in the equation (2) above, the required equation of the
cone is given by
(x*+/-fz*) sin* 0=iZ*+:F^ or(/+z*)(l-sin» 0)==x« sin* 0
tan* 0.
...(5)
or
3^0
Analytical Geometry 3-D
SOLVED EXAMPLES (G)
c«i
-i-u
*
(Agra 1977)
as (2 ~3 5i ^ *if
of the vertex A of the,cone are given
sinL the «il' V' '
"!'*
“●’*
““ °f ">e cone ‘ten
^ nee. the axis makes equal angles with the co-ordinate axes, we
±—”^JL „V(/»-hw°-i-)|S|
1
„
I
1
1
Vils-hf + lTj-yjd r W,h “ “
P°‘"'
y.
on the cone so that the
o .r. s of the generator AR are x-2, y+3 ^ r_5.
Now the semi-vertica!
between the
axis Of the conra^dfhfgVnr.or
ore. the
Jd“*‘fon of the right circular cone is giJen by
V3
or
ls%0^* ^e^
2
x+y+z-.4
vw{{x-2)^-\.(p+jjq:^^z:jyr^-
Squaring and cross-multiplying, we get
9 {x8~4x-l-4-i->;a^6;;+9-|-2*-10z+25}=-4 (x^+yt^z^
or
g
(.....; o^d its ::^Ltxr:i.''; ri
ejnat/y maimed to the ao-ordmate axes.
Sol.
j The co-ordinates of the
cone lls
Find the eguatiL of thd.
(1.0,1) .
vertex A of the cone are
the d'^r's
“■0point (i, 1, . 1) and hence
and (1, 1. 1)
Z i-l to^ flT^toTr
The axis of the
6 cone is equal ly inclined to the co-ordinate
axes and so its d.c.’s
»"= i/V-i. 1/V3, \/V3. [See Ex! 1 abovel.
Thus if 9 be the
semi-vcriical angle of the cone, then
cos d=Hl/v2)±HlV3i+0^V3j'
\fpTi»+Td>
^
1
V3'
.. (1)
The Cone
381
Now let R {x^ y, z) be a general point on the cone and so the
d.r.’s of the generator AR are x-1, y-0, z—1. The angle bet
ween the axis of the cone and the generator AR is also d and so
we have
cos f»-(j^~n.(l/V3)+J^.(l/A/3)-Hz-l).(l/^/3)
or
I
x-\~y~^Z‘—2
.
\/3“V3\/ix^ +J'*+z-2-2x-2z+2/
Squaring and cross-multiplying, we get
(x8+/+z2-2x-2z-f2)cp(:r+j;4.z_2)a
or
>>z-1-za:+jcy~;c—2;;—z-1-1=0.
This is the required equation of the right circular cone.
Ex.3(a). The axis of a right circular cone^ with vertex ar origin
O, makes equal angles with the co-ordinate axes and the cone passes
through the line drawn from O with direction cosines proportional to
* t —2, 2. Find the equation.
(Avadh 197$)
Sol. The vertex of the cone is the origin O (0, 0,0). The
axis of the cone is equally inclined to the co-ordinate axis and so
its d.c.’s are 1/V3, I/v^3, l/\/3[See Ex. I above]. Since the cone
is passing through a line with d.r.*s 1, -^2, 2 and so this line is a
generator of the cone. Let 9 be the semi-vertical angle of the cone
and so we have
. cos 6 (1/V3).l-}-(l/V3).(-2)+(l/V3).2 1
V{lH(-2)a+(2)»}“
...(I)
Now consider a general point R (x, y, z) on the cone and,
therefore, the d.r.’s of the generator OR are x-0, y-0.z-0 i.e.
X, y^ z. (fence 6 is also the angle between the axis of the cone
and the generator OR, Thus
cos 9 (l/V3).x-f(l/V3).>>+(l/V3).z
or
or
or
1
3V3
x-{-y-{-z
using (1)
(x»+j'Hz*).=9(x+;^+z)2
4x*-f-4y24- 4z2+9;>z+9zx-f
c=0.
This is the required equation of the cone.
Ex.3(b). Show that the equation of the right circular cone
which passes through the line 2x*=*3j;<=—5z and has x^y*=»z as its
axis also passes through the axes oj co-ordinates.
382
Analytical Geometry 3-D
Soli. The equatioDS of a generator of the cone are
2xc=3>>= —5z or
The axis of the cone is
jr/15=>»/10=z/(—6)
(1)
...(2)
xj\=yi\^zl\.
If 6 be the semi-vertical angle of the cone, then
19
1
15.1 +10.1 +(-6);i
cost?
V(15*+10*+(- b)*)V(i* +i‘+1*)“
V3 V3...(3)
Consider a general point R{x, y^ z) on the cone. Thie vertex
of the cone is the origin D. which is the point of intersection of
the lines (1) and (2). Therefore, the d.r.*s of the generator OR are
X—0,^—0,z—0 i.e. X, y^ z. We have
x.l-l-y.I-l-z.l
cos 6
or
or
or
1
x+y-\-z
, using (3)
V3 V3\/(xH.>'“+z>>
X*+;,;9^-22=(x+;;+Z)>
^z+zx+*>>=0.
This is the required equation of the cuOc which clearly passes
through the co-ordinate axes because 1, 0, 0; 0, 1, 0; and 0,0, 1
all satisfy the equation of the cone.
Ex. 4, Find the equation of the right circular cone whose vertex
is the origin and whose axis is the line x=t, y=>2t» z<=>3t and which
has a vertical angle of 60".
(Panjab 1982(S))
Sol. The vertex of the cone is 0(0, 0, 0) and the equations
of its axis are
...(1)
x«=f, j?=2/, Z‘=3/ i.e. x/I=>>/2*=»z/3=r.
the d.r.’s of the axis of the cone are 1, 2, 3.
Consider a general point R{x, y, z) on the cone. The d.r.’s of
the generator OR are x~0, y-0, z-0 i e. x, y, z. The vertical
angle of the cone is 60° and so its semi-vertical angle 30° is given
by
1 .x-H2.>>-|-3.z
cos 30°=
or
or
or
\/3_
x+2y+3z
2 V(l4;V(^*+>Hz*)
21 (xH>'*-fz*)=*2(x-|-2;;-|-3z)* ●
19x2-H J ^y2^2z*_24yz--I2zx-8x>'«0.
This is the required equation of the right circular cone.
383
The Cone
Ex. 5. Find the equation of the right circular cone whose axis
is x<=y^Zt vertex is the origin and whose semi-vertical angle Is 45®.
(Bordwao 1974)
Sol. Proceeding as in Ex. 4 above the required equation of
the cone is given by
x^+y^+z^-4 (yz-i-2X-i-xy)<=>0.
Ex. 6. Find the equation of the right circular cone with vertex
at {\, —2, —1), semi-vertical angle 60® and the axis
(Gorakhpur 1980)
(x= i)/3=.-(>'+2)/4=(z+1)/5.
Sol. The vertex of the cone is ^4(1, — 2,-1). The equations
of the axis of the cone are (x—1)/3*=»(;'+2)/(—4)=(z+l)/5.
The d.r.’s of the axis of the cone are 3» —4, 5. The semi
vertical angle of the cone is 60®.
Consider a general point R{x, y, z) on the cone and so the.
d.r.*s of the generator AR are x—1,y-\-2, z+l. Hence the requi
red equation of the right cone is given by
3.(x-l)+(-4).(y+2)+S.(z+l)
cos 00 =^{(3)a+(_4)a4.(5)a}^{(x-l)»+(;;+2r+(z+l)*}
■
3x-4;>+5z-6
or
’“■v/(50)V{^*+3'*+2»-2x + 4y+2z+6)
Squaring and cross-multiplying, we get
25 (x*-l-/+2*-2x+4y + 2z+6)=2(3x -4:V+5z-6)«
or 7x8 _ 7^8 _ 25z*+80>;z - 60zx -l-48xy-l-22x-l-4y-l-170z-l-78*=»0.
Ex. 7. Find the equation of the cone formed by rotating the line
2x+3y=6, z=0 about the y-axls.
Sol. A generator of the cone is given by
2x-l-3>'=»6, z=0 or 2 (x—3)=—3j», z=0
...(1)
or
● (x-3)/3=y/(-2)=z/0.
The axis of the cone is y-axis whose d.c.’s are 0,1, 0. The
vertex A of the cone is the point of intersection of (1) with the
y-axis {i.e. x=0, z =0) and is given by (0, 2, 0).
If 0 be the semi-vertical angle of the cone then it being the
angle between the generator (1) and the y-axis {i.e. the axis of the
cone) is given by
-2
3.0-K-2).l-f0.0
cos 0«=>
VU3)*+(^”+W}V{0“+l*+0*}
V(13) ...(2)
Consider a general point i?(x, y, z) on the cone and so the
d.r.’softhe generator AR are x—0, y—2, z—0 i.e. x, y—2, z.
Also the angle between the generator AR and the axis of the cone
{i.e. y-axis) is 0. Hence we have
384
Analytical Geometry 3-D
cos 6=
-2
or
VH3)
or
x.0+(y-2).\-{-z 0
V{x''‘-\-{y - 2)*+z*)V(0®+1*+0»)
y--2
.using (2)
V(x^+y^j^.z^-4y+4)
Squaring and cross-multiplying, we have
4 (x^+y^+z^-Ay+A)=]3 iy^-.4yA-4)
4x*—
-j-4z*4-36^—36=0.
This is the equation of the required cone.
Ex. 9. Find the equation of the cone generated by rotating the
line x/l^ylmc=>zln about the line xjar^^ylb^^zjc as axis,
Sol. The equations of the axis of the cone are
x/a=y/b=z/c.
...(1)
The equations of a. generator of the cone are
x!l==^ylm-^zjn.
...(2)
Let ^ be the semi-vertical angle of the cone then it is the
angle between the lines (1) and (2; and so we have
fl/+bm4cn
cos 0=
...(3)
Consider a general point R (x, y, z) on the cone. Now the
vertex A of the cone being the point of intersection of(1) and (2)
is given by (0. 0, 0). Thus the d.r.*s of the generator^/? are
jc—0, y—0,z~0 i.e. x, y, z. Also the angle between the axis (I)
and the generator AR is 0 and hence we have
ax-\-by-\-cz
cos 0=
^(0*46*4 c^Vix^ -'fj^+~zy
(4)
Equating the two values of cos 0 given by (3) and (4), the
required equation of the cone is given by
ax-hby^cz
alf-Jm-^cn
z^)V(/“4w‘4«"f
or
(ax-i-by-hczy (/*4 w*4n")-(xa4T®4z*)(a/46w4c/i)*.
Ex. 9(a). //* a right circular cone has three mutually perpendi
cular generators, then show that the semi-vertical angle is tan~^y/2.
tPunjab 1982 S; Madras 76)
Let Us consider the equation of the right circular cone
with the vertex at the origin, axis the z-axis and the semi-vertical
angle 0. Its equation is given by [See § 13(B)case II)
■x*4-THr* tan* 0=0.
...(1)
II the cone (1) has three rnutually perpendicular generators,
Sol.
The Cone
385
then the sum of the coeflScients of x^, and in the equation (1)
must be zero.
or tan* 0=2, or tan0= V2i
... l+H-(-tanM)-^
or
0=tan-/V2.
Ex. 9 (b). Ifa rlg^t circular cone has three mutually perpendi
cular tangent planes^ then show that the semi-vertical angle of the
cone is given by cof^ y/2.
Sol. Let us consider the equation of the right circular cone
with the vertex at the origin, axis the z-axis and the semi*vertical
angle 0. Its equation is given by [See § 13(B) case II]
...(1)
x^+y^-z'^ tan* 0=0.
The cone (1) will have three mutually perpendicular tangent
planes if
‘6c+ca+flb=/*+g*+A**
[See § 12]
or
l.(-tan* 0)+(-tan* 0).1+1.1=0+0+0
or
2tan*0*=l or cot* 0=2 or 0=cot“'V2.
Ex. 10. Lines are drawnfrom O with direction cosines propor
tional to (1, 2, 2);(2, 3, 6);(3, 4, 12). Show that the axis of the
right circular cone through them has direction cosines — 1/V3, l/V^*
l/\/3 and that the semi-vertical angle of the cone is cos~^ (1/V3).
(Meernt 1973)
Sol. Clearly the vertex of the cone is at the origin 0(0,0, 0).
Let the direction cosines of the axis of the right circular cone
be /, m, n. Let 0 be the semi*vertical angle.'
The cone is passing through the given lines drawn from O
and so these given lines are generators of the cone. Hence each
of these given lines with d.r.*s (1, 2, 2); (2, 3,6);(3, 4, 12) i.e,
d.c.’s (1/3, 2/3, 2/3);(2/7,3/7, 6/7);(3/13, 4/13, 12/13) is inclined
at an angle 0 to the axis of the cone whose d.c.*s are /, m, n.
We have
..(I)
cos 0=(I/3) /+(2/3) m+(2/3)n
...(2)
cos 0=(2/7)/+(3/7) m+(6/7) n
and
...(3)
cos 0=(3/13) /+(4/l3) W+C12/13) n.
Subtracting (2)from (I), we get
...(4)
0=(I/21)/+(5/2I)m-(4/21)n or /+5m-4n-:0.
Again subtracting (3)from (1), we get
4/+I4m-10n=0.
; Solving (4) and (5), we get
...(5)
386
Analytical Geometry 3-D
/
m
n
V(P-^m*+n»)
1
-I
1 “1
“^3The d.c.*s of the axis of the cone are
-1/V3, 1/^/3. 1/V3.
Putting the values of /, /w, n in (1), we get
3a/3+3^3T3-^3“3^3-^3
§ 14. The EoTeloping Cone.
or
«=C08-
I
Definition. The enveloping cone of a given surface is the locus
of the tangent lines drawn from a given point to the given surface. ●
The enveloping cone is also called the tangent cone to the surface
with the given point at its vertex.
(A) The equation of the enveloping cone. To.find the equa
tion of the enveloping cone of the surface(conicold) ax^^by*-\-cz^*=‘l
with the vertex at the point (jc,, yu z{).
(Kanpur 1976, 78)
The equation of the given surface is
...(1)
The equation of any line through (Xi, y^ z%) are
Xx—Xi)ll=^(y—yi)lm~iz-zi)ln^r (say).
...(2)
The co-ordinates of any point on the line (2) are
{Ir-^Xu mr+yu nr+Zi).
...(3)
Let the line (2) meet the given surface (1) at the point given
● by (3)* Then the co-ordinates of the point given by (3) will
satisfy the equation (I) and so we have
(mr+y,)*+c(«r-l-ri)a=l
or (o/*+hm*-|-c«*) r*+2(fl/;si-f hmyi+cnzi) r
+(oAf,HW+«i»-l)=0. ...(4)
If the line (2) is a tangent to the given surface (1), then it
will meet the surface (1) at two coincident point and hence the
two values of r obtained from (4)should be equal. The condition
for which is
or 4 (,alxi+bmyi.+ cnzi)*=4 (fl/“+hm?+cn*){axx^^byi*+cz^ — l)
or {alxi-\-bmyi’\‘cnzif={al^4-bm^-\-cn^)[axx*-\-byx^-{-czx^—l) ...(5)
the enveloping cone of the surface (1) /.c., the locus of the
tangent line (2) is obtained by eliminating /,
n between the
equations(2) and (5) and is therefore, given by
{a{x-xx)xx4-b{y-yx)yx^‘C{z-zx)Zxf
{x-^Xif-^-b (y-yxf^c {z-zxy)(axi>+i»yi»+C2i»~l)
387
The Cone
or {{axxx^hyyx+czzx -\)-{axx^+byx^-\-czx^-1)}*
c.{(flx»+^>/+«*-l)-2 (axxi-{^byyi-\-czzi-\)
Maxi^-{-byi^.+czx^-1)}X{axj^+byx^+czi*-l). ...(6)
For convenience let us set
Ssax^+by^-^cz^ — 1» Si^axi^+byi^+czi^— 1
..;(7)
and Tsaxxi+byyi+czzi—l,
.
Using the aboye notations, the equation (6) of the enveloping
cone becomes
(r-s',)2=(5-2r+5i).5i
or
T^^2TSt+Sx^=SSi-2TSx+Sx*
55,=r
or
where 5,5, and T are given by equations (7).
SOLVED EXAMPLES(H)
}
Ex. 1. Find the equation of the enveloping cone of the sphere
NX*+y*+z*=fl® with the vertex at the point (x,, yu z,).
(Panjab 1975; Allahabad 75)
Sol. Proceeding as in § 14(A), the enveloping cone is given
by
Ex. 2. Find the equation of the enveloping cone of the ellipsoid
1 with the vertex at the point(Xu yu Z\)»
Sol. Proceeding as in § 14(A), the equation of the enve
loping cone is given by
+v+a!
Ex. 3(a). Find the enveloping cone of the sphere
x^+y^-^z^-2y+6z-{-2=0 with its vertex at(1,1, 1).
Sol. We have 5=xa+>'»+2®-2j;4-6z+2c=0.
The vertex of the enveloping cone is (1, 1, !)■
/. 5,= l*+l'+P-2.H-6.l-b2=.9,
r=jc.l4->'.H-2.1-(y+I)+3 (z+l)-l-2«x+4z+4.
The required equation of the enveloping cone is given by
*55,«=r*
or
(x*+;;Hz’-2y+6z+2).9=(x+4z-b4)>
or 9x*+9>'2+9z*--i8:y+54z+18=»x2+16z*+16+8zx+8x-i-32z
or
8x*+9y*-7za-8zx-8x-18>;+22z+2=0.
Ex. 3 (b). Find the enveloping cone of the sphere x®+/-l-z*
—2x+4z= 1 with its vertex or (1, 1, 1).
Sol. Proceeding as in Ex. 3 (a) above, the required equa
tion of the enveloping cone is given by
388
Anafytfcal Geometry l-D
4x*+3;>*--.5z*~ 6>’2—8x+162—4«=d.
Ex. 4. Show that the lines drawn from the origin so as to touch
the sphere x^.+y*-\‘Z^+2ux-^2vy’h2wz-hd=0 He on the cone
d{x»+y^+z^)={ux-^vy+wz)K (Kanpur 1976)
Sol. Here it is required to find the enveloping cone of the
given sphere with the vertex at the origin.
The equation of the given sphere is.
2mx+2v>+2ivz+</=.0.
..(1)
We have 5',«0+0+0+6+0+0+^f=
r=x.0+>;.0+r.0+ti (JC+0)+v (;;+0)+w (z+0)+ rf
●=^ux-{-vy+wz+d.
Therefore the equation of the enveloping cone of the sphere
(1) with the vettex at the origin is given by
‘S'S'i=7'2»
or
(JC*+>'*+z*+2«x+2v;;+2wz+af}.rf=»(Mx4-v;;+H'z+rf)2
or
(x*+>>*+z*) d+2 (wx+vy+wz) d-^cfi
=(iix+vj>+m'z)H2 (uxi-vy+wz).d+d^
or
dix*-\‘y*+z^) = {ux+vy-\-wz)K
. Ex. 5. Prove that the plane z«0 cuts the enveloping cone of
the sphere x*+>'*+2*=!l which has its vertex at (2,4,1) in a
rectangular hyperbola,
Sol. Here
Ssx*+;;»+i*-H*0, 5'i«(2>*4-(4)»+(l)»-ll = 10,
r=x.2+>.4+z.l-ll«2x+4;/+z-ll.
therefore by the formula SS’i=J*, the equation of the
enveloping cone of the sphere S=0 with the vertex at (2, 4, 1) is
given by
(x*+.V‘+z»-11). 10c=.(2x-|-4;^+z-11
...(1)
The plane z=0 cuts the enveloping.cone (1) in the conic given
by
(x*+/-U).10=(2x+4:y-ll)*. z=0
or
6x>-6y>-16x>^+44x-88y-231 = 0, z=0.
...(2)
Now in the first of the equations (2),
the coeflBcient of x^+the coeflScient of y*=6+(—6)=0.
Hence the conic (2) represents a rectangular hyperbola.
Ex. .6, Find the locus of a luminous point which moves so that
the sphere x*+y*+z®—2oz«=«0 casts a parabolic shadow on the
plane z=aO.
Sol. Let the luminous point be P (Xi, y,, ii).
tibe equation of the given sphere is
389
the Cone
5=jc*4-^*+z*—
Q.
5'i«=i;if,2+;;,*+V—2fl^i and
fl («+Zi).
The equation of the enveloping cone of the sphere
with
the vertex at P (Xi, yi^ Zi)is given by iSS’i=r*
le. (x®+y*+z*-2atz)(Xi*+;;i*+Zi*~2flZi)
or
^{xxi-^yyx-Vzzx-a (z+Zi)}* ...0)
The plane z=0 cuts the cone (1) in the conic given by
W+yi*+zi®-2flZi)=*(xx,+yyi-flZi)>, z<=0
X* (yi^-\-Zx^—2azx)—2xxyxxy-^y^ {Xi*+Zi*—2azx)
+2aXxZiX-^2ayxZiy—a*zx*^0, z=0. ...(2)
Now we know that the conic
ax“+2Ax;;+h/+2gx+2/y+c=0, z=»0
represents a parabola if
Applying this condition the conic (2) will represent a parabola if
(-Jfi>'i)*=W+^i*-2flZi)(xi*+zi*-2flzi)
or
jc,*y,»=x,W+yi*Zi3-2flZiyiH^i*JCx*+2i*—2flZi»-2flZiXi*
—2azi®Hr4o*Zi*
or
Zi^ (Xx^+yi*+zx*-4qzi+4a^)-2azx
or
zi (Xx*+V+^i*-4fl2i+4fl*)-2a (Xi^+;;i»)=»0.
The locus of the point P (Xi, >>i, Zi) is
z {x^+y^+z^—4az+4a^)—2a (x*+y*)«=*0.
Ex. 7. The section of the enveloping cone of the ellipsoid
x^/a^+y^/b^=>z^lc^=\ whose vertex is P(Xx, yx, Zx) by the plane
z=0 is {i) a paraboia, (//) a rectangular hyperbola,(Hi) a circle.
(Meerut .t984R)
Find the locus ofP in the above three cases.
Sol. The equation of the given ellipsoid is
V+Z/h^+zVc*-1«0.
Therefore Sx-=Xi^/d»-\-yxVbHzxVc^-1,
and
T=xxxla^ '{●yyilb^+zzi/c*> -1.
The equation of the enveloping cone with the vertex at
SSx=T^
P (Xx, yi, zi) is
xxi .yyx .zzx
V*
or
c*
V
...(1) .
The section of the enveloping cone (1) by the plane z=0 is
7\a»
or
+^Ar
1U ^
6»
/ii!+
^/eELVV.5L"\
\a»
+c» /
zbO
1U 2x,x^2;;iy
\
a*
b*
0, z-0.
...(2j
390
Analytical Geometry 3-JP
_ . Case (i). If the conic (2) represents a parabola, then apply
ing the condition *h^=ab*, we have
or
a*b^ a* W
r b^ \a8
XiW_XxW
_yi^ . zW
a^b*.’ a^b^
b^c^
+c*
c®
8
c
‘/
a»
0
or
or
...(3)
But XiVa>+yi^/b^+zi^/c^-l¥>0 since the point P(xi,
Zi)
does not lie on the given ellipsoid.
Hence (3) gives ZiVc*—l=*0 or zi*=>c* or Zi«=±c.
.V The locus of the point P (Xi,
zj) is z=±c.
Case (ii). If the conic (2) represents a rectangular hyperbola,
then we should have
the coefficient of'**-i-the coefficient of;;2ca0
0
-or
or
a-^d*
+
or
c* (Xx*-|-;^i*)-Ha«+6») zx»=a»-l-h*.
The locus of the point P(Xu Pi, Zi) is
c* (x»-i-p®)+(aHh*)z®=a*-|-6».
Case (iii). If the conic (2) represents a circle then we should
have the coefficient of x^=the coefficient of p‘, and the coefficient
of;ep=0.
and
■■ i.(9’4-')4(S-‘4'-')
Xiyt’=0 i.e. Xi«=*0 or pi=0.
If xi=*0 then from (3), we'have
a^U*
or
or
V b^\c^
V
Pi“
a*h* c* \a« 6V a>* 6*
c*Pi*+(**-fl*)
c*.
/. in this case the locus ofP {Xu Pi, z,) is
c*p*rt-(6*—a*j z*«c* (h*—a*), x«=*0.
...(3)
391
The Cone
Kyi^O then from (3), we have
7^\c^ M d«V
or
.*.
'j
c*Xi*+(a*—i>®) 2i*=c*(a*- 6*).
in this case the locus of P
Zi) is
c^x^+ia^ - b^) 2a~c» (a*-6*).7^=q.
Ex. 8. Find the locus of a luminous point which moves so that
the ellipsoidx^ja^+y^lb^+z^Jc^=:\ casts a circular shadow on the
plane z=0.
Sol. Let P (Xi, yu Zi) be the luminous point. It is required
to find the locus of P when the section of the enveloping cone
of the given ellipsoid with the vertex at P(Xi, yu ^i) by Ibe
plane z=0 is a circle. Now proceed as in the case (iii) of Ex.7
above.
Ex. 9. Show that three mutually perpendicular tangent lines
can be drawn to the sphere
from any point on the
sphere x^+y^-\-z^=(312) r^.
Sol
The equations of the given spheres are
and xa+;y8+z»«(3/2) r>. ...(2)
...(1)
x^+y^+z^=r*
Let P(xi, yu zi) be.any point on the sphere (2); then
..(3)
Xi*+>’iH2i*='(3/2) r*.
Now three mutually perpendicular tangent lines can be drawn
to the sphere (1)from the point P if the enveloping cone of the
sphere (1) with the vertex at P has three mutually perpendicular
generators.
For the sphere (1), we have
2
and
T=xxi-i-yyi+zzi-r*.
The enveloping cone of the sphere (1) with the vertex at
‘5Si=2’*’
P(Xu yu zi) is
or
(X*+/
-r*)(Xi*+.Vi2+-r»)=(xxx +yyi+zzi-r«)».
It will have three mutually perpendicular generators if
the coeflF. of x*+lhe coefT. of j»*+the coeflf. of 2*=0
or
+(Xi»+Pi*»+2i‘-r*-2x“=0
or
2(Xi8+Pi*+2i*)«3r* or 2(3/2)
using (3)
Hence Proved.
or
f*=r>, which is true.
392
Analytical Geometry 3-D
Ex. 10, Find the locus of points from which three mutually
perpendicular tangent-lines can be drawn to the paraboloid
'
ax^i-by^^2cz.
Sol.' Let P (Xi, yu Zi) be any point whose locus is required.
Here
2cz,
-2czi,
and
T=»axxi+byyi-c(z+Zi).
A The enveloping cone of the given paraboloid with the
vertex at P(Xu yu zi) is
*5^5,=J2’,
i.e.
(ax^+by^—2cz)(aXiH W-2cz,)
...(1)
={axxi+byyi-c (z+Zi)}\
Now three mutually perpendicular tangent lines can be drawn
from P (Xi, ytt Zi) to S—0 if the enveloping cone (1) has three
mutually perpendicular generators, the condition for which is that
in the equation (I) the sum of the coefficients of x>,
and z»
should be equal to zero
i.e.
a {byj^-2czi)+b (axi^-2czi)--c^=0
or
ab {Xx^+yx^)--2c(a+6)Zi=c*.
The locus of P (x,, y^, z,) is
ab (X®
2c
zc=c>.
Ex. 11.
Find the locus of the pointsfrom which three mutually
perpendicular tangent lines can be drawn to the conicoid
ax^+by^-\-cz*=\.
Sal. Proceeding as in Ex. iO above the equation of the
required locus is
a (6+c) x»+^ (<?+fl)y+c (fl+Z>) 28=04.^4.^.
Exercises
1.
Fiijd the equation to the cone whose vertex is the origin
and whose base curve is the circle zs=>3, x^-^y^-=^9.
Ans. x*+j;*—z*e=0.
2. Findjbe equation of the cone with the vertex at the
origin and which passes through the curve
, flxH*>’Hc2>=l, ax*4-)3;;*=2z.
(Kanpur 1981)
Ans. 4z* (ax^-\-by^’j-cz*)=>(etx*-\-py^f.
3. Find the equation ol the coue with vertex at the origin
ana direction cosines of its generators satisfying the relation
3/*—4m*4-in*=0.
Ans. 3x*—4>'*4-5z*«=»0.
The Cone
393
4. Prove that the equation of a cone whose* vertex is the
point (a, j8, y) and base curve as
x=0 is
(yx—az)*«=4b(a—x)(ay—]8x).
S. Prove that the equation
jki2 _ 2;;8 4-3z2—^xy+5yz-6zx+8x—19y—2z—20*=»0
represents a cone. Show that the coordinates of its vertex are
(Lucknow 1982; Meernt 86 S)
6. Show that the vertex of the cone
7x»+2ya+2z*-10zx4*10xy+26x-2y*f2z-17=0
is (1.-2.2).
12
Central Conicoids
§1. latrodactioQ. In the chapter on ‘The Plane’ we have
shown that the general equation of first degree in x,;; and z
represents a plane surface. In a similar way the general equation
of second degree in x\ y and z namely .
* axa+^U'*+C2*+2/>'z+2gzx+2Ax;;+2wx+2v>»f2w2+J=0 (1)
always represents a conicoid or a quadratic surface.
. The equation (1) by a proper transformation of axes can be
reduced to the following standard form
ax*+4iy>+cz*=l.
...(2)
The standard form (2) is called the central conicoid.
Now we shall discuss below some important properties of the
central conicoid given by (2).
(a) The origin is the centre of the central conicoid
ox*+6>>*+cz*c=3
'
Let P(xj, yu Zi) be any point on the central conicoid (2), so
that we have
ax,a-t-6V+«i*=.l
which can also be written as .
'
(-xOHh i-yi)^+c (-zi)»=.1.
...(3)
The relation (3) shows that the point Q{—Xu ->i, -Zi) will
also lie on (2) if />(x„ y^zi) lies on (2). The middle J)oint of the
chord PQ Is (0,0, 0). This shows that all/chords of the surface
(2) which pass through the origin are bisected, at the origin.
Therefore the surface (2) has a centre and it is at the origin. Thus,
origin is the centre of the central conicoid (2) and due to this
property the surface (2) is called a central conicoid.
(b) To prove that the co-ordinate planes bisect all chords per
pendicular to them.
Let P(x,i yu zi) be a point on the central conicoid (2), then
clearly the point fi(—*i,yu Zi) also lies on it. The ^middle point
of the chord PQ is (0, yu Zi) which clearly lies on the YOZ plane
395
Centrral Contcolds
i.e. the plane x^O. The d.r.’s of the chord PQ are 2xi. 0, O^.c.
1,0. 0 showing that the chord PQ is perpendicular to the plane
YOZ Hence we conclude that the co-ordinate plane TOZ bisects
all chords perpendicular to it. Similarly we can prove that the
co ordinate planes ZOZand ZOFalso bisect all chords perpendi
cular to them respectively.
Hence the central conicoid (2) is symmetrical with respect to
all the three co-ordinate planes and these planes ar« called the
principal planes of the central conicoid.
The three principal planes {U. the co-ordinate planes in this
case) taken in pairs intersect in three lines {i.e. the co-ordinate
axes in this case) which are called the principal axes of the central
conicoid.
(c) The standard form (2) of the central conicoid can take
the following three forms each of which represents a surface with
a definite name assigned to it:
V*
(")
X*
V*
2*
=l
(Ellipsoid)
2^
..... x^ y^ 2
(Hyperboloid of one sheet)
1.
(Hyperboloid of two sheets)
Now in the following articles we shall discuss the above three
surfaces {i.e. central conicoids) in details.
§ 2. The Fltipsoid.
The standard equation ol the ellipsoid is given by
..(1)
(i) The origin is the centre of(he ellipsoid given by (1). Pro
ceeding exactly as in § 1 (d) we see that all the chords passing
through the origin are bisected at the origin. Therefore the surface
(1) has the origin as it centre,
(ii) The co-ordinate planes bisect all chords perpendicular to
them. Proceeding exactly as in § 1(b), we see that the co-ordinate
planes bisect all chords perpendicular to them. Hence the ellipsoid
(1) is symnictrical "ith respect to all the three co-ordinate planes
● and these planes arc called the principal planes of the ellipsoid.
The three principal planes taken in pairs intersect in. three
lines (i.e. the co-ordinate axes in this case) which are called the
principal axes of the ellipsoid.
396
Analytical Geometry 3-Z)
(iii) The ellipsoid is a closed surface.
be rewritten as
The equation (1) may
...(2)
If X is numerically greater than a, then from the equation (2)
we find that y^fb'*-\-z*lc^==a -ve quantity i.e. the sum of two
perfect squares is a negative quantity which shows that the least
one of;; and z should be imaginary. Hence it follows that X can<>
not be numerically greater than a and therefore, the surface exists
between the two parallel planes
and x —a. In a similar
manner it follows that the surface (I) lies between the parallel
planesyz=b, y~—b and
z= c.
Thus the ellipsoid (1) is a closed surface,
(iy) The intercepts on the co-ordinate axes. The equations of
the x-axis are y=0,
0. It meets the surface (1) in the points
A{Of 0,0) and A’{—a, 0, 0). Therefore, the surface (I) intercepts
a length 2a on the x>axis. In a similar manner the surface (1)
intercepts lengths 2b and 2c on the;; and z axes respectively.
These intercepts 2a, 26 and 2c are called the lengths of the axes of
the ellipsoid.
(v) The sections of the ellipsoid by the planes parallel to the
co-ordinate planes. Consider a plane z=A parallel to the XOY
plane.
The section of the ellipsoid (1) by the plane z=A is the ellipse
given by
xV^->●®/6^=*l~A*/c^^=A.
The lengths of the semi-axes of the ellipse (3) are
a V( 1 - A Ve*) and 6 V( I - A»/c*).
Now the following cases arise ;
...(3)
^
numerically greater than c. Wje have proved
in (iii) ot this article that the ellipsoid is a closed surface bounded
by the planes x=a, x=.-a; y=b, y=.-^b and z=c, z= -c. Hence
A cannot be numerically greater than c otherwise the section will
not be real /.e., the plane z«=>A will not cut the ellipsoid.
Case II. If A=c or A ●—c. In this case (3) becomes
showing that x=0,:y=.0. Therefore, the section is an infinitesinally small ellipse or in other words the section reduces to the
points (0, 0, c) and (0, 0, — c) and both the planes z=c and z= —c
are the tangent planes to the ellipsoid (1) at these points.
Central Contcotds
397
Case III. 7/’A«0. In this case(3) becomes
x^la*+y^lb*=»\, z^O
which is an ellipse in the XOY plane.
Case IV. IfX increasesfrom 0 to c. When A increases from
0 to Ct the quantity (l^A^c^) becomes smaller and smaller and
vanishes at A=c. Hence the lengths a'\/{l—\^lc^) and b\/il—X^Jc^)
of the semi-axes of the ellipse (3) become smaller >nd smaller and
vanish at A=>c. Hence the ellipse (3) diminish in size as we go
away from the XOY plane.
Now if we consider the section of the ellipsoid (1) by the
plane z= — A, all the above arguments hold good.
Similarly the sections of the ellipsoid (1) by the planes para
llel to the planes YOZ and ZQX can be considered,
(vi) Sketch. From the above discussion it is possible to draw
a rough sketch of the ellipsoid (I) and is given below :
§ 3. The Hyperboloid of one sheet.
The standard equation of the hyperboloid of one sheet is
given by
+z^jc*=1.
...(1)
(i) Proceeding exactly as in § 1(a) we find that all the chords
passing through the origin are bisected at the origin. Therefore,
the surface (1) has the origin as its centre,
(ii) Proceeding exactly as in § 1(b) we observe that the
co-ordinate planes bisect all chords perpendicular to them. Hence
we conclude that the hyperbofoid of one sheet (1) is \,symmetrical
398
Analytical Geometry 3-D
with respect to all the three co-ordinate planes and these planes
are called the principal planes of the surface (1).
The three principal planes taken in pairs intersect in three
lines (t.e. the co-ordinate axes in the present case) which are called
./
the principal axes of the surface.
(iii)
The intercepts on the co-ordinate axes. The equations of
the x-axis are ;^=0,z=0. It meets the surface(1) in the points
A{a, 0, 0) and
0, 0). Therefore, the surface (1) intercepts
a length la on the x-axis. In a similar manner the surface (1)
intercepts a length 2b on the j;-axis. The 2-axis i.e. x=0, y=0
meets the surface (1) in imaginary points [z-iVC—0]
z*axis does not meet the surface (1) in real points,
(iv) The sections of the hyperboloid of one sheet by the planes
parallel to the co-ordinate planes. The equation of any plane
parallel to
plane is given by 2^ A. The section of the sur
face (1) by the plane z=A is the ellipse given by
1+A*/c®, z=A.
...(2)
The lengths of the semi-axes of the.ellipse (2) are
oVd+AVc*) and hV(H-AV)
and its centre is (0, 0, A) i.e. centre lies on the z-axis at a distance
A from the origin. Since the quantity \/(l + A*/c^) remains real
for all real values of A, we. get an infinite number of elliptical
sections as A increases numerically from 0 to oo. Also the sizes
of these ellipses increase as the value of A increases i.e. as we go
away from the XOY plane.
2=
In a similar way the sections of th.e surface (1) by the planes
—A are the ellipses on the other Sides of the plane XOY.
The section of the surface (I) by the plane x<=*A (i.e. plane
parallel to YOZ plane) is the hyperbola given by
yW~zVc^=l~xya\ x=A.
Similarly the section of the surface (1) by the plane y=A
(i.e, a plane parallel ZOX plane) is the hyperbola given by
xya^-~zyc^=\-.\^lb\y=X.
<v) Sketch. From the above discussion it is possible to draw
a rough sketch of the hyperboloid of one sheet and is given
below :
399
Central Conlcotds
r
//
/
/
//
//
'
V
»/
I
y'
■●x
(
r—
\
»
I
X'
2o
X
I
I
J
I
-J—
I
I
I
J.
/
y
I
-n'\
\
T
t
//
4^5E'
§ 4. The Hyperboloid of two sheets.
is given by
Its standard equation
:ca/a»-//b*-2»/c2=»l.
...(1)
(i) Proceeding exactly as in § 1(a), we fi nd that all the chords
passing through the origin are bisected at the origin. Therefore,
the surface (1) has the origin as its centre.
(ii) Proceeding exactly as in § 1(b) we observe that the co
ordinate planes bisect all chords perpendicular to them. Hence
we conclude that the hyperboloid of two sheets given by the equa
tion (1) is symmetrical with respect to all the three co-ordinate
planes and these planes are called the principal planes of the
surface (1).
400
Analytical Geometry 3-D
The three principal planes taken in pairs intersect in three
lines (/ e, the co-ordinate axes in the present case) which are called
the principal axes of the surface (I).
(iii) The Intercepts on the co-ordinate axes. The x-axis (y=0t
r=»0) meets, the surface (1) in the points A (a, 0, 0) and
A* {—a,0,0). Therefore, the surface (I) intercepts a length 2a on
the x-axis. But the y and z axes do not meet the surface (1) in
real points.
(iv). The sections of the hyperboloid of two sheets by the planes
parallel to the co-ordinate planes. The equation of any plane para
llel to YOZ plane is given by x=A. The section of the surface (1)
by the plane x«=»A is the ellipse given by
x=A.
...(2)
The lengths of the semi-axes of the ellipse (2) are
b\/(X*/a*^\) and cv'(AVfl*‘-l)
and its centre is (A, 0,0) t.e. the centre lies on the x-axis at a
distance A from the origin. The ellipses (2) are real .when A is
numerically greater than a and increase in size as A increases
beyond a. If A is numerically less than a, then the ellipses given
by (2) become imaginary and hence no portion of the surface lies
between the planes x=a and x= — a.
The sections of the surface (1) by the planes y=A and z=A
are hyperbolas given by the equations
xya^-z*lc^=.1 -1-A8/6*. y=A
and
respectively,
(v) Sketch. Prom the above discussion it is possible to draw
a rough sketch of the hyperbolid of. two sheets and is given
ahead.
Remark. From the articles 1 to 4 discussed above it follows
that the standard equation ax^+by^+cz^=^l of the central conicoid represents
(i) an ellipsoid if a,6, c are all positive,
(ii) a hyperboloid of one sheet if any two of a, b, c are posi
tive and the remaining third is negative, and
(iii) a hyperboloid of two sheets if any two of a, b and c are
negative and the remaining third is positive.
401
Central Conicoids
2T
X'
o
X
y
2'
§5. The tangeiit plane.
(A) To find the equation of the tangent plane to the central
conkoid ax^-\-h^+cz^—\ at the point (Xi. yu «i)*
(Kanpur 1979, 81; Lucknow 78i 80)
The equation of the central conicoid is
...(1)
Let (Xi, yif Zi) be any given point on the surface (1).
The equations of any line through (Xi, yu n) a^d having
d.c.’s /, w,n are given by
...(2)
(x-xi)//=(y->'i)/w=(^-«i)/»*''(8ay)^
The co-ordinates of any point on the line (2) are (/r+^i,
wr4-y„ «r+zi). Let theline (2)meet the conicoid (1) at this point,
then this point will satisfy the equation (1) and so we have
fl (/r+Xi)»+*(mr+Vi)Hc(«r+^i)*=l
or
r» (fl/*+&m*+cn*)+2r {alXi-\-bmyi+cnzi)
-^(axi^+byi^+czi^-l)^0- ...(3)
Since the point (Xi, y^ z,) lies on the conicoid (1), we have
...(4)
axi^+byi^-\‘cz^*=\.
Using (4), the equation (3) becomes
...(5)
(al^+bm^+cn^)-{-2r (n/xi+hmyi-|-c«2,)=0
The equation (5) is a quadratic equation in r and hence gives
two values of r i.e. the line (2) meets the surface (1)in two points.
402
Analytical Geometry -3D
If the line (2) is a tangent line to the conicoid (1) at the point
(■^1*
^i) then the two values of r should be coincident. Clearly
one value of r given by (5) is zero and hence in order that the
line (2) is a tangent line to the conicoid at (jfi,
z,), the other
value of r should also be zero the condition for which is
alXi-\-bmyt-^cnzi=0.
...(6)
The tangent plane to the conicoid at (x“j, yi, zj) is the locus
of such lines through (xi, yi, Zi) that satisfy the condition (6) and
so its equation is obtained by eliminating /,/n, «, between the
equations (2) and (^). Hence the required-equation of the tangent
plane to the conicoid (1) at the point (jf,, y^ Zi) is given by
a (x-xi) xi-i-b (y~yi) yi-\-c (z-z,) zi=0
or
axxi-]rbyyi-\-czz^=aXi^-^byi^-\-cZi^
or
axXi-^byyi-hczzi=\.
using (4).
Rule. To write the tangent plane to the conicoid at the point
(^i»
Zj) replace x® by xx^ y^ by yyi and z* by zzi(B) To find the equation of the tangent plane to the ellipsoid
xVa*+:vV*®+z7ca= 1 at the point (xj, y^, Zi).
Proceeding exactly as in § 5 (A) and replacing a by ](a^, h by
\{b^ and c by 1/c® everywhere, the required equation of.the tan
gent plane is given by
XXila^-^yyxfb^-\-zzilc^=\,
§ 6. The. condition of tangency.
(A) To find the condition that the plane lx-\-my-\-nz=p may
touch the central conicoid ax^ -f
cz® = I.
(Agra 1980, 82 ; Allahabad 78; Lucknow 82; Kanpur 78;
Meerut 83S, 85; Rohilkhand 77; Punjab 7.8; Rajasthan 71)'
The equation of the given pla^e is
/x-fwy+nz=p
. .(1)
and the equation of the given conicoid is
a**-f-by®-i-cz*= 1.
...(2)
Let the plane (1) touch the conicoid (2) at the point (xj, yj, zi).
The equation of the tangent plane to the conicoid (2) at the point
(^1.
^i) is
axxi-\-byyx+czZx=\..
...(3)
If the plane (1) touches the conicoid (2) at the point (xj, yx^
z,), then the equations (1) and (3) represent the same plane and
hence comparing their coefficients, we have
n P
I
m
n
_T __ m
f
or
Xj=
axx~byx~~cz^\
bp'
cp
...(4)
Central Conicoids
403
Again since {Xu Vi,
iies on (2), we have
axi^+byi^-{-czi^= 1
using(4)
or
a (//flp)®+* (mlbpY-\-c {nlcpY-=\>
...(5)
or
l^la+m^lb+n^lc=p^
u
This is the required condition that the plane (1) touches the
conicoid (2).
● u
The co-ordinates of the point of contact of the plane (1) wit
the conicoid (2) are given by (4) U are {Ijap. mlbp, n}cp) where
p is given by (5).
Note. Substituting the value of p from (5) in (1), it immedi
ately follows that the planes
// /a ml nl\
Ix+my-i-m
always touch the central conicoid (2).
(B) Tofind the condition that the plane lx+mylrnz===P^^y
touch the ellipsoid x^Id^+y^Ib^-\-z^lc^= I.
-> oa
(Rajasthan 1974, 76, 78; Kurnkshetra 77; Bnrdwan 76,8^
Lucknow 86)
Proceeding exactly as in § 6(A) above and replacing fl by
l/fl»; b by l/h« and c by l/c^ everywhere, the required condition of
tangency is given by
...(1)
Note. From (1), we get
p=±V{a^P-\-blm^-\-c^n^)- '
Using these values of p,
obsej^e that t^be plaiies
—Ix^my+7tz^=^y/ia^ll+b'^m^^€lrPj
always touch the ellipsoid jPla^+y^llP+z^lc^^ 1.
§ 7. The Director Sphere.
Definition. The director sphere is the locus of the point ofinter”
section of three mutually perpendicular, tangent planes to a central
conicoid.
(Rohilkhand ’977; Allahabad 79)
The equation of the director sphere. Tofind the equation of the
director sphere of the central conicoid ax^+h)'®+«*= 1.
(Rohilkhand 1981; Lucknow 79; Kanpur 76; Meerut 84 S,89 S)
The equation of the central conicoid is
(1)
Let the equations of three tangent planes to(1) be
. liX+miy-jr «iZ=
l^x+miy-^ntz=
...(2)
404
Analytical Geometry '^-D
and
/3X+Way4-W3Z=V{/a*/®+
Wa*M
...(4)
where /j, Wj, /„
m and /g. Wg, «a ar® the actual direction
cosines of the normals to the above three tangent planes respectively.. If these tangent planes are mutually perpendicular, then
we have
A*+/2®+V=1, m,*4-W3Hm8*=I, «iHnaHw8*=l
.. (5)
^Wi+./8w,+/aW8=0, mi«i+m8W2+W8n3=0, nj/i-f Wa/a+Vs=0*
● -(6)
The director sphere of the surface (1), being the locus of the
point of intersection of the three tangent planes (2),(3) and (4) is
obtained by eliminating /j, mi, n,; /g, mj,- and /g,
Wg between
the equations? (2),(3) and (4) with the help of the relations (5)
and (6). Squaring and adding (2),(3) and (4^, we get
+«iz)2+
mjy+Wgz)?H-(/jX-l-Way H-Waz)*
, =(l/fl) r /iH(l/^)2 mi*H-(l/c) 2? «i2
or
y* TwiHz"^ »i"+2yz 2 Wi«i+2zjc.2’
●+2A:y 2/iWi=(l/fl)+(l/6)+(l/c),
using (5)
or
;«Hy**+za=(l/tr)+(I/A)-f(l/c),
using (5) and (6).
This is the required equation of the director sphere of the
central cpnicoid (1).
Corollary. If the central conicoid be the ellipsoid
then proceeding exactly as above, the equation of- its director
sphere is given by
1977; Rajasthan 71;_Goriiqanakdev 75)
SOLVED EXAMPLES (A)
Ex. 1. Find the equation of the tangent plane to the central
conicoid 3x8-5y«+z*+2=0 at the point (1,1, 0).
Sol. The equation of the given conicoid is
3x*-5y*-f z®+2=0.
...(1)
The equation of the tangent plane to the conicoid (1) at the
point (1, 1, 0) [See Rule to § 5] is given by
Ans.
3;e.l r^5y.l+z.0+2=0 or 3x—5y+2=0.
Ex. 2 (a). Find the equations to the tangent planes to the ellipsaid :^{cfi+y'^/b'^+z^fc‘= 1 which are parallel to the plane
(Punjab 1972; Rohilkhand 80)
Sol. The equation of the given ellipsoid is
...(1)
;c2/a8+ya/f,2+z2/ca=l.
Central Conicoids
405
Let the equation of a tangent plane parallel to th6 plane
lx-\-my’\-nz=ii be given by
...(2)
lx-^my-\-nz=^p.
If the plane (2) touches the ellipsoid (1) then we have [See
§ 6, (B) give complete proof.]
..(3)
Putting the values of/» from (3) in (2), the required equations
of the tangent planes are
/x4tm;;+/iz= ±
Ex. 2(b). Find the equations of two tangents planes of the
surface ax^+by^-]-cz^=\ which are parallel to the plane
lx-\-my-\-nz—0.
(M.U. 1990)
Sol. Proceeding as in Ex 2(a) above, the tangent planes are
lx-\-my+nz== ± ViFla+myb+n^/c)
Ex. 2(c). Show that the plane x+2y+3z==2 touches the conicoid
(Gnruoanakdev 1976, Meerut 83)
X®—2j;®+3z®=2.
Sol. The.equation of the given conicoid is
...(1)
;c2 -2v®+32®=2 or (1/2) x®-j^®+(3/2) 2®=1.
The equation of the given plane is
...(2)
x+2j>+32=2.
If the plane (2) touches the conicoid (1), then applying the
condition /®/a+m*/fe4 «*/<?=/»*[See § 6(A)], we have
1
172+^1
- 2-4+6=4 or 4=4.
which is true, Hence the given plane (2) touches the given conicoid.
Ex. 3. Find the equations to the tangent planes to the hyper
boloid 2x®-6:i;®+32®=5 which pass through the line
3x-3>»+62-5=0=x+9j?-32.
(Agra 1976; Rajasthan 74; Meerut86)
Sol. The equation of the given hyperboloid is
2x®-6;>®+3z®=5 or (2/5) x®—(6/5)/+(3/5) 2*«=»1. ...(1)
The equations of the given line are
...(2)
3;c-3;,+6z-5«0, x+9y~3z=0.
The equation of any plane through the line(2)is
3x-3;;+6z-5+A (x+9;^-3z)=0
...(3)
or
(3+A) x+(-3+9A):v+(6-3A)z=5.
If the plane (3) touches the hyperboloid (1), then applying
the condition /®/a+/M*/b+nVc=P*[See § 6(A)], we have
(5/2)(3+A)®+(-5/6)(-3+9A)®+(5/3)(6-3A)®=(5)»
or
15(9+6A+A*)-5(9-54A-f81A®)+10(36-3fA+9A®)=150
'406
Analytical Geometry 3-Z)
or
or
3(9+6A4-.A2)-(9-54A+81A2.)+2(36-36A+9Aa)=30
-60A*+60=0 or A*=l or A=±l.,
When A=I,from (3) the equation of the tangent plane is
4;c4-6;'+3z=5.
When A= — 1, from (3) the equation of the.Urgent plane is
2x—l2y-\-9z=5.
Ex^ 4. A tangerit plane to the ellipsoid x^la^+y^lb^+z^lc^=^
meets the co-ordinate axes in points PyQandR. Prove that the
centroid of the triangle PQR lies on the surface
(Rohilkband I978j Kanpur 77; Agra 75; Meerut 85S, 89)
Sol. The equation of the given ellipsoid is
...0)
x^.la^+y^lb^ArZ^!c^=°\.
The equation of any tangent plane to the ellipsoid (1) is
[See Note to § 0(B)]
...(2)
/x+/wy+/iz— y/fcPP-\-lfrn:’-\-c^r^).
The plane(2) meets the co-ordinate axes in the points given
by
P W(aH-+b^mHc^n^)ll, 0,0}; Q (0, v'(«®/HA®mHcV)/m,0}
and
. R {0,0,
Let (a,)?, y) be the co-ordinates of the centroid of the triangle
PQR,then
a _ j/(ay+ft^m«+c»«»)//+0+0 ^1
1
P^ 3m
V(a’/‘+A*m*+cV), y=^ v'(«‘''‘+iW+c%»).
Thus 3fe=
or 9/^a-=a®/®-l-i»®mHc®«%
. fl*
●●
9/2fl*
9a*/8
0,8 *-9/8fle8 “a8/a^^2;„a^.c8„8 ●
IP
96*m^
c*
9c2/i*
Similarly p2=a*/H6®m*-fc»/|2 ’
Adding the relations in (3), we get
fl* 6® c® 9
6®m*-l-c*rt2) ^
...(3)
The locus of the centroid (a, p, y) is
Proved.
fl®/jc*-l-W+^/z*=9.
Ex.5(a). Tangent planes are drawn to the ellipsoid ●
through the point (ot, p, y). Prove that the
perpendiculars to them from the origin generate the cone
(Allahabad (980)
v.«x+p3'+7z)®=(a-x®-f 6®y®-}-c-z®).
401
Central Conlcolds
Sol. The equation of the ellipsoid is
...(1)
The equation of any plane through the point (a, ji, y) is
/(x-a)+w (.v-^)+»(^-y)=o
...(2)
or
/x++nz=/a+m/i+ijy.
If the plane (2) touches the ellipsoid (1), then we know that
c»/i®=(/a+
[See § 6(B)].
.● .(3)
Also the equations of the perpendicular from the origin to
xjl-=ylm^zln,
...(4)
the plane (2) are given by
Eliminating /, m, n between (3) and (4), the required equation
of the cone generated by the line (4) is given by
Proved.
a^-x^+bY+c'z^Hix+my+nz)K
Ex. 5 (b). Tangent planes are drawn to the ellipsoid
xVd^+y^lb'^+z^lc^—^ through the point (a,^, y). Prove that the
perpendiculars to them through (a, y) generate the cone
(jc_«)+j9 {y
(z~-y))*<^a^ix -
(z-r)"*
Sol. Rewrite all steps of Ex. 5(a) upto the equation.(3).
Also the equations of the perpendicular to the plane (2) from
the point (a, fi, y) are given by
(X - U)ll =>(y-P)lm=(z - y)ln.
... (4)
Eliminating
between (3) and (4), the required equa
tion of the cone generated by (4) is given by
flS fx-oLy^b^ iy—fiy-^c^ (z-y)H{« (x-a)-\-/} (y-P)+r (z-y)}*
Proved.
Ex. 5 (c), Tangent planes are drawn to the conicoid
nx2+hy*+cz*=l through the point
Prove that the per¬
pendiculars to them from the origin generate the cone
(ax+j9y+yz)««(x*/n)+(3'W4-'z*/c). (Kaapw 1976^ 81)
Also show that the reciprocal cone of the above cone is
.(ax»+hyHcz»)(flaHft)9Mcy»-l)-T(aax+^ri5y+cyz)2=0..
and hence show that the tangent planes envelope the cone
(flxHhy*+cz«~l) {aa^+bfi^-{-cy*-})=‘{a9.x-\rbpy^cyz-1)»,
Solution i Part. Proceeding exactly as in Ex.. 5 (a) above and
replacing a* by 1/n, b^ by 1/6 and c* by 1/c, the equation of the
required cone is given by
...(1)
(otx+fiy-{-yzy^{xVa)+(yVb)MzVcy
Proved.
II Part. To find the equation of tie Reciprocal, cone to (1).
The equation (1) may be rewritten as
X* (a*-l/fl)+/(/?*--l/6)+z* (y*-l/c)+2^yyz+2yazx4-2a^xy=0.
408
Analytical Geometry 3-D
Let the equation of the cone reciprocal to the cone (2) be
^x^+By'^-^Cz^-\-lFyz-\-2Gzx-\-lHxy=^Q.
...(3)
Then
t-- JL\ _
a(b?^-\-cy^-\)
abc
ax^ (a9.^+b0^+cy^-l)-'a^a?x^
■
abc
a.nd Cz®.
"U + c "be )~
ax^ (6^2-fcyg—1)
abc
with similar expressions for
Ax^=
Also F>=>gh - af=(ya).{0Lfi)-(a^-1(a)(py)=fiyla,
C=yajb, H=afilc.
Putting these values in (3), the required equation
reciprocal cone is given by
I
abc [{flx*(aa*+*/?*+ey^-l)~aW}
of the
+{by»(uaH
_ i)_
}
+{cz2
1)-eVz^)}]
, ^yyz . 2yxzx 2afi
a
“T"+T
or
.
(flxH^3'*+c23)(aaa+6pHcy2-.l) -{a^ofix^+b^fiYi.cY2^
■-2bcpyyz-2cay9izx—2abctfixy]^0
or (ax--f y+cz^) {ax*-]rl>P^-\ cy^--\)=(a(xx-\-bPy-\-cyz)K Proved.
Ill Part. To find the equation of the enveloping cone.
The equation of the enveloping cone is given by SSibsT^.
Hence S—ax^+by^+cz^~\. The point is (a, /?, y).
Si<=aa^^-^bfi’^-\-cy^ — l
and
T^—aax+bPy-^cyz—\.,
Hence the required equation of the enveloping cone is
(flxa-[-V+02*-l) {au^+bpHcy^ - i)={aoLX+bfiy+cyz-i)K
Ex. 5 (d). Tangent planes are drawn to the central conicoid
ax?+£»>*+cz*==l through the point (a, ;5, y). Show that ikeperpendiculars from the centre of the conicoid to these planes generate
the cone
(ax f py+yzY==x^la+yyb+z^jc.
(Rohilkhaod l!W9; Kanpur 81; Madras 77)
Sol. The centre of the conicoid ax^+by^-{-cz^=\ is the
« 'gin. Hence it is the same problem as Ex. 5 (a) above.
Ex. 6. Find the locus of the foot of the centre perpendicular
on varying tangent planes to the ellipsoid
(Meerut 1984; Rajasthan 75; Kanpur 81)
409
Central Conicoids
Sol. The equation of the given ellipsoid is
...(1)
The equation of any tangent plane to the ellipsoid (1) is
...(2)
ix-\-my-\-nz~y/{a^l^+
●
The centre of the ellipsoid (I) is tlje origin and hence the
equations of the normal to the plane (2) from the centre (0, 0,0)
of(1) are
x//=y/m==z/«.
The foot of the perpendicular from the centre on the tangent
plane (2) is the point where the line(3) meets the plane (2) and
hence its locus is obtained by eliminating /, m, n between (2) and
(3) and, therefore, is given by
x.x +y.y+z.z=^y/(a^x^+b^y^+c^z^)’
Squaring, the required locus is
(;c2+j,24.22)a=,a2;cH*V4 cV.
Ex. 7. Find the locus of the perpendiculars from the origin to
the tangent planes to Ihe surface )i‘la>+yW+^lc‘=^ y>hiehcutoff
from its axes intereeptSt the Sunt of whose reciprocals is eguai to a
constant k.
Sol. The equation of the given ellipsoid is
...(1)
The equation of any tangent plane to the ellipsoid (1) is
...(2) .
/;c+ my+nz=y/{a^F+b^m^+
●
The intercepts made by the plane (2) on the co-ordinate axes
are
and
respectively.
The sum of the reciprocals of these intercepts is given to be k.
l+m+n
...(3)
^^m+nf=k^(aV^-^b^m^i-c^n^).
Now the equations of the perpendicular from the origin to
the tangent plane (2) are
...(4)
x//=7//H=z/rt.
or
Eliminating /, m, n between (3) and (4), the locus of the per
pendicular (4) is given by
Ans.
ix+yi z)^=k^-(fl2v«+W+c«22).
Ex. 8. Prove that the equation to the two tangent pllmes to
the centralconicoidax^^-by^-\-cz^-=-\ which pass through the line
w.=liX-\-miy-\-niZ -pi=0, «a=V^^'”2>'+”2^"P2=0 is
410
Analytical Geometry 3-/)
2
2/
W,2
//i/a . »U/7i3 ,«iWa
\
Vfl"^ b
Sol. The equation of the given conicoid is
...(1)
ax^’i-by'^-^cz^=\.
The equation of any plane through the given line Mi=0=Ma is
...(2)
«i+Am2="0
or
(/i.v+miy+«iZ-/;i)-|-A (/aX+may+«aZ-pa)=0
OF
...(3)
(/i+A/a) x-|-(n7i+Aw8) 37+(Wi+A«a)
If the plane (3) touches the conicoid (1) then applying the
condition "'l^ja+m^lb -f nyc=p^*\ wc have
(/i+A/a)*/n +(mi+Ama)V6+(ni+A«a)7c=(Pi+Apa)^
..●(4)
Now from (2), we have A= —ujuz. Putting this value of A in
(4), the required equation of the two tangent planes is given by
2
(/i - uJzWIa+imi’-UimMVb 4- («i—«i«2/'«2)
or
-Wi/al^/n +■ (/wi«2—Uimz)*lb+(niMj—Utn^^jc^ipiUz-«ipa)*
2
or
«lW9
Ml*
c
+Ma2 /V, »*i*, «i“_.
PiPi
)
2
Proved.
)=0.
Ex. 9. Iflr is the distance between two parallel tangent planes
to the ellipsoid x*/fl*4-y*/^*+2*/c*= 1, prove that a line through the
origin perpendicular to the planes lies on the cone
(Kanpur 1977)
X* {d'^-r^)J^yi (^a - .r*)+z* (c*-r*)=0.
Sol. The equation of the given ellipsoid is
..(1 )
xV+y*/^Hz*/c*=l.
The equations of the two parallel tangent planes to the ellip
soid (I) are
...(2)
/.r 4-my+«2=±
c*n*).
The distance between the two parallel tangent planes given
by (2)
«The difference of the perpendicular distances from
the origin to the planes given by (2)
la+T+T
i.e.
or
or
-(^(/*4-m*4"W*)
a*/*4-d*m*4-c*rt*«=r* (/*4-w*+«*) [on squaring]
/a (a2_,2)^^3 (^2_r*)4-n* (c*-r*)=0.
...(3)
Central Conicoids
411
Now the equations ofthe lines through the origin and perpen
dicular to the planes(2) are
...(4)
xll=ylm=^zln.
Eliminating /, m. n between(3)and (4), the required locus of
the perpendicular (4) is given by
Proved.
^2(a2__r2)+y (f,2_r2)+z2(c*-r8)=0.
Ex 10. Show that the tangent planes at the extremities of any
(Rajasthan 1978)
diameter of an ellipsoid are parallel.
Sol Let the equation of an ellipsoid be
...(1)
We know that the centre of the ellipsoid (1) is the origin.
Hence every line passing through (0, 0,0) and intersecting the
ellipsiod (1) is a diameter bf(l). Thus let the equations of a
diameter of(1) be
...(2)
xfl=ylm=^zln.
Any point on (2) is (/r, wr, wr). If it lies on (1), then
r=±l/V(/VaHm2/frH«W=
(say).
Now the extremities of a diameter are
(M, mX, nX) and (- /A, — mA, — «A).
The equations of the.tangent planes to (1) at these points are
iXx mXy nXz^ =1 Le.y Ax ,^.iil L
15" +“l2* "T
+C* ” A
...(3)
6*
a^
1
lx . my nz
—/Ajc mXy nXz_
and
..,(4)
c-“1 i.e., ^-l-p-+72-«-
or
The equations (3) and (4) are the equations of two parallel
planes.
Ex. 11. IfP be the point of contact of a tangent plane to the
ellipsoid which meets the co-ordinate axes in A, B and C and PL,
PM,PN are the perpendicularsfrom P on the axes, prove that
OLOA^a\OM.OB^bKON.OC^cK
Sol. Let the equation of the ellipsoid be
...(1)
Let the co-ordinates of P be (a, P, y), so that the equation of
the tangent plane to the ellipsoid (.1) afP (oc, fi , y) is
...(2)
The plane (2) meets Ih'e co-ordinate axes in A, B, and C and
so OA,OB and OC are the intercepts made by (2) on the co-ordi
nate axes hence we have
OA=d^lix, OB^b^ifi, OC^cAy.
A\1
Analytical Geometry i-D
Again it is given that PL is the perpendicular from the. point
P (a, y) on the ;c-axis so that OL=ec.
OM=fi and ON—y.
Similarly
Thus
OL,OA=a.{aya)=a^- ; OM,OB=
and
ON.OC=y.{cVy)^c^.
Ex. 11., If the line of intersection of perpendicular tangent
planes to the ellipsoid whose equation referred to rectangular axes is
;}^la^+y^lb^+z^/c^=\ passes through thefixed point (0, 0, k)y show
that it lies on the cone
x^
i.c^-\-a^-k^)+iz-ky (a^+b^)=0.
(Meerut 90 ; Eurukshetra 71)
Sol. The equation of the given ellipsoid is
...(1)
x^]a^-\-y^lb^-\-zVc^==l.
The equation of any plane through the point (0, 0, k) is
I(x — 0)+m (y—0)+/i(z—A:)=0 or lx-\-my+nz=nk. ...(2)
If the plane (2) touches the ellipsoid (1), then we have
aH^-{-bW+c^tP==n%^ [See § 6(B)].
or
m2=o.
...(3)
Now the equations of any line through the'point (0, 0, k) are
x—0 y—0 z—k
X
y z-k
A
.. (4)
P
A “ f* “ V
If the line (4) lies in the plane (2), we have
IX+mp+m—O.
...(5)
The direction cosines ly W,nof the normal to the tanget
plane (2) are related by the relations (3) and (.5) and since (3) is a
quadratic relation in /, /m, n,. it shows that there are two sets of
values of /, w, n and hence for every line (4) through the fixed
point (0, 0,/c) there are two tangent, planes to the ellipsoid (1)
which intersect in the line (4).
Eliminating n between (3) and (5), we have
or
or
/2+2//W A/» {c^-k^)
(fl8vHA2c2-A2A:2)(//w)H2Aj»
Let its roots be
(//m)
la/m^^
Then the product of the roots=—.
mi ma <7V+A’*c»—
413
Central Conicoids
*●
liU
6V+(C* -^2)^2 -(c2^
”l«9
[by symmetry]
Since the tangent planes are to be perpendicular, we have
/i/a+/«im3+Wina=0
or
(bh^+ic^-k^) i«®}+{(c2-A:2) AHaV}.+ {aV+6aA*)«0
...(6)
or
A*+(a"+C®-fc2) ^9^(fl2 + ^2) ^2^0.
Eliminating A, /e, v between (4) and (6), the equation of the
required cone generated by (4) is given by
x2+(fl®+c2- A:9) /+(u®+6®) iz-ky=0.
Ex. 13. Through a fixed point {k, 0, 0) pairs ofperpendicular
tangent lines are drawn to the conicoid ax^+by^+cz^==l. Show that
the plane through any pair touches the cone
y^
(x—k)^
^
0.
(Allahabad 1979)
Sol.
The equation of the given conicoid is
...(1)
ax^-\-by^’\-cz^=\.
The equations of any line through the given point {k, 0, 0)
are
...(2)
{x-k)ll=-ylm=zln^r (say).
The co-ordinates of any point on the line (2) are (/r-f fc, mr,
nr). If the line (2) meets the conicoid (1), then this point will
lie on (1) and so we have
a (Ir^-kf^-b (mrf+c {nrfi=l
.. (3)
or
{al^+bwT^+cn^) ^-2kalr+{ak^-\)^0.
If the line (2) touches the conicoid (1), then the two roots of
(3) must be equal and hence applying the condition
-4AC~0\
we have
{2kal)^-A
{ak^~\)=^0
...(4)
or
{al^+bm^-\-cn’'^)(.ak^-\)=kW.
Thus for any tangent line (2) to (1) through the point (A, 0, 0),
the values of /,
« should satisfy the relation (4).
For any two perpendicular tangent lines to (1) through the
point (A:, 0, 0), let the two sets of values of /, w, n be lu Wi, «i
and %, Wa, Wa so that the equations of these perpendicular tangent
lines through (/c, 0, 0) are given by
x—k
and
...(5)
Va
;«a Wg
/i
mi Wi
414
Analytical Geometry 3-D
Therefore, in view of(4), we have the following two relations
(ali~ -1- bmi'+crti')(ak^— 1)=
V
and
(n/a*+h/Wa"+e«a")(ak- — 1)=k-a-li'.
Adding, we get
{a Ui-+UW{mi^+m,^)-\ cin,^-hn,ys (ak^-\)=k^a^(/iHV)
...(6)
Let the line {x—k)ll2~y\m^=zln^ through the point (/:, 0,0)
be the normal to the plane containing the two tangent lines given
by (5).
Thus we have three mutually perpendicular lines. Hence we
have the relations
Similarly mi2-f-ma*=V+^8^»
Using these relations,(6) becomes
● {o(/W32+w8")+&(V+V)+c(V+0}(fl^®-l)
=W (waHwaO
or /gS (b^c)(ak^-l)^ms^ {c(nA;2-l)-a}+na*[b (nk2-l)-fl}=0
...(7)
The equation of the cone generated by the normal to the
plane containing the perpendicular tangent lines given by (5) is
obtained by eliminating /g, /Wg, Wg between the equations (7) and
{x—k)ll^=yfm^=zln^ and is therefore given by
(x-ky{b+c)iak^--l)-^y^{c{ak^-\)-a}
-\-z^{b(ak^-l)-a}^Q. ...(8)
The cone reciprocal to the cone (8) is clearly given by
(x~kr .
y
£_i- =0
(A+c){ak^-l)'^c iak^-l)-a b{ak^-\)-a
which is the required equation of the cone which is touched by
the plane through the lines given by()
Ex. 14. Provfi that the locus of'points from which three mutu
ally perpendicular planes can be drawn to touch the ellipse
z=0
is the sphere x^+j-+z2=c2+6®.
(Punjab 1973; Lucknow S6)
Sol. The equations of the given ellipse are
x'^fa^+y^lb2. 1, z=0.
...(1)
Let the equation of one of the three mutually perpendicular
planes be
lix+miy-jrrtiz^pi.
...(2)
The plane (2) meets the plane 2=0 in the line given by
Central Conicoids
415
...(3)
kx-^miy=P\y
Now if the plane (2) touches the ellipse (1) then the line (3)
will touch the ellipse (1).in the jc;>-plane and the condition for
the same is
Putting this value of in (2), the equation (2) becomes
...(4)
kx+m^y-\-nxZ=y/{a‘l{--Vb'm{%
Similarly the equations of the other two tangent planes to
the ellipse (1) are
...(5)
niZ=y/{aH^-\-b-m^)
and
...(6)
hx+ni^y-\-niZ^y/{cPl^-\-b'^m^).
Squaring (4),(5) and (6) and adding, we get
Jc'*
r/Mi«i+2zx
or
x\I +y-.1 +Z-.1 -f 2;'z.O -1-2ZX.04-'ay^.0=fl'M +^M
or
tv r/x2=l,2:/i/wi==0etc.]
x^ y^
a--\-b^
which is the required locus.
(Kanpur 1982, Meerut 84 R)
§ 8. The Polar Plane.
...(1)
Definition. Let
ax^+by-'^cz^—X
be the equation of a conicoid and. ^(a, /5, y) be any point. Draw lines
APQ to meet the conicoid(1) /« the points P and Q. The locus of the
point R such that AR is the harmonic mean of AP and AQ (i.e., AP,
AR and AQ are in harmonic progression) is called the polar plane of
the point A with respect to the given conicoid.
(A) The equation of the polar plane.
The equations of any line through the point y4(a,|5, y) are
..(2)
(x-aj//=(y -fi)/m={z- y)/n^r (say),
where /, m, n are the actual d.c.’s of the line.
The co-ordinates of any point on (2) distant r from A{v.yPy y)
are.(/r-l-a, mr+fiy nr+y). Therefore, the distances of the points
P and Q where the line (2) meets the given conicoid (1) are giveh
by the quadratic in r®
a (/r+«)®+6
(nr-|-y)2=l
or
r® {al'A- bm^-{-cn~)+2r (u/a-t-h/w^+cny)
+(flaHfej5Hcy2-l)=0.
...(3)
Let the two values of r be rj.. and r2 and let ri=/lP and
r^^AQ. Now APy AR, AQ are in harmonic progression.
.
2
1
1
1 1 rx+f2
r^rj
“ AR AP~^AQ Ti -rg
416
Analytical Geometry 3-D
2 —2(fl/g4 bmfi-j- cny)
AR^ ax^-^bfi--\-cy‘*‘—\
or a&.-+bp'^+cy'-~i=—{ax (l:AR)-{-b^ (m.AR)-\’Cy
or
...(4)
Now let (x, y, z) be the- co-ordinates of the point R on the
line (1) whose distance from the point A (a, fi,y) is AR and so
we have
{x-oi)llMy~fi)lm={z~y)ln=^AR.
...(5)
/. I AR^x—a, m AR—y—fii n.AR=z—y.
Now in order to find the locus of R, we are to eliminate /,
m, n between (4) and (5) and so it is given by
act^+b/i^+cy\-\=-{ax (x~oi)+bfi(v-^)4 cy (z-y)}
...(6)
or
aax+bfiy-{-cyz=\.
This is the required equation of the polar plane of the point
4(a, /?, y) with respect to the conicoid (1).
Note. The equation (6) of the polar plane of the point A (a,
fii y) w.r.t. the conicoid (1) is of the. same form as that of the
tangent plane of the conicoid (1) at the point A (a, y) on the
conicoid.
(B) The pole of a given plane.
Tofind the pole of the plane lx-\-myArnz^p w.r.t.. the conicoid
(Meerut 1983)
ox*-fhy®+C2*=l.
The equation of the given conicoid is
...(1)
ax^-^by‘^-\-cz^=\.
Let the pole of the plane
...(2)
lX'\-my+nz=p
w.r,t. the conicoid (I) be the point
y). But the polar
plane of P(a, y) w.r.t. to (1) is
...(3)
aoLx-^b^y4rCV2=^\.
Therefore the equations (2) and (3) represent the same plane
and so comparing their coefficients, we get
n
m
/
1
a^_b^ cy
n
I ~m
:. The required pole is (ll(ap)^ ntlibp), n/{cp)).
§ 9. Properties of the polar planes and the polar lines.
(A) Let P (ai,
yj and Qia^, y?a, J'a) be any two points. It
clearly follows that if the polar plane of the point P with respect
to a given conicoid passes through the point Q then the polar
plane of O w.r.t. the same conicoid will pass through P. The
Central Conicoids
points P and Q satisfying this property are
points.
(B) Let
417
called the Conjugate
bMhe equations of thftWanet.%7?&ly follows that if the.
pole of the plane «i=0 lies on the plane «8=0 then the pole of tne
plane Ma=6 will lie on the plane «i=0. The two such planes «i-u
and Ma=0 are called the conjugate planes.
(C) The polar lines. The polar line or simply polar of a given
line AB is another line PQ such that the polar planes of all points on
AB pass through PQ.
Let the equation of the conicoid be
...(1)
ax®
Let the equations of the line AB he
(2)
{x- c»)ll={y-^)lm={z-y)ln^r (say).
,
Any point on (2) is R (/r+a, mr+)5, iir+y). The equation ot
the polar plane of the point R with respect to the conicoid (1) is
ax (/r+a)+h;;(mr-l-/?)+cx (/ir+y)=l
or
{aax^-bfiy+cyz- l)+r (alx-\-bmyArcnz)^^^
This plane for all values of r clearly passes through the line
of intersection of the fixed planes
aax+
cyz—1=0, a/x+
=0.
Let this line be the line P<2'
.
aj>
Thus we observe that the polar plane of any point R on AB
passes through every point of the line PQ and hence the polar
plane of any point on the line PQ must pass through R. But R is
an arbitrary point on AB and so the polar plane of every point on
PQ will pass through every point of the line AB.
Thus we conclude that the lines AB and PQ are such that the
polar plane of every point on
passes through PQ aod vice
versa and the two such lines are called the polar lines^. ,
.
(D) Tofind the condition that the two given lines be the polar
lines.
Let the equation of the conicoid be
ax®+h;;®+cz®=l.
Let the two given lines be denoted by AB and PQ and let
their equations be given by
X—
..4
n
m
/
respectively.
418
Analytical Geometry 3-D
Proceeding as in § 9(C) above the polar. line of AB with
respect to the conicoid (1) is the line of intersection of the two
planes given by
aax-\‘b^^y-\-cyz— 1 =0
(2)
and
alx-{-bmy-{-cnz=0.
■(3)
Now if the polar line of AB be the given line PQ then PQ will
lie on the planes (2)and (3) both. The conditions for which are
flaa'+fcjffS'+cyy'—1=0 1
a(xr+bpm'+cyn'=0 ]
and
ala' bmp'-^cny'—O
all'+bmm'-{-cnn'=0 j
Hence the required conditions are given by (4) and (5).
..(4)
...(5)
(E) Conjugate lines. Two lines AB and PQ are called the con
jugate lines if when the line PQ intersects the polar line of the line
A B then the line AB also intersects the polar line of the line PQ.
To find the condition for the two given lines to be conjugate lines.
First rewrite the steps upto equation (3) of § 9 (D) above.
Any point on the line Pg is (/'r+a', w'r+jB', n'r-\-y'). If the
line PQ intersects the polar line of AB [given by (2) and (3)] then
for some value of r this point will satisfy both the equations of
the planes (2) and (3) and so we get
aa {Vr-\-a')+bfi (mV+|5')+cy (n'r+/) -1 =0
and
al (/V+a')4-ftm {m'r-{-p')-\-cn {n'r-\-y') = 0
or
...(4)
(aaa,'-\-b^*+cyy'—\)= -r (aal'+bfim'-^cyn')
...(5)
and
(qa'l+bp'm+cy'n)=>—r (alT-\-bmm'-\-cnn').
Eliminating r between (4) and (5), we get
(aaa'+bfifi'-\- cyy' —.1) (all'+bmm' -\-cnn')
...(6)
^-(aal'-i-bpm'A-cyn'yiah'+bmp'+cny').
This is the condition that the line Pg intersects the polar
line of AB. The symmetry of the result (6) shows that the line AB
will also intersect the polar line of PQ. Hence (6) is the required
condition.
SOLVED EXAMPLES (B)
Ex. 1. Prove that the locus of the poles of the tangent planes
}f ax^-\-by--\-cz^={ with respect to a'x-+b'y^-\-c'z^=\ is the coni(Allahabad 1982)
ioid (a'xy/a-{-(b'yfi/b+(e'z)'^lc—\.
Sol. The equations of the given conicoids are
a'x^-hb'y^+c'z^=\.
ax‘-\-by^+cz^=\.
...(1)
...(2)
419
Central Conicoids
Let the equation of a tangent plane of the conicoid (1) be
...(3)
lx-\-my-^nz^p.
Then
(/Vfl)+(/nW+(nV<?)=/>^
Now let the pole of the plane (3) w.r.t. the conicoid (2) be
(●^i»
^i) so that (3) is the same plane as
...(5)
a!xxx+b'yyi+c'zzi=>\.
Comparing the coefficients of the equation (3) and (5),
we get
/
m
^
n _P
a’Xx b'yi
c%
1
or
l=a!xxPy m=b'yip, n=c%p.
Putting these values of /, m, n in (4), we have
(a'xipfla+(*>ip)V.i+(c'ziP)'“/e=p’“
or
(a'x,)Va+miYIbHc-z^tVc^ 1 The locus of the pole (jci,
Zi) of the plane (3) w.r.t. (2)
is given by
(a'xflaHby)Vb+(c'z)Vcx=l.
.Proved.
Ex. 2. Prove that the locus of the poles of the tangent planes
of a^x^-\-by^-C‘z^==\ with respect to a®x®+PV^4y*2*=>l is the
(Eurukshetra 1974)
hyperboloid of one sheet.
Sol. Proceeding exactly as in Ex. 1 above and replacing
a by
b by
c by — c?, a' by a*, b* by /?*, c' by y®, the required
locus is given by
a«x2
yy
which is the equation of the hyperboloid of one sheet.
Ex. 3. Show that the locus of the pole of the plane
lx-\-my-\-nz=p with respect to the system of conicoids
S
=
is a straight line perpendicular to the given plane
(Kanpur 1976 ; Bnodelkhand 72)
where k is a parameter.
Sol.
The equation of.the given system of conicoids is
Z-
1.
...(1)
The equation of the given plane is
...(2)
/x+m.v+/iz=/?.
Let (a, fiy y) be the pole of (2) w.r.t. the conicoid (1). The
polar plane of (a, /3, y) w.r.t. the conicoid (1) is
ax
yz
1.
I
a"+k~^t>’-hk^c^-i‘/c
...(3)
420
Analytical Geometry 3-D
The equations (2) and (3) should represent the same plane
and hence comparing their coefficients, we have
a/(a^’hk)_ff/(b^4-k^ yl{c^-\-k)
/
m
n
p'
«=/(fl8+A:)/p,
{b^+k)lp, y~n {c^+k)!p
mb^ mk
nc^ nk
a
=— 3_
p
p’ P p
p y—p P
Therefore, we clearly we have
et-aV/p fi-m^jp y-ncVp k
I
' m
n
P
A The locus of the pole (a, yff, y) is
x~(^lfp y—mV^lp z—nc^fp
I ●
m
n
This is a straight line with its direction cosines proportional
to /, m, n which are also the direction ratios of the normal to the
plane (2). Hence the locus is a straight line perpendicular to the
given plane (2).
©
f p
or
Ex. 4. Find the locus of straight lines drawn through a fixed
point {(z,^y) at right angles to their polar \^ith respect to the
comcoid ax^-\-by^A‘Cz^=y.
(Lucknow 1976, 80; Kanpur 82; Meerut 89)
Sol. The equation of the given conicoid is
ax^+by^+cz^^l.
...(1)
The equations of any straight line through the point (a, y)
are
{x—a)ll=(y-.fi)lm=(z-y)ln=r (say).
...(2)
The co-ordinates of any point on (2) are(/r+a, mrifi, nr {-y).
The equation of the polar plane of this point w.r.t. the conicoid
(l)is
ax {lr+(i)-^by (mr+fi)-\-cz(ur-fy)=l
or
, (aax-^b^y+cyz’-l)+r (alx-hbmy-{-cnz)=Q.
This plane for all values of r will pass through the line
aax-^bpy-\-cyz—[=0, alx-hbmy-\-cnz=0.
...(3)
This is the polar of the line (2) w.r.t. the conicoid (I).
Let A,ji, V be the direction ratios of the polar line (3) so that
we have
aotX-^-bfip-^-cp—O, alXj-bmix-\-cn^—0.
Solving,
V
y
= —-jJ^.—
be(nfi—my) ca {ly —noi) ab {mot.—Ip)'
...(5)
Since the lines (2) and (3) are given to be perpendicular, we
have
/A+m/t-t-nv=0
or
fibc (np—}ny)-\-mca {ly - na)+«fl6 {ma—ip)^0.
Dividing by abclmn throughout, we get
Central Conicoids
421
^
am
a
a
an^ bn bl ● cl
JL,
cm
or
...(5)
The locus of the line (2) is obtained by eliminating /, w, n
between (2) and (5) and hence is given by
4.(i-4)+,4,&4)t,-^(K9->Ex. 5. IfP {xi, yi, Zy) and Q{Xi,y^,z^ are any two points
thenfind the equations of the polar of FQ with respect to the conicoid ax^-\-by^-\-cz^=\.
Sol. The equation of the given conicoid is
ax^+.by^-^- cz^=],
Now the equations of the line joining P (atj,
^2)are
X—Xi
Xi-Xx
y-yi
y^—yx
Z-2i
z^-zx
j,)
●●●(!)
and
r (say).
The co-ordinates of any point on this line PQ are
(^i+r (xa-xi), yx->rr (yg-yj, Zx-^r{z^-zx)).
The polar plane of this point w.r.t. the conicoid (1) is
(x2-Xx)}-{-by (y^-j-r (y,-yi)}-|-cz (z^+r (Za-Zi»=l
or (fl^Xi+byyi+cz?i—l)+r (ox (xa--Xi)+^y (ya~yi)
-\-cz (za-2i)}=0.
This plane for values of r will pass through the line given by
the planes
nxxi+6yyi+czzi= 1
...(2)
and
ax (.r2-x,) + by (ya—yi)+cz (^a-^i)=“0
i,e.
flxxa -f byy»-l- czz^ ~ axxx+byyx+czzi
i.e.
axxa4-6yy84-czza=l. [using (2)].
.(?)
The equations (2) and (3) are the required equations of fhepolar of the line PQ.
Ex. 6. If P (xi, yi, zi) and Q (Xg, yg. z^) be any two points on
a conicoid ox'^+by^+cz*— 1, show that the polar of the line PQ is
the line of intersection of the tangent planes at P and Q. (Alld. 1978)
Sol. Proceeding as in Ex. 5 above the polar of the line PQ
is the line of intersection of the planes given by the equations (2)
and (3). But (2) and (3) are the equations of the tangent planes
to the conicoid (1) at the points P (xj, y^, Zj) and Q (xg, yj, Zj).
Hence the polar of the line PQ is the line of intersection of the
tangent planes at P and Q.
422
Analytical Geometr}/3-jD
Ex. 7. Find the locus of straight lines through afixed point
y) whose polar lines with respect to the conicoids ax^-\-by^
-\-cz^=\ and a’x^-\-b'y^-\-c'z^== \ are coplanar.
Sol. The equations of the two conicoids are
■●●(1),
a'x^+by^-\-c'z^ = \. , .,.(2)
The equations of any line through the point (a,
y) are
' {x—a)Jl=i{y -fi)jm—{z -y)jn—r (say).
...(3)
.
The equations of the polar line of the line (3) w.r.t. the conicoid (1) [See § 9 (C)] are
flax+h^y+cyz= 1, alx-^bmy-^cnz=\).
(4)
Also the equations of the polar line of the line (3) w.r.t. the
conicoid (2) are
a’ox-\-b[Py+c'yz^\,a’lx^-h>my+c’ni==(i._
...(5)
The two polar lines given by (4) and (5) are given to be
coplanar and hence we have
ax
bP
cy
1 «0 (o-a> (b-b^)P {c- c')y 0 =0
or
al
km
cn
0
al
bm
cn
0
a'«
b'P
c'y
1
a'(X
fl7
Vm
c'n
0
a'l
or
or
(a -a') X
ib-b') p
al
bm
a'l
b'm
b'm
(c—c') y
cn
c'n
c‘y
1
c'n
1
=0, expanding
the
determinant
along the fourth
column
{a-a') x {bc'-b'c) mn^{b~b') P {ca'-c'a) nl
+ {c-c')y{ab'-a'b)lni=^0.
Dividing by /wn throughout, we get
{a-a') {bc'-b'c) {ail)-\-{b-b') {ca'-c'a) (/5/m)
■\-{c-c'){ab'-a'b){yln)==Q.
...(6)
The locus of the straight line (3) is obtained by eliminating
/, m, n between (3) and (6), and therefore, it is given by
(a-a') (be'-yc) (^^j+(6_4-) ^a,■-c^a)^^^^)
Ex. 8.
■V.
0.
+(c-e')(ab'-a'b)^jl.^'j
Prove that the lines through (a,
y) at right angles to
423
Central Conieoids
theirpolars with respect to xy(a+b)-{-y^/i2a)-jrZ^I(2J))~l generate
the cone
iy—^)(az- yx)+(z-y)(a;;-^x)=0.
The equation of the given conicoid is
jc2/(fl+6)+//(2a)+z2/(26)= 1.
For convenience let us put a'= l/(a+A), h'=l/(2a), c'=1/(26);
Using these values the given equation of the conicoid becomes
...(1)
a'x^+h'y^-\-c'z^=\.
Sol.
Now proceeding exactly as in Ex. 4 above the equation of the
required locus is
OL
or
or
&4)+
x-ct
y-B
a
x —v. -■ (2a-26)-l y -P (26-0-6)+^^ (a+6-2a)=0
[Putting the values of a', 6', c']
2a
2a (g-6) d {a—b) y (fl-6)
L=o
0, or
X-ct
X
-rCt
y -^ z-y
- y-d
z-y
or
or
or
§ 10.
P
(—
\x-«.
y-p
(ay-fix)
az—yx
0,
(x-a)(y-P)~^(x-ct) (z-y)
(y-P) (az-yx)-\-(z-y) (ay-fix)=^Q.
Locus of chords bisected at a given point.
Proved.
(Punjab 1976)
Let the equation of the conicoid be
ax^-\-by^-{-cz^=\.
...(1)
Let (a, p, y) be the middle point of a chord of (1) so that its
equations may be given by
(x-a)//= - ^)//M=(z~ y)/n = r (say),
...(2)
where /, m, n are the actual d.c.’s of the line.
Now the co-ordinates of any point on (2) at a distance r
from (a, p, y) are (/r-|-a, mr+p^ nr+y). Let this point lie on (1)
so that the distances of the points of intersection of the chord (2)
with the conicoid (1) from the point (a, /?, y) are given by the
quadratic in r*
a (/r+a)2+6 (/wr-h^)2+c (»r-fy)2=l
or
r^'(al^-^bm^4-cn^)-¥2r (ala-\-bmp+ony)
-f(aaH6;5Hcy=‘-l)=0.
...(3);
But (a, p, y) being the middle point of the chord (2), the' i distances of the points of intersection from (a, j3, y) should be |.
I ●
424
Analytical Geometry 3rD
equal in magnitude but opposite in sign i.e. the sura of the two
values of r should.be zero and so we should have
—2(ah-\-bmfi+cny) 0
fj+r.
aP-\-bm^+cn»
or
ah-{-bmfi+cny^O.
...(4)
Therefore, the locus of the chord (2) for varying values of
/, w, « is obtained by eliminating /, m, n between (2) and (4) and
is thus given by
aoL (x - a)+^p (y -fi)+cy (z- y)c=iO
or
a«x4-bfiy+cyz=a(K^-^ bfi^+cy^or
aaX+bfiy+cyz— 1 =aa®f bfi^+cy^— 1
or
..(5)
T=Si
where T and
have their usual meanings.
Pro|ierty. The section of the conicoid by the plane (5) is a
conic with its centre at (a, y?, y).
We know that the section of a conicoid by a plane is a conic.
Here the given plane (5) is the locus of the chords of the coni
coid with their middle point as (a, y9, y). Therefore, all the
chords of the conic section of the conicoid by the plane (5) and
passing through the point (a, y?, y) are bisected at the point
(a, y?, y) and hence by definition of centre the centre of the conic
SOLVED EXAMPLES(C)
Ex. 1. Find the equation to the plane which cuts 3x*+2y®
— I5z*=4 in a conic whose centre is at the point(—2, 3, —1).
Sol. The equation of the conicoid is
15£>
.
4 +2 - 4 r-*-0
and the centre of the conic is(-2, 3, -1).
We have
„ 3(-2)2 .(3)2 . 15(-1)2
1=3+ -_ii —1=1
Si-—4
+
^2 4
2’
and r=?-*(-2)+t (3)-i|£.(_l)-l= _|-*+|.
/. The required equation of the plane is r=5i
or
or 2x-2y-5z+2=0.
Ex. 2. Find the centre of the conic
^*V^+y2/16+22/4c=l, 2x+2y~z=3.
Ans.
425
Central Conicoids
Sol. Thfi given conic is the section of the conicoid
x^/9i-y^l\6+z^l4=\
...0)
.:.(2)
by the plane
2x+7.y—z—3.
Let (a, fiy y) be the centre of the given conic. Then the given
conic is the section of the conicoid by the plane *r=iSi’
3
-1
i.e.
9 +16
4
9+16+4
J-21 4-?^.
...(3)
T + b6+“4 ~ 9+16 + 4
The equations (2) and (3) must represent the same planes and
hence comparing their coefficients, we have
(a/9) 0/16) (y/4) ■(aV9)-l-(/?V16)-f(yV4)
. .
3"
-y=—=“1 =
^
...(4)
/. a=18A:, ^=32/:, 7= —
or
a2
and
9+1T+ 4"-^^
...(5)
Puttin, for a, p, y from (4) in (5), we get
36)fc2-l-64)k2J-4Jfe=*=3A: or /:«=3/104 (since/c#0).
Putting for k in (4), we get
a:^27/52, yff= 12/13, y=-3/26.
Ans.
The required centre is (21152, 12/13, —3/26).
Ex. 3. Prove that the centre of the conic
lx-{-my+nz—p, ax^ \-by^-rCz^= 1 is the point
I Ip
mp
np \
where
Sol.
and:(l^Ja)-{-(m^lb)-\- (rPlc)=p^.
The given conic is the section of the conicoid
...(1)
ax^-\-by^-\^cz^=\
...(2)
by the plane
lx-\-my-\rnz~p.
Let (a, /?, y) be the centre of the given conic. Then the given
conic is the section of the conicoid by the plane ‘T=Si*
...(3)
i.e.
a7.x-\-bpy^-cyz=a^:'-\-bp^-]rcy'^>
The equations (>) and ^3) should represent the same planes
and hence comparing their coefficients, we have
aa.
6/3
cy
(say).
n
P
v.^klja, p^kmlb, y=knlc
i ""w ”
...(4),
and
a%^+bp^+cy^-=^pk
or a. [kV^ja^]+b.[k^m^lb'^]-\-c.[k'^n^lc‘]=pk, using the relations (4)
426
or
Analytical Geometry 3-i)
or Po^-^pk or k=plpo^.
Putting for k in (4), we get o.=lp!p^\ fi=mplPo^ y==np(p^\ .
The required centre is {lpip^\ mplPo\ np/po^).
l^/a-\~m^lb-\-n^lc~pk
Ex. 4. Prove that the centre of the section of the ellipsoid
x^ja^-]ry'^lb^-\-z^jc^= \ hy the plane ABC whose equation is
xla-{-ylb-{-zjc= 1 is the centroid of the triangle ABC.
Sol. Let (a,
y) be the centre of the section of the ellipsoid
●● (1)
xla~\-ylb-rzjc=\.
...(2)
Now the conic section of the ellipsoid (1) having its centre as
the point (a, y) is given by the plane
by the plane
i.e.
a‘
b^
~fl2
+^
...(3)
The equations (2) and (3) represent the sartie plane and hence
comparing their coefficients, we have
(«/fl2) (j3/t>2) (j,/c2) (a2/fl2)_j_(38//,2)_|_(y2/c2)
=k (say).
1
a.=aky P—bky y=ck
and
●v(4)
o.ya^+fi^/b^+y^jc^=k
C^k^ . ,
or
+
or
3k^=k
or
^
. .
using the relations (4)
/c=^
[V MO].
Putting this value of k in (4),.we get a=io, p=^by y=^c
The centre of the section is (^a, ^b, Jc).
...(5)
Again the equation of the plane ABC is given by (2) and so
the vertices of the triangle ABC arc'A (a, 0, 0), B (0, by 0) and C
(0, 0, c). Hence the co-ordinates of the centroid of the triangle
ABC are OJa, Jft,
which are same as the co-ordinates of the
centre given by (:)●
Proved.
Ex. 5. Prove that the cenires of the sections ofax^-\-by^-{-cz^= l
by the planes which are at a constant distance p from the origin lie
on the surface
{ax^+by^-\-'cz^-)--r-.p-i
Sol.
The equation of the given conicoid is
ax'-\-by^-'fCz-={.
...(1)
Let (a, .8, y) be the centre of a plane section of the conicoid
(1). Then'the equation of the plane givnig this section of the
conicoid is
i.e.y
aoLx f b^y -\-cyz=act- -f bfi^-f cy-.
..:(2)
427
Central Conicoids
If the plane (2) is at a constant distance p from the origin,
then
,ua.^+b^^+cy^
(aa2+ bp^+cyY=P^
Hence the locus of(a,|3, y) is the surface
(flxM-*/+cz2)*=/ (a^xHW+c'^z^).
Ex. 6. Show that the locus of the middle points of the chords
ofthe conicoid ax^-\-by^+cz^—\ which pass through a fixed point
(x^ y\ z') is
ax (x-x')+by (y~y)-\-cz(z-z')=0.
(Lucknow 1979, 82 ; Kanpur 79 ; Allahabad 81)
Sol. The equations of the given conicoid is
...(1)
ax^^by^+cz‘*-=l.
Let (oc, fit y) be the middle point of a chord of the conicoid
(1). Then this chord lies in the plane ^T—Sf i.e.,
aax-\-bfiy.+cyz -1 =aa^+
cy® — I
or
...(2)
ao.x+b^y-}-cyz=saa^-i-bfi'^+cy^.
Now if the chords of the conicoid (1) having (a,f, y) as their
middle point pass through the fixed point (x', y\ z'), then the
point (x', y\ z') should lie on.the plane (2). Therefore, we have
aax'+h^y'+cyz'= aa2+i>)52+cy*
or
fla (a-x')+^>y?(/?—/)+cy (y-z')=0.
Hence the equati.->n of the required locus of the middle point
(a,)?, y) of the chords of the conicoid (1) which pass through the
fixed point (x'» y\ z') is
ax (x—x')+by (y—y')-\-cz(z—z')=0:
Ex'. 7. Prove that the section of the ellipsoid
x7a*+yV^®4-z*/c®= 1
whose centre is at the point (in, ^6., ^c) passes through the extremi-.
ties of the axes.
(Rohilkhand 1981)
Sol. The equation of the given ellipsoid is
S=x^/a^+yW+z-lc^~i==0
.. The centre of the section is (Ja, \b, ^c) and hence the equa
tion of the plane giving this section is ‘7’=Si’
' x.^a
iW .(^6)2 . acf-1
i.e.
a * b^
a*
c2
or
xla+ylb+zfc=l. ' '
●●■(I)
The extremities of the axes of the given ellipsoid are {a, 0, 0),
(0, 6, 0) and (0, 0, c) and the plane (1) clearly passes through these
three points. Hence the section whose centre is at the point
428
Analytical Geometry 3-Z)
(Jfl, 16,|f) passes through the extremities of the axes.
Ex. 8. Prove that a line joining a point P to the centre of a
conicoid passes through the centre of the section of the conicoid by
the polar plane of P.
Sol. Liet the equation of the conicoid be
...0)
ax^-^by^'\‘cz^=\.
Let the co-ordinates of the point P be (x',')'', z’) so that the
polar plane of P
z') w.r.t. the conicoid (1) is
...(2)
axx'-f byy'-f- czz'— 1.
Let (a,
y) be the centre of the section of the conicoid (1) by
the plane (2) so that the equation of the plane giving this section
IS
i.e..
or
axa-{-byfi^czy— 1 =ad*-l-6/5^4*
1
aax -I- b^y-\-cyz—act^-{-bP-^cy^.
...(3)
The equations (2) and (3)represent the same planes and hence
comparing their coefficients, we get
oa _^_c7 or
ax*~~by'~~^'
x' y' z'
...(4)
Again the centre of the conicoid (1) is {0, 0, 0) so that the
equations of the line joining P (;x\ y\ z') and (0, 0, 0) are
x-O j>-0 z—0 or X _y
z
jc'—0"“;;'—0~z'—0
x'~y~z'
...(5)
If the line (5; is to pass through the point (a, jff, y) then the
co-ordinates of the point (a, /?, y) should satisfy (5) and so we
y
should have - ^
which is true by virtue of(4). Hence
x’ y 7‘
proved.
Ex. 9. Triads of tangent planes at right angles are drawn to the
. ellipsoid x^fa^-{-y^fb^-^z^lc^=[. Show that the locus of the centre of
section of the surface by the plane through the points ofcontact is
(.v2fy-+22)=^flH6Hc2)(x*/aH.v76*-l-2W.
Sol. The equation of the given ellipsoid is
x7a«+;;,762^zVc2=l.
●●(I)
Let (x'» y\ z') be the co-ordinates of the point from' whiph
three tangent planes at right angles are drawn to the given ellip
soid (1). Then by definition of the director sphere the point
(x', y,z') will lie on the director sphere of the ellipsoid (1)i.e. on
x^+y^+z^=a^yb^+c\ [See §7}
x'^-f;^'=*+2'2=fl2+62-l-c2.
...(2)
429
Central Conicoids
Now the plane through the points of contact is the plane of
contact of the point (x\y\z') with respect to the* ellipsoid <1)
and its equation is given by
...(3)
x'x(a^+yy/b^+z'zjc^=\.
Again let (a, .3, y) be the centre of the section of the ellipsoid
(I) by the plane (3), so that the equation of the plane giving this
section is *7'=iSi*
i.e.
axla^-\‘fiylb'^-\-yz!c^ -1
1
or
...(4)
-fy2/c*=a2/fl8+j5*/6»+f/c"The equations (3) and (4)represent the same plane and hence
comparing their coefficients, we have
1
JC' v' z'
-.(5)
a
= y—
+
Putting the values of x', y', z' from (5) in (2), we get
a2-|-^Hy
=084^2+^2.
[(a2/a2)-f(y?W+(yW?
The locus of the centre (a, y?, y) is given by
(x8+j;2+z2)=(a24b8+ c2)[ixVa^)HyW-)+izVc^)yEx. 10. Show that the centres of the sections of- a central
conicoid that are (i) parallel to a given line lie on a fixed plane and
(ii) that pass through a given line lie on a conic.
Sol. Let the equation of the central conicoid be
...(1)
ax^+by^-^cz^=i\. .
Let (a, p. y) be the centre of a section of the given central
. conicoid (1). Then the equation of the plane giving this section
isTc=5i’
ie.y
<7ax+ bpy \-cyz— I =ad?-\-bfi^-\-cy^ ■ 1
or
...(2)
aax^bpy-Vcyz=a:^-\-bfi^+cy®.
Again let the equations of a given line be
...(3)
(X - x,)//=(:F-yi)/m=(z ~Zi)/rt.
(i). If the plane (2) is parallel to the line (3) then wc have
(aa)./+(fe^).m+(c7).«=0.
The locus of the centre (a, p, y) is
alx+bmy-^cnz=0
which is the equation of a fixed plane since a, b, c, /, m, n are all
constants being given numbers,
(ii). If the plane (2) passes through the line (3), we have
a<xXi+bPyi+cyZi=av.^+bp^-\-cy^, ,
and
(act).]f(bp).m 4-(cy).«=0.
/
430
Analytical Geometry 3-2)
The locus of(a, /?, y) is the curve of intersection of the
surfaces
ax^-\-by^-\-cz^~-axiX -hyiy —cziZ~0
aIx-]-bmy-i-cnz=0
...(4)
(
The two equations given by (4) together represent a conic.
Hence the locus of the centre is a conic.
and
Ex. 11. Find the locus of the centre of the section of the conicotdax^A-by^-\-cz^=^ which touches Ax^A-By--\-Cz^=\.
(Madras 1978 ; Allahabad 76)
Sol. Let (a,/?, y) be the co-ordinates of the centre of a.
plane section of the conicoid ax^-\‘by^+cz^=\.
Then the equation of the plane giving this section is ‘T’mS'i*.
i.e..
aoLxA‘bfy-\-cyz -1 =ad^->rbf^+cy^~ 1
or
..:(1)
aa.x-\-bfiy -j- cyz—act^-^bp^-f cy*.
Now if the plane (1) touches the conicoid /ljc'-l-5y*-l-C2*= 1,
then we have
[usin g § 6(A)]
The locus of(a, f, y) is
a^x^lA+bYIB+c^zyC:=(ax^-\-by^-+cz^f.
Ex. 12. Prove that the middle points of the chords of
ax^-hby^-hcz^=-~l
which are parallel to x=0 and touch
He on the surface
by^(bx^-}-by^+cz^—br^)-\-cz^(cx^A“by^-{-cz“ cr^)=Q.
(Kanpur 1982)
Sol. The equation of the conicoid is
ax^+by^-y-cz^--1.
●● (I)
Let (a, {3, y) be the middle point of a chord. The equations
of any straight line passing through the point (a, /S, y) and parallel to the plane x=0 (i.e. YOZ.plane) are given by
(x--a)/0=(y—y?)/w=(2—y)/«=;\ (say),
..(2)
where 0, m, n are the actual d.c.’r, of the line.
Any point on the line (2) at a distance A from (a, y) is
(a, wA+/?, /tA4 y). If it lies on the conicoid (!) then the distances
A of the points of intersectien of (I) and (2) from (a, fi y) are
given by
or
07.^+h (mAH-/3)2+c (wA+y)’*=-1
A* (2»m®-f c/72)-|-2A {bmfi-\-cny)- (r/'/.’+^.S^+cy*—1)=0.
Since (a, //, y) is the middle point of the chord intercepted by
Central Conicoids
431
(2) on (1), therefore the sum of the roots (i.e. Ai+A») of the above
.equation should be zero and for it vvc have
...(3)
bmfi+cny~0 or {mfn)~- {cy!b^).
Now according to the given question if the chord (2) touches
the sphere x^-hy“+z^=r\ then we have
the length of the perpendicular from the centre (0, 0,0) of the
sphere to the line (2)=thc radius r of the sphere
0~a 0--y
i.e.
m
or
o.
or
or
or
or
■
+
0 -y 0-a
n
n
2
+
0
0-a 0-fi
0
2
m
(my—p/7)2+rt2a*+«i2a2=7*
[V 02+m2+«*=l]
{my{m^-^n^)
(-rHa2)(m2-{-«2^+(wy-p«)2=0
(;-r2+«2){] H-(m,'«)2}-!-{/?-m7//i}2=0
2
cy
0, using (3)
(_r2+«2) (Z,2y?2-|-c2y“)-i-(^»/?24-cy2)2=..0.
The locus of(a, fi, y) is
(—r--\rx^) {b^y^+c^z^)+{by^+cz^)^=0
or
by^ (bx^-\-by^ -\-cz'—br^)+cz^ {ax^-^by^-\-cz~ - cr2)=0.
Proved.
§ 11.
Normal to a Conicoid.
(A) To find the equations of the normal to a central conicoid at
the point (a, fi^y).
(Agra 1982; Rohilkhand 80; Utkal 70)
Let the equation of the central conicoid be
ax^+by^+cz^=l.
...(1)
The equation of the tangent plane at (a,/5, y) of the conicoid
(1) is
...(2)
aax-{-bpy-\-cyz=h
The normal to the conicoid (1) at (a,(i, y) is the straight line
perpendicular to the tangent plane (2) and passing through
(a, j?, y) and hence the required equations of the normal are
given by
...(3){x-a.)/{aoi)^(y-^)l{bp)^iz-y)licy)
where ay., bP, cy are the direction ratios of the normal (3).
Now let p be the length of the perpendicular from the origin
. to the tangent plane (2) so that
432
Analytical Geometry 3-D
1
...(4)
In view of(4) the actual direction cosines of the normal(3)
are
cyp and hence the equations (3) of the normal to the
conicoid (1) at (a, 3, y) in terms of actual direction cosines are
given by
y~fi z-y
aap afip ~~ayp
(B) To find the equations of the normal to the ellipsoid
...(5)
(Kanpur 1980; Lncknow 81; Korushetra 73)
Proceeding exactly as.in § 11 (A), the equations of the normal
to the given ellipsoid at (a,
y) are
and
x-a y-f z~y
ccpla^ ^pjb^ ypJcV
1 a2
y2
,
~T=~T+^-r--r
...(1)
where xp/a^^ ?p/b\ yplc^ are direction cosines
X-OL y-fi z-V
or
a/fl2 - [i/^2 - yjc2’
where afa^ f/b-, yjc^a.TQ direction .ratios.
SOLVED EXAMPLES(D)
Ex. 1. The normal at P(a, yS, y) of a central conicoidmeets the
three principal planes at C?i, (7j, Gg ; show that PG^, PGg, PG^ are in
a constant ratio.
(Agra 1980; Meerut 84S; Kanpur 78)
Again if PG^-\-PG^-\-PGz^—k^^ then find the locus of P.
(Rohilkhaiid 1981)
Sol. Let the equation of the central conicoid be
ax^+by'^-\-cz^~\.
The equations of the normal to the conicoid (I) at the point
(a, ji, y) are
X—a y—f
aap
bfip
●..(2)
cyp
where/? = l/v'(a2a2^^2^2_|_^2j,2j
where r denotes the actual
distance of any point on the normal (2) from P (a, y?, y).
We know that the principal planes of the central conicoid (I)
are the co-ordinate planes. The normal (2) meets the plane x=0
in Gi. So we have r=PG^.
0-a
from (2)
~PGi or PG,= \l(ap).
'aap
433
Central Conkoids
Similarly if the normal(2) meets the plane y«0 and z=0 in
the points Ga and Ga respectively, we have
PGa=-\Kbp) and PGs=-llicp).
PGi; FGi; PGa«-l/(ap); -Hibp); -l/(cp)
:(1/h):(1/c)
which are in a constant ratio since b, h, c are fixed for the given
conicoid.
Second Part.
Now it is given that
or
(-bw-a*(-a-
k*
Or
or
(aV+»>;8>+cV) (!/<'''+>/»*+●/«*)=**
[V
l/p*«a*«*+6*/5*+cVl
the point P(a, /3, y) also lies on the surface
...(3)
(aax*+hV+c*z*) (1 /fl*+1/6®+ l/c*)=fe*.
But the point P(«, /3, y) lies on the given central conicoid (1)
and, therefore, the locus of P is the curve of intersection of the
conicoid (1) and the surface (3).
Ex. 2. Find the distance of the points of intersection of the
normal at P(«, jS, y) to a central conicoid with the co-ordinate planes.
Sol. It is the same as the first part of Ex. I above.
Particotor case when the given conicoid is an ellipsoid. Let the
central conicoid be the ellipsoid
...(1)
x®/a*+;;®/6®M z®/c*=l.
The equations of the normal to (1) at P(«, /S, y) are
c V
«
V— fl
Z—V
/5p/6®
yp/c®
r(say)
...(2)
where \jp^ssa*/a*+p^lb*+y^lc* and where r .denotes the actual
distance of any point on the norma! (2) from P(«, P, y). If the
normal (2) meets the plane x=0 in Gi then putting r=aPGi in (2),
we get
0-a
PGi or PGi^—a*fp.
apla*
■
●
Similarly if the normal (2) meets the co-ordinate planes y=»0
and z=0 in Ga and Gs respectively then
PGg= —6*/p and PGa= —c*{p.
j
434
Analytical Geometry 3-D .
Ex. 3. If Q Is any point on the normal at P to the ellipsoid
x*la*+yW-\rzV.c^-=l such that 3PQ=^PG,+PGu+PGs where Gu
Gii -Oz are the points where the normal at P.meets the co-ordinate
planes respectively, then the locus of the point Q.is
a^x^
c»z«
1
(2fe*-ca-a«)8
(Kaopnr 1977; Bnndelkband 79)
Sol. The equation of the ellipsoid is
. xVa^-\-yy^.+2^lc^a=,U
Let the co-ordinates of a point P on (1) be (a, p, y) so that
the equations of the normal at P(a, p, y) are
X—CK,
V —fl
Z—y
...(2)
1
p* a*^b*-c*
where
(3)
Let Q be the point(^„ y^ r,). Since Q lies on (2), we have
Xi~-<t_yx-p_Zi-y
poifg^. pPlb^ py(c^”'
A x^^uMparfa% y^^p-\.{pprlb»), Zi=y-l-(pyr/c*).
Now /*(«, p, y) lies on the ellipsoid (1) so that
...(4)
«Va*+i8*/*HyVc'*='l
...(5)
Again it is given that 3P0=/>Gi-hPGa-|-PG8.
a«
/. 3r=P P P [V
and putting
the values of PGi etc. as
found in Ex. 2 above]
or
r«-(fl8+fts+c>)/(3p).
Putting this value of r in (4), we get
2fl8-b*-C»
g (<2^+d^+c^)
JC jcaa—
3fl8
3fl»
)●
=“(
so that
Similarly
a
'
3oxi
a 2(P-b*-c^
.
3czi
2b8-c^-a*’ c“2c«-a»-/>8
Putting these values in (5), we get
9a^xi^
9bW
9cW
a 1.
{2a*—b*—c*f ' (2b*-c“-a»)a ’ (2c>—fl»—b*)*
the locus of Q (xi, yu Zi) is
435
Central Conicolds
1
cV
b^y*
'(2c»-o’‘-Z>*)*”9‘ Proved.
(2aa-6a_ci]S^(2/,aEx. 4. F(W the length of the normal chord through P of the
ellipsoid x*/fl*+)'*/6*+2®/c*=l and prove that If it is equal to 4?Ga,
where Gs is the point where the normal chord through P meets the
plane z=0, then P lies on the cone
X* (2c2-fl«)/fl«+/(2c®-6*)/6HzVc*=0.
Sol. The equation of the ellipsoid is
...(1)
Let the co-ordinates of a point P on (1) b^(a.^„. y)» so tha^
(2)
The equations of the normal to the ellipsoid (1) at P (oc,^, y)
are
X—a
pa/fl*
=£=:Z.=r fsavl
pplb^ py!<P
where
...(3)
.(4)
p\
The co-ordinates of any point Q (say) on the normal (3) are
a+fpar/a»),.i8-|-(pi3r/6®), y+fpyr/c») where r=Pfi.
If PQ\s the normal chord of the ellipsoid (1) then Q will lie
on (1) and so we get
or
p®r*
or p*r*
-2
or
=Pfi, the length of the normal chord. ...(5)
or
Now if Pg—4PG8(as given) then Pfi =—4c*/p
[See Ex. 2 above for the value of PGs]
-2
4c«
p8 l(a»/a«)+(
or
2c®
or
a®
^ P
[Putting the value of PQ from (5)]
436
Analytical Geometry 3-D
“
The locus of P (a. j5, y) is
X*(2ca-o8>/aH3'* (2c*~6aj/6« +2Vc* =0, which is a cone.
Ex. 5. If a length PQ be taken on the normal at any point P of
the ellipsoid x*/flH:FV^®+2Vc*= I such that PQ^X^fp where \ is
constant and p is the length of the perpendicular from the origin to
the tangent plane at >. the locus of Q is
C®2*
1.
(fl*+
'(h* f A2)2 (>+
Sol. The equation of the given ellipsoid is
zVc*==> 1.
. (1)
^-Xet the co-ordinates of a point /»on (1) be (a,)5, y), so that
aVa»+i5V^*-fy«/r*-l.
are
...(2)
The equations of the normal to the ellipsoid (1) at P(a,j8, y)
X—
. * ■ y~^ -^g-y
r (say).
pa/a> piS/P py/c^^
..(3)
where pctfa^.p^lbl, py{(P are the actual d.c/s of the normal.
Let(Xi. yi, zj) be the co-ordinates of a point Q on the normal
(3)such that PQ^X^fp (as given),
then putting x<=x„ y- yi. z=z, and r==AVp in (3), we get
*i=a+~, yi«^-{-^,Zi«y-H^ata
fl”Xi
_P2il
a
C»Z|
h»+A8* ’'"’c»+A»'
Putting these values of(a, j3, y) in (2), we get
cV
=1.
(fla-f-AY (**+A8)3 (c8-f^a)»
The locus of the point g (xi,
z,) is given by
fl*X*
C»2*
h*y2
=1.
(a»-+AY'(hHAY'
Proved.
Ex. 6. The normal at a variable point P of the ellipsoid
xVoHyV^*+z*/c“=l meets the plane 2=0 (/.«. the xy y/ane) in Gg
Gsg fs drawn parallel to z-axis and equal to GaP. Show that
the locus of Q is given by
x*/(a*-e*)-|-yV(6*-c>)-i-2Vc»= I.
Also find the locus of R if OR is drawn from the centre equal
and parallel to GaP,
Sol. The equation of the ellipsoid is
...(1)
437'
Central Conkoids
Let (a, jS, y) be the co-ordinates of a point P on the ellipsoid
(I), so that
...(2)
aV+/5Vft*+y*/c»=l.
The equations of the normal to the ellipsoid (1) at P(a, jS, y)
are
x-tt
r (say),,..(3)
'^pplb^ pyIc*
where pa/a®, pj8/6®, py/c® are the actual d.c.’s of the normaUso that
r denotes the actual distance of any point on the normal(3) from
the point P(a,)3, y). If the normal (3) meets the plane zaO in
Gs then putting z==0 and r<=PGiin (3), we get
c>
0-y
PGa, so that PGa*=>—P
py/c®
Putting r=PGa=-- c^lp in (3), the co-ordinates of G^ are
(a^c»a/a^i8-qc®i5/tVO).
Now Gs is a point in the xp-plane and G^Q is a line drawn
parallel to z-axis such that GzQ=>GaP=‘—c*/p. If (xi, pi, zi) are
the co-ordinates of |2* then
XiaX-co-ordinate of Ga^a—(c*a/a*),
Pi=y-co-ordinate of Ga-jS—(c*i3/6®), ’
and
Z\^GaQ=—c*lp.
a-ax,
/. a=»a“- c® ’
'6a-c«*c
P
...(4)
From Zi/c«—c/p, we|have
Zl8
or -4* =sc®
c*
c® p*
a* 4* ● c* J
r«*_+ei+.rl]
[Putting for 1/p® from § H (B)]
or
y*
Zl(5)
Putting the values of a,
we get
a®x,®
W
or
or
or
and y from (4) and (5) in (2),
\
8
a®Xi® . ^®Pi*
C®Xi>
cW _ 1
(a*—c*;® (6»-c®)*”c® (a»~c*)® (b*-c*)»
[Again using (4) for a and jSJ
Xi® (o®—c*) , Vi® (6®—c®) , Zl®
I
(a®-c®)®
(P»—c»)® ’ c®
a
Zl*
Xxa
Px
\
i:
a»~c® b*-c® c*
438
Analytical Geometry 3>D
The locus of Q (;ci, yi. Zi) is
2
X" -
2
y*
.
Proved.
'c>
Second part. The equations of tbe line QR passing through
O (0, 0, 0) and drawn parallel to rhe normal at P are
X
y
(pa/fl*) (Pi3/h»;
Taking k=OR=‘GaP.‘=-~c^, the co-ordinates of R are
■ jSc8
/
OC*
I
a* * “■ fl*
*
Also let the co-ordinates of
be (jfg, y^, Za). so that we have
Xa=,—ac*/fl®,ya=—Pc^la^» z^^—y
or
a=a
P=
y
-2i>
Putting these values of oc, jS, y in (2), we get
a® Jfa®/c*+
.V
The locus of
=1 ●
(Xa, yz, Za) is fl*x*+bV+cV«=c*.
Ans.
Ex. 7.
The normals at P and P\ points of the ellipsoid
x*/fl*4-yV^“+2*/c*=»l» meet the plane z=0 in Gz and G'^and make
angles 6, d’ with PP\ Show that
PGz cos
4-P'G'a cos fl'=0.
(Agra 1976)
Sol. Let (a, p, y) and (a', p\ y') be tfie co-ordinates *of. tbe
points P and P' respectively. Now proceeding exactly as in Ex. 2
above, we have
P(7a= -cVp. P'G'3=-cV.
where p and p' have their meanings as given in § 11.
The d.c.’s of the normal at P are pa/a% p/3/6®, py/c* and the
d c.*8 of the normal at P' are p'aVfl*, p')376®, pV/c®. The direction
ratios of PP* are a'—a, P'—p, y'—y and so the d.c.*s of PP' are
(a'-a)ZPP', i/3'-/5j/PP', (y'-y)ZPP'.
Since B is the angle between the normal at P and the line PP',
we have
cos 0
pa a'— « ,
P'-P .p^ y- y.
“a®* PP '
6® ' PP'
V PP'
A P(?8 cos 0=3 — £l P. r« («'-«) ,./3 (/3'-jS) . y (y'-y)
6®
c®
■p*PP'L fl®
a
PP'lo® .
6® ^c®
-1-+^
U® ^6®
43$
Centra! Conlcotds
c
c»
't
since P(a, jS, y) lies on the given ellipsoid.
laxt!
Similarly P'G's cos B'^"~pp> ^”I^ b9
+1^
..
PGa cos 0
c»
.
+^a T^a j
=» — PGb cos
cos
or
Ws cos
Proven.
Ex. 8. If the normals at P and Q» points on the ellipsoid
x»/fl«+;/a/6a+z*/c*«l, intersect then prove that PQ is at right angles
to Its polar with respect to the ellipsoid.
Sol. The equation of the given ellipsoid is
x*/a* +y^jb^.+z*/c®=1.
...(1)
Let the co-ordinates of the points P and Q on (1). be
(«i, Pu n) and («8, pi. ya) respectively.
The equations of the normals to the ellipsoid(1) at the poin^
P and Q are respectively given by
x-cti_y-rPi_z-yi
:.(2)
^
^
x~tta_p-fia_z-ya
and
.(3)
oiila^ Pzlb* Yilc^
If the normals (2) and (3) intersect, then we have
i
●
or
yi-ya
ai-aa
Pi—Pi
tti/fl*
)3x/6*
yi/c“
aa/fl*
Pilb^
y«/c*'
PiYi—piYi
bH*
(««—«2> I
c^u*
Pa) ^yitta—ya«ij
(4)
Now the equations of the line PQ are
X-ai _y-Pi ^z-Yi .
«i—«a Pi—Pi yi—ya
...(5)
The polar of PG I.6* the polar of the line (5) [See § 9 ^C))
with respect to the ellipsoid (1) is given by
...(6)
aiX/aHPiP +yiz/c*—1*=0
440
Analytical Geometry 3-D
(«i-a,)(x/fl*)+03,-j8a)(yfl^)^(Y^^y^)(^/c>)=0. ...(7)
Subtracting (7)from (6), we get
...(8)
Let lu mu «i be the d.c.’s of the polar line of PQ i.e, of the
line of intersection of the plane(6) and (8). Then we have
h
„
mi
_
^lya—^aVi
yiaa-yaai
b*c*
c*a^
tti
...(9)
aj/Sa-agjS,*
Therefore, if the line PQ [given by (5)] and its polar [whose
d.c/s are given by (9)] are to be perpendicular then applying the
condition for perpendicularity, we have
0
yia,—yja,
which is true by virtue of(4). Hence the result.
§ 12. Number of normals.
To prove that six normals can be drawn to an ellipsoid from a
given point(x„ y,, Zy).
(Allahabad 1975, 78; Lucknow 71, 74, 77; Madras 76;
Meerut86; Robilkhand 78; Punjab 72j
Let the equation of the ellipsoid be
x^laU-y^lb^+z^lc^c=.\,
...(1)
The equations of the normal to the ellipsoid (1) at P (a, /5, y)
are
X—oc_y—jS z—
«/fl* i8/b» ,,^/=A(say).
...(2)
If the nomal (2) passes through the given point(*i, vi, n),
then this point will satisfy (2) aud hence we have
«/a*
fjb*
<=A.
y/C*
.
yA
or
a o«+A* b i»+A’ r ^’A
But the point P(«,j8, y) lies on the ellipsoid (1).
A «V«*+j8Vi'‘+y*/c“=r
441
Central Conicoids
or
C*Zi®
(fl>+A)>
*=al
I
...(4)
(6HA)» ■^Ac»+A)*
[Putting the values from (3)]
This equation is of sixth degree in A and hence gives six
values of A. Putting these six values of A one by one in (3), we
get six values of a, jS, y i.e. we get six points on the ellipsoid (1)
the normals at which pass through a given point (Xi, yi, Zi).
Hence from a given point six normals can be drawn to an ellipsoid.
Corollary. Proceeding exactly as above, we can prove that,
from a given point (xi, yi, Zi) six normals can be drawn to any
9
central cohicoid ax^+by^+cz 1.
C=3
§ 13. Cubic curve through the feet of the normals.
To prove that the six feet of normals {drawn from a given point
to an ellipsoid) are the intersection of a cubic curve with the given
ellipsoid.
Let the Equation of the ellipsoid be
...(1)
Let (Xi, yi, Zi) be the given point. Now if the normai at
P(a, j5. y)to the ellipsoid (1) passes through (Xi, yi, Zi) then
proceeding as in § 12, we get
...(2)
Now consider the curve whose parametric equations are
fl*Xi
6Vi
...(3)
where A is a parameter. We have proved in § 12 that A has six
ykiues [given by equation (4) of § 12] and hence corresponding to
= ieach of these six values of A, we shall get a point on the curve (3).
...(4)
Again consider a p\en^ Ax-\-By+Cz-{-D=>0.
ThClffOlnTrof iniemeij'tiQp qf ^1^ curve (3) with the plane (4)
are given by
...(5)
This equation being a cubic in A gives three values of A and
shows that the curve (3) meets the plane (4) in three points and
hence the curve (3) is a cubic curve.
Hence we conclude that the six feet of the normals that are ,
442
Analytical Geometry 3-2)
drawn from a given point(xuyu z\) to the ellipsoid (I) are the
intersection of the ellipsoid (1) with the cubic curve (3).
§ 4, To find the equation of the cone through six concurrent
normals (the six normals drawn from a point to an ellipsoid).
(Lucknow 1971;81; Rohilkband 79; Meerut 84)
Let the equation of the ellipsoid be
jc Vfl® +y^ll>^+2*/c®=1.
(I)
Let (xi, yu Zi) be any point(not on the ellipsoid). We know
that from (Xi, yi, zi) six normals can be drawn to the ellipsoid (1)
[See §12).
Now the equations of any line through the point (Xi, yu Zi)
are
x-x, y-yi zi-z,
m
n
—T
...(2)
If the line(2) is the norma! to the ellipsoid (1) at the point
(«»Pt y)» then we haye
l=*pcci'a^, m=p$lb^t m=pylc\
...(3)
Now if the normal at 2^(a, J3, y) to the ellipsoid (1) passes
through (xj, yx, Zi) then proceeding as in § 12, we get
- d»xx
C*Zi
i Y
c»+A
«HA’
...(4)
Putting the values of a, j8, y from (4)in (3), we get
/=£.
or
^-P
c*' cHA
5f‘=a»+A,^‘=6>+A,^=c»+A.
/
m
n
From the above relations, we see that
==(a“+A)(i«._c8)+(/,HA)(cVaV+(«*+A)(a*-**)
caO.
».(5)
Eliminating /, m,n between (2) and (5), the equation of the
cone on which the normal(2) lies is given by^
pXi(b^-{-2c^) , pyi (c^—a^) , pzi (a^—b^)CSlO
X—Xi
2-Zx
y-yx
x;,
(b^—c^)
yt
(c»q»)
Zi (a^-b^)
.
or
x-x,
y—yi
z-zi “ *
...(6)
This is the required equation of the cone on which lie all the
six normals drawn from (Xi, yi, zx) to the ellipsoid (1).
443
Central Conicolds
Corollary. To prove that the curve through the six feet of the
normals drawnfrom a point (xi, pu ^i)
ellipsoid lies on the
cone through the six concurrent normals.
The parametric equations of the cubic curve through the feet
of six normals are given by [See § 13J
ii
b^yi
a^xi
,. ^ - C‘Zi
...(7)
c2+A
where A is a parameter.
If this curve is to lie on the cone (6) then the values of Xty$z
from (7) should satisfy the equation (6).
X, y,z from (7) in (6), we get
Putting the values of
=0 or 2(oHA)(»’-c*)=0
or (fl»+A)(6?-c»)+(;fc*+A)(c2-aa)+(c*+A)
or
0=0, which IS true. Hence the required result is proved.
SOLVED EXAMPLES(E)
Ex. i.
Prove that the feet of the six normals drawn to the
ellipsoid x*ja^-{-y^/b^-\-z^lc^= 1 from ar^y point (Xi,
zi) lie on the
curve of intetsection of the ellipsoid and the cone
a^ (b^—c^) Xi b^ic^-a^)y^ cUa^-b^)zx^y
X
^
y
Z
[Rohilkband 1980; Madras 77; Gurunanakdev 75;
Lucknow 74; Meerut 85, 89, 90(P)]
Sol.
The equation of the ellipsoid is
X*/a®
+2*/c*= 1
...(1)
Proceed as in § 12, the co-ordinates (a, /5, y) of the six feet
of six normals drawn to the ellipsoid (1) from the given point
(Xi, yu Zi) are given by
f>_Ey±.
P“f)>*+A’ ^ c“+A
where A is a parameter and is
given by an equation of sixth degree.
Solving each of the above relations for A, we get
8
. b'^yi
-c«.
A=
—fl®, A=
a
V
/3
Now we clearly have
=A (63-c2>+A (c2-fl2;+A (fla-6*)=0
444
or
Analytical Geometry 3-D
a*x^ (b*-c^) b*yi (c*—a*) c*zj (a*—b^)
a
i8
y
=0® (b^-c»)-hb» (c>-fl8)+c«
=0.
The six feet (a, j8, y) of the normals to (1) drawn from
(Xi,yi, Zt) lie on the surface
a*xi
fa*-h«)
=0
X
"*■
y
z
...(2)
The equation (2) is a homogeneous equation of second degree
and hence it represents a cone. Bui since the six feet (a, jS, y) of
these normals lie on the given ellipsoid (I), therefore they lie on
the curve of intersection of the ellipsoid (i) and.the cone (2).
Ex, 2. If A, B, C; A\ B\ C are the feet of the six normals
from a given point to thi? ellipsoid x^/a^-\-y^lb^-\-z^fc^=:\ and the
plane ABC is given by lx+my+nz<=>p, prove that the plane A' B' C
is given by {xla^l)f(ylb^m)+{zlc^n) + {\fp)^0.
(Punjab 1977j
Sol. The equation of the given ellipsoid is
x^/fl* 4-y
2 Vc* == i
-■(1|
Let the co-ordinates of the given point be (Xi, yi, Zi). Now
the co-ordinates (a, jS, y) of the feet -of six normals irom (xi, yi, Zi)
to (I) are given
a=d®xi/(fl*+A), P‘=*b^yil{b*Ar\), y=c*Zi/(c*-l-A)
...(2)
where A is a parameter and its six values are given by the
equation
fl-x, 2
3
*=»1
(a“+A)» +(i>*+A)>
...(3)
Now the equation of the plane ABC is given to be
lxA-my\-nz~p
and three of the six feet of normals given by (2) lie on this plane,
so that we have
/fl*Xi mb^yx
ncHx
fla^A+hHA +c»-i-A
peaO
...(4)
This being a cubic equation in A gives us three values of A.
Again let the equation of the plane A* B' C be
i'x+
-1-«'z —p'*=0.
According to the equation the remaining three of the six feet
of normals given by (2) i>e on this plane and so we have
ro>Xi
m'b^yi
n'e^zx
h^-l-A + c“-t-A -p'=0.
445
Central Conicoids
This equation is of third degree in A and hence gives us the
remaining three values of A.
Clearly the product of the equations (4) and (5) must give the
same equation as is equation (3) Hence comparing the like
terniis, we get
I
or /'=
an
(fl“+A)»
(fl*+A)*
Similarly, we have
1
1
I
m*<=
b^m ’
c*n^
p
Putting these values in
n'z--p'«=0, the required
equation of the plane A' B’C is given by
iL. JL
Proved.
an^b^m
Ex. 3. If the feet of the three normalsfrom P to the ellipsoid
ll^
p/flwe xla+yjb+zlc^ I, prove that
thefeet of the other three lie in the plane
42/c-f 1=0 and
P lies on the line
a
y -r(a^-h*) z.
(Allahabad 1979)
Sol. The equation of the ellipsoid is
...(1)
The first part is exactly similar to Ex. 2 above.
Now clearly the six feet of the normals lie on
{xfa \-y/b+zlc—l)(x/a+:F/b+z/cfl)=0
or
(x/o-f
z/c)*—1=0
or (x*/n‘+//^»Hz»/c*-l)+2 {(yzlbc)-\-(zxlca)-\-(xylab)}^0 ...(2)
Since these six feet lie on the ellipsoid (1), therefore, using (1)
the equation (2) becomes
bc ■*"
~^ab~
0
or
ayz+bzx-^cxy—0
..(3)
Now if the n .rmals be drawn from P(Xi, >i, zi) to the ellip
soid (I) then these six feet of the normals lie,on the cone.
[See Ex. 1 (a) above]
fl®Xi (fc‘—c’) , b^yi (c®—a*) , c^zi ja^—b'^) =0
z
X
y
or
a^xi (6*-c*) yz-\-b^yi (c*-a*) zx+c*zi (o*-b*) xy=0
Comparing (3) and (4), we get
o*Xi
c*)/a=bVi
(u*—b*)/c
...(4)
446
Analytical Geometry 3-D
or
. fl (6*—c®);ci«=6
7ie=c.(a*—6^) Zi.
/, The locus of P(x,, y^, Zi) is the line given by
a
x—b (c®—o®) y=c (fl®--h®) z.
Ex. 4. Prove that the lines drawn from the origin parallel to
the normals to flx®4-6>^®+fz®=>l at its points of intersection with the
plane lx-\-my-)rnz’=p generate the cone
/x®
V®
z®\
Ilx
my
nz
'’‘\7+b+7)=\a+S+-c )■
Sol.
(Kanpur 1977)
The equation of the given conicoid is
flx®+6y®+cz®«=l.
...(1)
the equation of the given plane is
...(2)
/x+w^+nz=p.
Let(a, j3, y) be a point of intersection of the conicoid (1) with
the plane (2), so that we have
...(3)
oa®+ft/3®+cy®=l and /a-l-wjSI-l-«y=jp.
Now the equations of the normal to the conicoid (1) at
X—a y-p z—y
(a, p, y) are
...(4)
aa° bp
cy
The d.r.*s of this normal are aot, bp^ cy.
The equations of the line from the origin parallel to the
X
z
y
normal (4) are
...(5)
fla bp ^cy
From the relations (3), we have
or
p® (fla®-hi>j8®+cy®)=(/a-l-/wj3-f-ny)®
or
,^\a
(W , (gy)H
'^b '^c \
]/ (Q«) w (bP)
\ a '^
b
n(o0|*.
...(6)
Eliminating aa, hjS, cy between (5) aiid (6), we see that the
line (5) generates the cone
2 Jx* y® z® 1 ilx my nz \®
Proved.
" F+r+T|“(r+T+T) ●
Ex. 5. Prove that the points on the ellipsoid the normals at
which intersect a given straight line lie on the curve of intersection
of the ellipsoid and a conicoid.
Sol.
Let the equation of the ellipsoid be
x®/fl®+y2/6®+z®/c®c=l.
Let (a, p, y) be a point on the ellipsoid (1).
of the normal at (a, p, y) to the ellipsoid (1) are
...(1)
The equations
447
Central Conlcoids
X—g y—p z^Y
afa^ ~PJb^
Let the equations of the given line be
...(2)
...(3)
(X—Xi)//«=
(2—2i)/n
If the normal(2) intersects the given line (3), then the lines
(2) and (3) are coplanar so that we have
Xi-g
>i-j8
=0
y/c^
g/fl*
/
2i-y
n
m
the locus of(g, j8, y) is
This is the equation of a conicoid. But the point (g, /3, y)
lies on the ellipsoid (1) as well. Hence the locus of (g, )3, y)is
the curve of intersection of the ellipsoid (1) and the conicoid (4).
Ex. 6. Two planes are drawn through the six feet of the
normals drawn to the, ellipsoid x®/a*+;/*/^®+z*/c*«=l from a given
points each plane containing three. Prove that if Aj and Aa be the
poles of these planes with respect to the ellipsoid then
AxAa*-OAi^~-OAa^c=>2
Sol.
The equation of the ellipsoid is
...(1)
Let V (Xi, Pit Zi) be the given point. Proceeding as in § 12,
the values of A giving the feet of six normals drawn from
(Xi, yi, Zi) to the ellipsoid (1) are given by
fl*Xi*
bW
-lc=0.
I
...(2)
(fl*+A)>“^ (6*+A)2 "^(ca+A)*
Let the co-ordinates of the poles Ai and Aa of the two planes
be (xg, ya* Z2) and (xs, ya* za) respectively, so that the equations of
the polar planes of Ai and Aa are respectively given by
XXa . yya . zza
1=0 and xxa :yya . zzs —1=0.
■ fl> +
+ c®
fl® "* b®
c®
...(3)
The feet of the six normals, namely
fl^Xi
bVi
c^Zi
fl*-l- A * h*+A * c®-|-A )
(
where A has six values, lie on the two planes given by (3) and so
we have
448
Analytical Geometry 3-X^
£
leaO
XjXj
or
a^-\-h
and
I
S /_a^, ^3\__ I«=0
and
\aHA* W
y^yi
Zl'Zi
fts+A
cHA
XjXs
y^ya
IraO
...(4) ,
ZiZa
“ -1=0.
...(5)
6*+A ‘‘"cHA
Obviously (2) is tbe same as !he product of (4) and (S), so
comparing them, we have
1
XaXa yaya
z%z%
A
Now
XiXa^ —fl®, yaya^ —6®i
-c\
...(6)
^i^a®-0^i®-0^a«={(X2-X3)H(:F8--)'8)®+(za~^s)*)
—(*a®+J'a®+^2®)-W+^'sH^a*)
—2(X2Xa^jryay^+ZaZa)
—2(—a*—
c*),
using (6)
Proved.
2(fla+b®+c*).
Ex. 7. A is afixed point and P a variable point such that its
polar plane is at right angles to A P. Show that the locus ofP is a
cubic curve through the feet of the normalsfrom A to an ellipsoid.
Sol. Let tbe equation of the ellipsoid be
.X*/fl®+>»*//>*+z®/c*= I.
...(1)
Let tbe co-ordinates of the points A and P be (Xi, y\, Zi) and
(«, p, y) respectively.
'
●
The equation of the polar plane of P (a, /3, y) w.r.t. the
ellipsoid (1) is
ax/o®+j8^/6*+ yz/c^= 1.
Now the d.r.’s of the line AP arc Xi- a, j'j —]8, Z|-r y.
If the polar plane^2) of P is at right angles to AP then the
normal to the plape.(2) is parallel to the line
and so we have
Xi—</ ,yt — P
+ /3/b*
aA
i>i
Similarly
The locus of P(a,
Zi'-y
y/r“
=--A (say).
8
or
fl*X,
a=-i— *
fl*+A
2
c-Zi
y) is
t‘y.
C*Zi
^ dHA ’ y°b!^+X- Z— c*-t A\
which are the parametric equations of the,cubic curve thrj!)ugh the
feet of the six normals drawn from A (XxJyu Zi) to the ellipsoid
(1). [See § 13].
^
Central Conicoids
449
§ 15.. Diametral plane.
Definition. For a central conicoid, a diametral plane is the
locus of the middle points of a system of parallel chords.
(A) To find the equation of a diametral plane.
Let the equation of a central conicoid be
Let the equations of a system of pafaiiel chords be
...(2)
{X-a)ll^{y- P)lm-^-iz—y)!n=r (say),
where /, m, n are the direction cosines of each chord and so are
fixed, and r Is the actual distance of any point(a+/r,P+wr,y+nr)
on (2)from (a, )3, y). If(2) meets(1) then the distances of the
points of intersection of(1) and (2)from (a, jS, y) are the roots
of the quadratic in r given by
● a (a+/r)»+ft(j8+wr)2+c(y+nr)«=l
or r® (fl/*-fhm*+c«2)+2r(ala+bmp+cny)
+(na»+hi3a+ey*-.l)=0
...(3)
If(a, jS, y) be the middle f>ditlt oTthe chord, tbarn its distances
from the two points of intersection should be equal in magnitude
and opposite in sign and hence the sum of the two values of r
should be zero, so that we have ah-^bmp+cny=0
The locus of the middle points (a, j8, y) of the system of
parallel chords with d.c.*s /, m, n /.e. the equation of a diametral
plane for the conicoid (1) is
alx+bmy-\-cnz=‘0.
...(4)
Clearly the diametral plane (4) passes' through the centre
(0, 0,0) of the conicoid (1).
If /, m, n are the d.r.*s of a direction, even then the equation
of the corresponding diametral plane is given by (4).
Corollary. In case of tli^e ellipsoid
proceeding as above the equation of a diametral plane is
7x/fl* +● my/b®+nz/c®=0.
...(5)
This plane bisects a system of parallel chords of the ellipsoid
witli direction cosines/, m, n.
(B) To prove that every plane through the centre is a diametral
plane of the central conicoid corresponding to some direction.
Let the equation of the central conicoid be
flx®H-/>y®H-cz®=l.
...(1)
Consider a plane Ax-{-By+Cz=30
● ● . (2)
through the centre (.0, 0, 0) of the conicoid (1).
■ -V.
V
450
Analytical Geometry 3-D
We are to prove that the plane (2) is a diametral plane of(1)
corresponding to soine direction /, m,n (say).
Now the equation of the diametral plane of <4) corresponding
to the direction I; m, n is
...(3)
alx+bmy+cnz^O.
The planes(2)and (3) should ne the same and hence com¬
paring them, we get
/
m
n
al bm cn
or
A “T C
...(4)
Ala""Bfb^C{c\
Hence the plane(2) is a diametral plane of the conicoid (1)
corresponding to the direction /, m, n given-by (4).
§ 16. Conjugate diameters and conjugate diametral planes.
Let the equation of the ellipsoid be
...(1)
xVa*+y*lb*+z*fc^=‘l
with its centre at O (0,0, 0).
Let P fx,, yi. Z|) be any point on the ellipsoid (1) so that
d.r.*8 of the line OP are x{, yu 2. The equation of the diametral
plane [See § 15(A)] of the line OP is
...(2)
xx,/fl*-1-yy,/ft*+zzi/c*=0.
Now take a point Q (xa, ya, Za) on the section of the ellipsoid
(I)by the plane (2). This point Q will satisfy both the equations
of the ellipsoid (1) and the plane (2).
...(3)
X8*i/a*-f’yayi/b*+-ZaZi/c*:pO.
Thus the relation (3)is the condition that the diametral plane
of the line OP passes through another point Q on the ellipsoid
(1). But the result (3) is symmetrical in Xi, yu Zi sod Xa, ya. Za
and hence it shows that the diametral plane of the line OQ will
also pass through the point P.
Thus we conclude that if the diametral plane ofa line OP passes
through the point Q then the diametral plane of the line OQ will also
pass through the point P.
Now let the line of intersection of the diametral planes of OP
and OQ cut the surface of the given ellipsoid in the point
R(xu yn* z»)t so that'P lies on both the diametral planes of OP
and OQ and consequently the points.P and Q lie on the diametral
plane of the line OR i.e. on the plane
xxs/fl*-fyys/A*-l-zza/c*=0.
...(4)
Hence it shov;s that the diametral plane of OR is the plane
POQ. In a similar way the diametral planes of OP and OQ are
the planes QOR and POP respectively.
451
Central Conicotds
Now we shall define conjugate semi-diameters and conjugate
planes.
Conjogate semi-diameters. The three semi-diameters OP, OQ
and OR of an ellipsoid which are such that the plane containing any
two is the diametral plane of the third, are called the conjugate semidiameters.
Conjogate planes. The three planes POQ,QOR and ROP which
are such that each is the diametral plane of the line of Intersection of
the other two, are called the conjugate,planes or conjugate diametral
planes.
§ 17. The relationship between the co-ordinates of the points P, Q,
R where OP,OQ and OR are the conjogate semi-diameters
of an ellipsoid.
Let the equation of the ellipsoid be
...(1).
Let the co-ordinates of the points P, Q, R- be (xi, pi, Zi),
(X2,
(^3. Ja. Zi) respectively and OP, OQ,OR be the conju¬
gate semi-diameters of the ellipsoid (1) with 0(0,0,0)
centre..
Since the points P, Q, R lie on the ellipsoid (1), we have
jCiVfl*
1 "I
I
...(2)
I
x,Va^-{-y,Vb^-{-z,^lc*^l
J
We know that the diametral plane of any one of OP, OQ and
OR passes through the extremities of the other two and therefore,
we have
^
XiXifa*+yiyillP+ZiZilc*=Q 1
I
XiX»la*+yiy9lb*-^ZtZslc*^0
XaXi/flH.F8)'l/6*+^82l/c®='0
...(3)
I
J
If /, m, n be the direction cosines of any tine ^en we know
that/*4-w®-fn*=l. In view of this relation, tM relations (2)
imply that Xi/a, yijb, zjc\ Xz(a, yz/b, Zzfc and xJa, yzfb, Zzfc are
the direction cosines of some three lines.
Also if lx, mi, and Iz, m%, nz be the direcuon cosines of two
perpendicular lines then we know that Ixlz-Yfnimz-^nxnz—O. Tn
view of this relation, the relations (3) imply thiit the above three
lines are mutually perpendicular to each other.
452
Analytical Geometry 3-D
Again if A, mi, Hi; A, ma, n^; A, ma, »8 are direction cosines of
three mutually perpendicular lines then we know that At A» h*
mi ma, m^; ni, »a. fia are also the direction cosines of three lines
which are mutually perpendicular so that
A*+A*+A*“l etc. and Ami+Ama+Ams^O etc.
In view of these relations, we have
Le.
and
Similarly
Also
● ●●(4)
i.e.
and
e=0
a-
^i;'i+*a:Fa+X8y8‘=0.
y\Zt-hyaZi+ysZ9’=0,
ZiOCi+^a^a+^8*8“0*
Similarly
...(5)
Solving the first and third equations of(3)for Jfi/fl, yi/h, Zi/c,
we get
^i/g
ynZa—ysZt
—
yi/b
ZilC
ZaXa—ZftXa
x^y^—x^yt
be
ca
ab
cm
Vl-2?{(;'aZ8.-J'8^a)/hc}«J
●●
.:KaZa~;'8Z8 y\
F^^~Fc—
. ZiXy-~Z9X9 Z,
ca
±sinI905=±1.
JCaya-JTaJ^a
ah
..,(6)
Note. The results (2),(3),(4)(5) and (6) given above should
be committed to memory.
§ 18. Properties of coqjdgate semi-diameters of an ellipsoid.
Here we shall give some properties of conjugate semi-dia● meters of an ellipsoid. The meanings of OP, OQ and OR are the
same as already defined in § 16 and § 17 above.
I. The sum of the squares of any three conjugate semi-diameters
of an ellipsoid is constant.
(Allahabad 1978, 80; Agra 82; Madras 76; Koroksfaetra 77)
We have
OP»+Ofi»+OiP
(jc,*+y,a+zi^)+(Afa2+y8*+2a»)+(*a»+y3*+Z8»)
-=(jfi*+Xa>+*8*)+(>'i*+ya*+V)+(2iH2a*+Z3*)
«»=o*+h®-|-c*
[using relations (4) of § 17)
which is a constant. Hence proved^
Central Conicotds
45S
11. The volume of the paralleloptped having three conjugate
semUdtameters ofan ellipsoid as coterminous edges Is constant,
(RobilkbM 1981)'
Let QP, OQ, OR be the three conjugate semi'diameters of an
ellipsoid. The volume V of the parallelopiped having OP, OQ and
OR as three coterminous edges»6x(the volume.of the tetrahe
dron OPQR)
te„
F«=6xi ^1
7i
>a
Za
*8
Pa
Zz
*1
Pi
Zl
Xi
Pi
Zl
Xz
Pz
Zz
Xz
Pz
Zi
Xz
Pz
Zz
Xz
Pz
Zz
Xi
Xi
^8
Xi
Xa
Xa
Pi
Pz
Pz
Pi
Pz
Pz
Zl
Zz
Zz
Zl
Za
Zz
j
Xi*+Xz^+Xz^
*1>'i+JICi^8+W8
XiZi+XzZz-^XzZz
XiPi+Xa^^a+XzPz
Pi*+Pz*+Pz*
PiZi-\rPzZz'^PzZz
ZlXi-\-ZzXz'{-ZzXz
PlZl+PzZz-^-PzZz
Zl*+Zz*’\-Zz*
a*
0
0 ,
0
nr
Xt
V*<=*(Pb*c*
0
[using relations(4) and (S) of § 17]
0
0
or
V=abc which is constant.
III. The sum of the squares of the projections of three conjugate semi-diameters on a line is constant.
Let the direction cosines of a given line be I, m,n. Let OP,
OQ and OR be the conjugate semi-diameters of an ellipsoid.
The projection of OP on the line whose d.c.*s are I, m, n
*=.(x,-0) /+(;'i-0) w-i-(zi-O) n*=^lxi’^myi^nzi.
454
Analyticai Geometry 3-D
Similarly the projection of OQ and OR in the given line are
/xa+Wa^+n^a and /Xa+mya+nza respectively.
/. The sum of the squares of the projections of OPy OQ and
OR on the given line
=.(/xi+wyi+nXi)?+(/Xa+m>;a +»^a)H(/JCa+/wy3+«^a)*
(XiH*8*+*8*)+w* W+n* Szi*+2lm Sxiyi
~{-2mnSyiZi-\-2nlSziXi
=Z*o*-l-m*hH«V+0+0+0
[using the relations(4) and (5) of § 17 above]
which is cbnstanti
IV. The sum of the squares ofprojections of three conjugate
semi-diameters (of an eilipsoid) on any plane is constant.
Let the direction cosines of the normal to a given plane be
I, my Uy so that the square of the projection of OP on the given
plane is equal to
OP^-{IXifmyi+nzi)\
Similarly the squares of the projections of DQ and OR on the
given plane are
OQ^-ilXi+my^+nzt)* or
(/xa+m^^a+Wa)*
respectively.
The sum of the squares of projections of OP, OQ and OR
on the given plane
=OP^40fi*40/t®—{(/xi4w;'!4«^i)*4(/JCa4
4»«a)*
4(/'*84«rj'a4 «^a)*}
=(a> 4i>*4c*>~
4c*n»)
[Putting the values from properties 1 and 111 above]
=a« (l-/»)4Z>» (l~m*)4c*(l-n“)
=constant» since I, m, n are constants for a given plane.
V. If OP, OQ and OR are the conjugate semi-diameters of an
ellipsoid, then the sum of the squares of the areas of the faces QOR,
ROP and POQ of the paralielopiped having OP, OQ, OR as coterminous edges is constant.
(Rohilkband 1979)
Let li, mi, Hi; k, m^, n% and k, m^, Ha he the direction cosines
of the normals to the planes QOR,ROP and POQ respectively.
Also let Ai, Ai and Az be the areas of the faces QOR, ROP and
POQ respectively.
Let (Xi» yu 2ih {x», ya, Za) dnd (Xa, ya, za) be the co-ordinates
of the points P, Q and R respectively. Now if we project the
area QOR on the yz-plane i.e, on the plane xc=0, then we get a
triangle whose vertices are {0,y2, za),(0, 0,0) and (0, y^, Zg) and
the area of.this triangle.
455
Central Conicoids
=
casing relations (6) of § 17 above].
But the projection of the area Ai on the ^z-pIane«A4*
Aili=‘±{bcxi)l{2a),
Similarly if we project the area fiOiJ f.e. A on the planes
ZOJTand ATOT, weget
Aimi=^d:{cayi)/(2b) and i4ini«±(a^Zi)/(2c).
Squaring and addiug these three relations, we get
Arr{li^+mi*+nx^)=^[ibcxt)^l(2ay+{cayi)*l{2b)^^^^
a*b^
c*fl*
'
‘
b^c^
2
Zi .
or
4c*
...(1)
4^
+46«
In a similar way, the areas. A^ and Az are given by
■
^2»=
4o*
cSfl*
and
2,
Xa*+4p-^« +4C*
.43*=4a*
4b*
Za‘
...(2)
2
+4? Za".
...(3)
Adding (1),(2) and (3), we get
c*a*
6*c*
^l*+^2*+V
2
4b“
b*c*
4a*
9.
t9 .
-2
+4?-
.
[See relations(4) of § 17 above]
=i(bV+c*o*+a*b*) which is constant.
SOLVED EXAMPLES(F)
Note. In all the exercises given below we shall denote OP\
OQ and OR as conjugate semi-diameters of the ellipsoid
x*/a*+;>®/b*+z*/c*=» 1
and
(*1. yu Zl)» (^2..^2. Za) aod (*S» J^SI Za)
as cp-ordinates of the points P,Q and R respectively unless otherwise stated.
Ex. 1. Let OP^ OQ9 OR be the conjugate semhdiameters of
3x*+Ay*+z*=l. FwdAi/OP*+ Ofi*+OP»=0.
^ ^ ^^
(Allahabad 1979)
Sol. The equation of the given ellipsoid is
3x*+Ay*4'Z*‘=l.
.*, a*«=»i» b*= 1/A, c*«=»l.
We know that OP*+Ofi*+OP*=a*+b*+c*.
[See property 1,§ 18 above]
. 2=i+l/A+l or A«=3/2,
456
Analytical Geometry.3-D
Ex. 2. Show that the sum of the squares of the projections of
three conjugate diameters of an ellipsoid on the x^axis is constant.
(Allahabad 1979)
Sol. The direction cosines of x-axis are 1, 0, 0. From
property III of § 18 above, putting /= 1,
n=0, we get that
the required sum—fl*.(l)*-j-i«.(0)8-f.ca.(0)*=:a* which is constant.
Ex. 3. Find the equation of the plane PQR.
(Allahabad 1975, 81; Agra 75, 79,81; Bundelbhand 78;
Madras 76; Rohilkhand 77,80; Ravlshanker 70, 71)
Sol. Let the equation of the plane PQR be
lx-{-my+nz—p.
...(1)
Since the plane (I) passes through P{Xu Pu Zi), Q(Xi, yg, z^)
and J?(xs, yi, Za), we have
/*i+wyi4-nzi=p,
..(2)
/xa+mya+nza^p,
.(3)
and
/Xa+mys-l:/iZ3=p.
...(4)
Multiplying the equations (2),(3) and (4) by Xi.xa and Xa
respectively and adding, we have
or
or
/(Xi*-fXa*+Xa*)-l-m (Xiyi+Xaya+Xayal+n (XiZi-fXaZa+XaZa)
==/> (Xi+Xa+Xa)
/.fl»-l-iff.0-f«.0=p(Xi-|-Xa+JC8),
using the relations (4) and (5) of § 17
7-p(Xi+Xa+Xs)/fla.
Similarly we have
m=p (yt+ya+ya)/*®.
(Zi-l-Za+Za)/c«.
Putting the above vaiut s of /, m, n in (1), the required equa
tion of the plane PQR is given by
(x/a*)(xi-l-xa+xa)-f(y/bM (y,-i-ya f yal-Hz/c®)(z,-f-Za+Za)= 1.
Ex. 4. Prove that if the plane lxmy ●\-nz- ~p passes through
the extremities P^ Qt R of the three conjugate semi-diameterSy then
(Kanpur 1980; Lucknow 75)
Sol. Rewritting the steps oi Ex. 3 above upto equation (4)
and then squaring (2). (3) and (4) adding, we get
(/Xj-l-myi-i-«Zi)a+(/xa+OTya-l-«Za)*+(/JCa+mya-b/iza)*« 3p*
or P 2 Xi*+m* 2 yiH«* 27 ZyJ+2mn 2 y^z^+lnl 2 ZiX,
.
+2//H Xiyi= 3p*
or
Ex. 5,
[using relations (4) and (5) of § 17 above].
Prove that the plane PQR touches the ellipsoid
457
Central Conicolds
*»/fla-|-//6*+2Vc*«=i
(Kornkshetra 1977)
at the centroid of the triangle PQR.
Sol. Proceeding as in Ex. 3 above, the equation of the plane
PQR is
(x/fl3)(jr,+Xa+Xa)+(j^/6*)(;^i+J'2+3'3)+U/c*)(Zi+^a+^s)*!
Now clearly the co-ordinates of the centroid G of the triangle
PQR are
/Xj-l-Xa+Xs yi+3'2+J^8 X|+Za~{~X8\
[
3
^
.
It can be ea*=iily seen that this point lies on the ellipsoid
The equation of the tangent plane to the ellipsoid
at the point G is
X.(Xt+Xa4-Xs) ,yAyx + Va+J's) Z.fxi-I-Za+Xs)
3o*
or
3c*
36*
X (X,4-X8-l-X8)_Ly (3'i+J'2+3>8)j_2(Zi+Z2+Zs)
—
^3
"»■
p
c*
1/3
1.
-.(2)
The equation (2) comes out to be the same as the equation (1)
and hence it shows that the plane PQR touches the ellipsoid
2 (x*/a®)~ 1 /3 at the centroid of the APQR,
Ex. 6. Prove that the pole of the plane PQR lies on the ellipsoid
(Agra 1978; Rohilkhand 88)
OR
P: ^ve that the locus of the pole of the plane PQR ts
Sol. Proceeding as in Ex. 3 above the equation of the
plane PQR is
(x/fl*) (Xi+Xa+Xa)-f-(;//6*) (;'i+;'a+3'8)+(2/c*) (Zi+2a+Xa)“ 1
Let the pole of the plane (1) with respect to the given ellip
soid. x*/fl*+3>*/6*+2*/c**=«l be (a, j3, y).
The equation of the polar plane of (a, /3, y) with respect to
the ellipsoid
x*/o*-l-y*/6*"l-z*/^**=»l is
...(2)
ax/a*-lrj3y/6*-fyz/c*«= 1
Comparing (i) and (2), we get
1
a
Y
Xi-hXa+Xa
yi-l-ys+A'a
Zi+Zs+Xa
1
458
Analytical Geometry 3-/)
/. a=Xi+Xa+X8,P=yv^yt+yzf y=-2ri+Z2+^8.
Now» we have
«^l PA:y^^ixi+xa-{-xa? , (.yi+:ya+;^8)”_LUi+z8+g»)^
a^ b*
+ c2)
a*
+ I a*
fl*
^, Z2Za\ , /
■*■ h« ^ c»
c* )
fl2
b^
'=-5+;5+-aQ
D
C
[Using relations (4) and (3) of § 17 above]
e=3.
.*. The locus of the pole (a, j8, y) is x^/a^-{-y^fb^+z^lc^^3.
Ex. 7. Prove that the locus of the centre of the section of the
ellipsoid x^la^-\-y^lb^-\-z^lc^=>\ by the plane PQxi is the ellipsoid
x*(a^+y'^lb^-\^z^lc^=\. Prove further that this is also the locus of
the centroid of the triangle PQR.
Sol. The equation of the given ellipsoid is
S=xVa’>+.vW+zVc*-l‘='0.
...(1)
Proceeding exactly as in Ex. 3 above the equation of the plane
is given by
(x/fl») (xi+Xi+Xs)+{y/b^) iyi+yi+y9)-\'izlc*) (xi-f-2a+^8)=l.
...(2)
Let (a, jS, y) be the co*or^inates of the centre of the section
of the ellipsoid (1) by the plane PQR so that the equation of the
plane PQR is given by
y£
i.e.
**_L
2
. >L
...(3)
The equations (2) and (3) represent the same plane and hence
comparing them, we get
a
y
i
Xi-\-Xi+Xi yi+yi+yfi zi+za+za
.
●●
and
«-(xi+Xj+xAlct^ .^^ .yH P^(yi-\-y2±yz
b
\“~a
\
a
y--/z.+z2+z8\/«%j3^yM
c \
c
l\a^'^b^'^c^ y
Squaring and adding the above three relations, we get
<*-^4.^^jL.y^-.l^\§l+yl\^ /(^i+^a+Xa)^ . (>>i+3>a+:F8)^
(Zi+Za+Zs)*
c*
}
459
Central Conicoids
as in Ex. 6 above]
or
or
(4)
.*. The locus of(a, j8, y) is x^la^+y^lb^+zVc^’=h
Second part Let the co-ordinates of the centroid of the
triangle PQR be (a, jS, y), so that we have
at=j(xi-hxa+xs),)3=i (yi+yi-\-yzh y=i(zi+22+23).
a-8 . . V® 1 nx.-t-Xa-l-Xa)^ . (.yi+y2+J>8)^ . (2i-fZa+Zsj*
J
=i.3
(Proceeding as in Ex.6 above]
The locus of(a, j8, y) is xVa®+:vV**+z*/c*=i which is
Proved.
the same as (4).
Ex. 8. Find the equation of plane through the points (Xi, Xai ^fa)*
(A. y»» ^s) owirf (^J»
^s) where (Xr, y,, Zr)» r=l» 2, 3 are the extre¬
mities of the three conjugate semi-diameters and show that it touches,
afixed sphere.
Sol. Let the equation of the required plane be
ix-\-my-\-nz=p.
●●●(!)
The plane (1) passes through the points (Xi, x#, Xi), (yi, yai y»)
and (zi, Za, Zs), so we have
...(2)
/Xi+wiXa-f’^^a™/^*
...(3)
/yiH-mya+«y3=i>,
...(4)
and
/zi+mza+nzs—y.
Now multiplying the equations (2), (3) and (4) by Xi/fl*, yifb*
and Zj/c* respectively and adding, we get
xiXa. yiya, zi22\ , _ l’XiXs_^yiy9_i,^\
/
c^l
b^
or
/=P (Xi/a*-l-yi/6Hzi/c*)
[using relations (2), (3) of § I7j.
Similarly /n=y (X2/o*+ya/i>*+Za/c*), n=P {XzItP+yzIb^+ZaJi^)Putting the values of /, m, n in (1), the equation of the requi
red plane is
)+»(?+;:+?■)+=
...(5)
Again we are to show that the plane (5) touches a fixed
sphere.
460
Analytical Geometry 3-D
The length of the perpendicular from the origin to the plane
(5)
. J'a , £a
1/ /
Isi I a*
b* ^ c* ^
, 2£yiZi, 2UziXi
^
'/7K*S+7+»««j
[using the relations(4) and (S) of§ 17 above]
abc
which is constant.
s/{b*e*4-c*a“+a^b*)*
Hence the plane (S) touches the fixed sphere
o»Z>«c“
Ex. 9. Prove that the focus of the foot of the perpendicular
from the centre of the ellipsoid x*/a*4->'*/6*+2V«®=»l to the plane
through the extremities of three conjugate diameters ts
-j-6V®
●= 3
4*2*)®.
(Allahabad 1976; Meerot 83S)
Sol: The equation of the given ellipsoid is
x^ja^Jry*}b*-\-z*lc^^\.
...(1)
Proceeding exactly as in Ex. 3 above, the equation of the
plane PQR is
(x/a*) (Xx + jfa+X8)+(;;/fc*j (yi+;;a+ys)+(2/c8) (Zx+Za+Za)= 1.
...(2)
Let N (a, )3, y) be the foot of the perpendicular from the
centre O (0. 0, 0) of the ellipsoid (1) to the plane PQR.
Let p be the length of the perpendicular O^so that
p=OAf^the distance between O (0. 0, 0) and Af (a, j8, y)
= V(«®+^®+y*).
...(3)
The d.c.’s of ON^ the normal to the plane PQR, are
«/V(a»+iS®+y®). /3/v'(«®+i8®+y®>, y/V(«®+j8* +y»)
Ue.,
«//». ^Ip» yIp, using (3).
Hence the equation of the plane PQR is also given by
(a/p) x+{^/p) y+(ylp) z-p or ax+)3y+yzc=p*.
...(4)
Comparing the equations (2) and (4), we get
«Xl4-Xa+X8_yi+P8-fJ's _,^l+X8 + Z8_. 1
tP
c*
P'
461
Central Conicotds
x,±£fa±^
aot J>i+J>a4-y8
“p8»
6«
bp Zi+Za+gs- cy
- c*
P*
”
Squaring and adding, we get
(^i+Xa+Xs)* , C^i+J'a+ys)* ,(^i+Za+Zs)® ,
fl*
c»
or
"?■ + 6« +
c»
+[22;(?^*)+2r(^«)+2r(^«);
<*V*
P*
or (1 +1 + 1+0+0+0).p‘==flV+6*i32+cV
or
Lusiog (4) and (3) of § 17]
3 (a*+i3*+y*)*==(a*a*-|-6*iS»+c*y‘). from (3).
.*.
The locus of (a, /3, y) is
Proved.
3 (jc*+>>2+z*)2=fl*x®+hV+«‘^*Ex. 10. If Xf 11, 'i are the angles between a set of equal conjugate diameters of the ellipsoid x*/a*+y^{b^-\-z^{c*=\, then show
that
C05* A+cas* p+cos* vi
Sol.
ZSib^-c^)^
2 (fl»+h«+o*)»
The equation of the given ellipsoid is
xVfl*+yV^»H2Vc*=i.
...(1)
Let OP, OQ, OR be a set of equal conjugate semi-diameters
of the ellipsoid (1) so that
OPf=OQ^OR’=^r (say).
We know that O?*+0eH<7i?®=*fl*+6*+c*.
3r*=fla+hHc*.
...(2)
If the co-ordinates of P, Q, R are (Xi,
Zi)» (^a» Pa# ^a)#
(*3fP8. ^a) respectively, theu the d.r.’s of OQ and OR are
Xa# Pa# z» and Xa, y^, Zs respectively.
If A be the angle between
OQ and OR, then
yays+PaPa-l-ZaZa
^aXs+PaPa+gags
cos A<
r*
\/W +Pa“+Za®) 1/(^8*+Ps*+V)
.*.
cos* A = (X2X8+P2P8+^82a)V^*-
...(3)
From Lagrange’s identity, we have
(*»*+Pa*+Za*) (^8*+P8®+Z8*)-(«23fa+P2p3+raZ8)*“2? (PaZa-yaZ,)*.
Putting from (3) on the L.H.S. and from the relations (6) of
§ 17 on the R.H.S. of this identity, we have
a*bW
c*fl*yi*
r* .r*—r* cos* As=» ft*c*x,*
c*
462
Analytical Geometry 3-2)
Dividing by r* and transposing, we get
cos A-1:- r* (^+“fc4+c«
aW
Similarly cos® /x= 1
r*
and
cos® VC3 1 aW M.yz^.Zz^
r‘
)
Adding (4), (5) and (6), we have
cos® A-h cos* /It-1-cos® V
a
fl®6»c® fSxi . Sy^^ 2?2i®1 , fl*fc«c® ffl* . ^
...(4)
...(5)
...(6)
9(6*c®-i-c®a®-f a®6®)
[See (2)]
!●« (6®c®+cV+fl®fc®)=3(fl®+b»-i-c»)®
3 ((fla+68+c2)9_3 (6»c®H-c»n®-i-fl®ft*)r
“
(a»-i-6®-i-c®)*
c= 3 [fl* -h 6®+—ft*c* c®a®—fl®6®]/<a® -f ft® -j- c®)®
3 f2flH2ft«-l-2c«-2ft®c®-2c®o®-2fl®ft®]
~
2 (a®-|-ft®+c®j®
3 [(ft«-2ft«c®+c«)-l-(c«-^2c®fl®-|-fl®)-f^fa«-2fl®ft*-|^ft®)i
“““
^
2 (fl8 + 6«-|-c®)®
3 ((ft*-c®.)®+(c®-a®)®-f (a®.-ft®)®l
2 (a®H-ft*H-c®)®
Proved.
Ex. 11. Find the locus of the equal conjugate diameters of the;
(Meerot 1985)
ellipsoid x*/fl®4-//ft*+^7c*=> 1.
Sol.
The equation of the given ellipsoid is
x®/fl*-l-//ft®+z*/c®=l.
...(1)
Let OPt OQ and OR be the equal conjugate semi-diameters
of the ellipsoid (1), so that OP^^^- OQ-OR—r^ (say).
J-et (xi, yu Zx), {Xi, yg. z^) and (jts, >^3, 23) be the coordinates
of the points P, Q and R. We have ●
OP=r=> <9P®=r®=>JCj®+jf>i®+Zi®=r®.
...(2)
Similarly
ac2H>’2"+V«=r®
...(3)
and
...(4)
Adding (2), (3) and (4), we get
a®+ft8+c2=3r*.
.. (5)
[V 27A:i®=o», 2;j;x®=fta, i;zi*=c*]
Now let /, m, n be the d.c.’s of OP so that the coordinates of
the point P are {Ir, mr, nr). Since P(/r, mr, nr) lies on the ellipsoid
(1), we have
Pr^la^-\- m^r^lb^+n-r^[c^f=l
463
Central Conicotds
or
/a
/*
„9 1
_ m ■«a
_ 3 (/»+m»+«®)
...(6)
^+5T+^“;.a
flS+^a +c2
(Putting the value of r* from (5) and using the fact that
The relation (6) shows that the semi-diameter OP generates
the cone /.e., the locus of OP is the cone
x« y*
3 (x^+y^+2^)_t,
a*+6>+c»““ fl^+h^+c* “
or
+(zVc>) (2c*-fl»-6*)=0.
Exercises
1.
Show that the plane x+2;'+3z=2 touches the conicoid
x*-2j;»+3z®=2 and find the point of contact. (Meerut 1983)
APS,
(1. -1. 1).
2.
Find the equations of the tangent planes to the ellipsoid
7x* + 5;V*+3z*‘=60 which pass through the. line
(Kanpur 1974)
3z=0=7x+ 10.y—30.
Ans. 7x+5H’3^-=30, 1.4x+10;^4-9z=50.
3.
A tangent plane to the conicoid ax^-\-by'^^cz^*=»\ meets the
coordinate axes in P, Q and R. Prove that the centroid of
the triangle PQR lies on the surface
J- .J_+±=9
ax^^ay^^az*
4. Show that the locus of the point of intersection of three
mutually perpendicular tangent planes to a central conicoid
is a sphere concentric with the conoicid.
(Meerut 1984 S)
[Hint. Refer § 7.]
5. Show that the locus of the centres of all sections of the
conicoid flx*+frj;®+cz*=l by planes which pass through a
fixed point (x\ y\ z') is
ax (x-x')+by (y~y')+cz (z-z')=0.
6. Show that the centres of the sections of the ellipsoid
that are (i) parallel to a given line
lie on a fixed plane and (ii) that pass through a given line lie
on a conic.
464
Analytical Geometry 3-D
7. Prove that the six normals drawn from any point to a central
conicoid are the generators of a quadric cone.
(Meernt 1984/89 S)
[Hint. Proceed as in § 14.1
8. Prove that the feet of the six normals drawn to the central
conicoid'ox*4- -Ifrom any point (Xi, yi, Zi) lie on
the curve of intersection of the given ellipsoid and the cone
a(b—c) Xi .bjc-^al^ c
z,=0.
X
y
z
(Rohilkband 1977)
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