~~!
~iiir[l.L_~~~~~~~~~~~~~~~~~~~~~~~~-----,
~
-
Vectors in two and three dimensions
Content
3.1.1
Vectors in two and three dimensions
Learning Outcomes
Include:
(a)
Addition and subtraction of vectors, multiplication of a vector by a scalar, and their
geometrical interpretations
-?
xi + yj , xi + yj + zk , AB , a
(c)
Position and displacement vectors
(d)
Magnitude of a vector
(e)
Unit vectors
(f)
Distance between two points
(g)
Angle between a vector and the x - , y - or z - axis
(h)
Use of the ratio theorem in geometrical applications
3.1
vectors in two and three dimensions
3-1
Vectors in two and three dimensions
Define
Scalar
A scalar is a quantity that is completely defined by its magnitude only
and can be represented by a real number.
I Examples I
CD
@
a>
length (say, 100 centimetres)
monetary unit (say, $75)
time (say, 20 minutes)
Vector
A vector is a quantity that has magnitude and a definite direction in
space.
O A vector can be denoted by the following symbols:
•
AB , where the letters represent the end-points of the vector.
•
a small single bold-type letter such as a or p , where the
magnitude of a and of p is denoted by
lal and IPI respectively.
a or p.
•
a small single lightface letter with an arrow cap such as
•
a small single lightface underlined letter such as 9. or p .
(preferred handwritten form)
•
O Geometrically, a vector can be represented by a
B
directed line segment, i.e., a line segment
having a specific direction as indicated by its
arrowhead. The length of the line segment
proportionately represents the magnitude of the
vector (in appropriate units), while its arrowhead
represents the direction of the vector. Thus, if the starting point of a
vector is A and the ending point B, with the direction going from A to B,
then the directed line segment is denoted by AB , or alternatively by a
small letter, say, a.
A/
3-2
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O cosmic
z
O A vector which is specified in a three
dimensional plane can be conveniently
denoted using a column vector, say,
[
~)
3.-------- --- ---------::-:1
ryj ~
/-::-: ___ __________)(i_
zk
• where x, y and z are orthogonal
//y
components of the vector respectively.
y
: ,/
OR
x
Alternatively, in terms of its 3 basic
orthogonal unit vectors in the direction of the
positive x, y and z-axis respectively, i.e.,
md+yj+zk
•
If the two points are not fixed in space, it is a displacement
vector from the initial point to the final point. For such a vector, it
is not changed when if it is shifted parallel to itself as long as its
magnitude and direction are not changed.
•
If a vectors initial point is fixed at the origin 0, then the vector is
the position vector of the final point relative to the origin 0.
a = OA is the position vector of point A.
O The magnitude of the vector is the length of the directed line segment
representing the vector. It is represented by placing the symbol in an
absolute (modulus) sign , i.e.,
Magnitude of
AB = IABI
Magnitude of a= !al
Example
I
~
displacement (say, 10 km west)
velocity (say, 26 km/hr in the north-east direction)
Gl
acceleration (say, 35 ms-2 upwards)
CD
Zero (null) vector
Zero vector or null vector is any vector with zero magnitude and is
denoted as 0.
Unit vector
A unit vector is a vector with magnitude of one.
o The unit vector of a (=a) is always parallel to a and is denoted by
A
a
a=i;r·
3.1
vectors in two and three dimensions
3-3
Negative vector
~B
The negative vector of a ( = -a) has the
same magnitude but opposite direction as a.
A
A
Equal vectors
Vectors that have the same magnitude and direction are known as
equal vectors.
O
Equal vectors need not have the same starting point and ending point.
Multiplying a vector by a scalar
The scalar multiplication of a vector refers to the multiplication of a
vector by a scalar, say, k, so that the resultant vector has the same
direction (for k > 0) or opposite direction (for k < 0) as its original
vector and a different magnitude (for k
* ±1).
O Mathematically, the scalar multiplication of a vector, say, a, and a
scalar, k, is denoted by ka, where:
If a is not a zero vector and
0
o
k > 0, then ka is a vector with the same direction as a and a
magnitude lkllal
o
k < 0, then ka is a vector with a direction opposite to a and a
magnitude of lkl !al
o
k =0, ka is a zero vector, so we have Oa =0.
If a is a zero vector
f)
o
then ka is also a zero vector, so we have kO = 0.
O Multiplying a scalar to a vector is distributive .
•
k(a+b)= ka+kb
•
k(ma)=(km)a
Example
CD
3-4
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I
The vectors 2a and -2a are as
shown.
Cl cosmic
Add it ion of vectors
Vector addition is the operation of adding two or more vectors together
to give a resultant vector.
O
In the case where there are only two vectors, say, a and b , vector
addition can be carried out using:
0
Triangle law
The sum of a and b is obtained by
placing the starting point of b on the
ending point of a and then joining the
starting point of a to the ending point of
b . The sum is written as: r = a+ b .
The vector triangle method can be
extended and used to find the sum of
3 or more vectors through the
construction of vector parallelograms .
6
Parallelogram law
The sum of a and b is the diagonal
(denoted by r) of the parallelogram for
which a and b are adjacent sides
(outwardly directed).
a
O Addition of vectors is commutative and associative.
•
a+ b = b +a (commutative)
•
a+ b + c =(a+ b) + c =a+ (b + c) (associative)
Subtraction of vectors
Vector subtraction is the operation of subtracting one vector from
another.
O
The difference of two vectors, say, a and b , is
denoted by a - b. It is equivalent to vector addition,
with the orientation of the second vector reversed.
Thus, the difference of a and b can also be written
as a -b = a+ (-b).
a-b
O The commutative and associative law applies to subtraction of vectors .
3.1
vectors in two and three dimensions
3-5
Parallel vectors
Two non-zero vectors which have the same or opposite direction such
that each is a non-zero scalar product of the other are known as
parallel vectors.
O Mathematically, we can write:
a is parallel to b <=> ha = kb , where a * 0 and b * 0 and h and
k are scalars such at h * 0 and k * 0 .
0
If
•
!5._ > 0 , then a and b move in the same direction
•
!5._ < 0, then a and b move in the opposite direction .
h
~-
h
O Conversely, if a and b are non-parallel vectors , then either
•
h =0 and k =0, or
•
a = 0 and b = 0.
O If three points A, Band Care collinear points, we have
AB = kBC, or AB = kAC, or BC = kAC, where k is a nonzero scalar.
•
Conversely, prove that the vectors formed by the points are
parallel and they share a common point.
Vectors in two dimensions
O
Vectors can be resolved into two or more components.
O In two dimensional space, vectors are written as (;) or xi+ yj, where
i and j are the unit vectors of the positive x-axis and the positive y-axis
respectively.
Example
I
<D A vector a can be denoted in the
y
following ways:
•
in terms of unit vectors i and j
x1i + Vij where X1 and Y1 are
cartesian components of a.
a
i
x
0
----~--
•
3·6
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in terms of ordered pairs (;) or (x y) .
0 cosmic
0
For two points A and B with position vectors a= x1i + Vij or (;:
J and
b = x) + y 2 j or (;:] respectively:
lal = ~x/ +y/
ra =r(x1i + Y1j)
or
r(;:J,
for r E IR
Vectors in three dimensions
O
In three-dimensional space, vectors are
~ J xi + Yi + zk
written as [
or
z
z.. ························.'"
where i, j
,::<:.. . ........J<i_ -1;.-yj zk
-!<
and k are the unit vectors of the positive
x-axis, positive y-axis and the positive zaxis respectively and the scalars x, y and
z are the rectangular components.
•
i
~ ~J
•
j
~
[
k
y
x
x
or ( 1, 0, O)
m
or (0, \ 0) aod k
~
m
or (0, 0, 1)
3.1
vectors in two and three dimensions
3-7
CJ For two points A and 8 with position vectors a · x,i + y,j + z,k or (
and b • x,i + y ,j + z,k or (
~:]
~;]
respectively
lal = ~x/ +y/ +z/
a = b <=> x1 =x2 and Y1 =Y2 and z1 = z2
AB = OB - OA
3-8
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=b- a
C cosmic
Angle between a vector and the x-, y- and
z- axis
(j
Angle Bx between vector v = v1i + v 2 j + v3 k
and the x-axis:
z
V3. ....................... :
1
(<-__ ........ Y!if·-~~j +
V3k
e,
(j
: Bx
Angle By between vector v = v1i + v 2 j + v3 k
8y
0
,./v2
Y
..........................:./
and the y-axis:
v,
(j
Angle Bz between vector v = v1i + v2 j + v3 k
and the z-axis:
(j
The cosines are called the direction cosines of the vector and are
equal to the components of the unit vector.
2
cos 2 Bx + cos 2 By + cos Bz = 1
Ratio theorem
A-a+µb
P = - -A-+µ
where p, a and b represent
OP, OA and OB respectively.
0
(j
If Mis the midpoint of AB, then
I m=~=~( a+ b)
I·
Derivation :
OP= OA+AP
=QA+_µ_ A8 = QA+_µ_ (08 - QA)= A-QA+ µoa
µ+A,
µ+ A,
µ+A,
3.1
vectors In two and three dimensions
3-9
Worked Examples
IExample 1 j
Solve the following:
(c) Let a and b be the
2position vectors [ ~ ] and [-:2J.Determine
1
the distance between these two position vectors.
Solution:
(a)
[~+UJU3 J
(ans)
2
(b)
(c)
[~+[-~ ]{6~]
(ans)
UJ[~+[_!3J
Distance between a and b = ~2 2 + 4 2 + (-3) 2 = ,J29
(ans)
IExample 21
Find the angle that the vector 3i - j + 2k makes with the y-axis.
Solution:
Let
e be the required angle.
-1
cos e = --=====
~32+12 + 22
1
ffe
fJ = 105.5°
3-10
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(ans)
C cosmic
I Example 31
The points A, B have position vectors a, b respectively, referred to an
origin 0. The point C lies on AB between A and B such that AC = 2, and
CB
Dis the mid-point of OC. The line AD produced meets OB at E. Find in
terms of a and b,
(a) the position of vector of C (referred to 0)
-->
(b) the vector AD
Solution:
B
0
-->
-->
.
__,
20B+OA
(a) By ratio theorem, OC =
=!(a+ 2b) (ans)
2+1
(b) Dis the midpoint of OC.
-->
-->
-t
: . OD= 1oc = (a+ 2b)
-->
-->
-->
:.AD=OD-OA=i(a+2b)-a= i(2b-5a) (ans)
A
IExample 41
In a triangle ABC, the points D, E, F divide
the sides BC, CA, AB internally in the ratio
k:1 as shown. Prove that the sum of vectors
represented by AD, BE and CF is zero.
Solution:
-->
-->
-->
We have to show AD+ BE+ CF= 0
a
By ratio theorem,
k
1
1
1
AD = - -(k AC+ AB)' BE= - -(k BA+ Be)' CF = - - (k CB+
k+1
k+1
k+1
cA)
Adding,
-->-->-->
1
AD+
BE+ CF= k+
=0 (QED)
1
[(-->
--> CB
-->) + (-->-->
-->)] =-1-(0)
k AC+ BA+
AB+ BC+ CA
k+ 1
(ans)
3.1
vectors in two and three dimensions
3-1 1
Worked Problems
IExample i I
The position vectors of the points A, Band Care given by a = 4i - 9j - k ,
b=i+3j+5k, c=pi-j+3k. Find
(a) The unit vector parallel to AB.
(b) The position vector of the midpoint AB.
(c) Find the value of p such that A, B, Care collinear.
(d) If p = -2, find the position vector of D such that ABCD is a
parallelogram.
Solution:
- -
~
(a) AB= OB-OA =-3i + 12j + 6k = 3(-i + 4j + 2k)
Unit vector= -
1
1
- x3(-i + 4j + 2k) = - -(-i + 4j + 2k) (ans)
3./21
./21
~
(b)
t (a+ b) =Midpoint of AB = t (5i - 6j + 4k)
~
(ans)
~
(c) AB=A-AC
08-0A = A-(oc -0A)
-3i + 12j + 6k = A- [(p - 4 )i + 8j + 4k)]
6k
A- =-=> A-=t
4k
... -3 = t(p- 4)
p=2
(ans)
(d) If ABCD is a parallelogram, let OD= xi+ yj + zk .
~~
----
AB = DC => OB - OA = OC - OD
=(-2 - x)i + (-1 - y)j + (3 - z)k
: . Position vector of D =i - 13j - 3k (ans)
-3i + 12j + 6k
3-12
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Cl cosmic
J
Example
21
ABCO is a square and M, N are the midpoints of BC and CO respectively.
-->
Let p
-->
-->
-->
-->
= AM and q = AN . Express AB , AC and BO in terms of p and q.
Solution:
-->
-->
Let AB = a, BC = b
0
a+-tb=p
b+-ta=q
b
From 0 and 6,
b + -t (p - -t b) = q
fb=q--tp
3b = 4q - 2p
b=!(4q-2p)
a=
-->
:. AB=
t (2p - q)
t (2p - q) (ans)
-->
:.AC= a+ b =
t (2p - q) + ! (4q - 2p) = t (p + q)
(ans)
t (2p - q) = 2(q - p)
(ans)
-->
:.BO = b - a= ! (4q - 2p) -
3.1
vectors in two and three dimensions
3-13
The scalar and vector products of
vectors
Content
The scalar and vector products of vectors
3.2.1
Learning Outcomes
Include:
(a)
Concepts of scalar product and vector product of vectors
(b)
Calculation of the magnitude of a vector and the angle between two directions
(c)
Calculation of the area of triangle or parallelogram
(d)
Geometrical meanings of la·bl and lax bl, where b is a unit vector
Exclude
•
3-14
triple products a·b x c and ax bx c
AL eGuide Mathematics
Cl cosmic
The scalar and vector products of
vectors
Define
Scalar product
L
The scalar product of two non-zero vectors a and b,
I
written as a·b is given by a·b =
la! jbl cos BI where B is
the angle between a and b.
•
a
a·b is positive when cos() is positive, and a·b is negative when
cose is negative.
!j
!j
If a, b and c are non-zero vectors, the laws of scalar product are:
•
Commutative:
a·b = b·a
•
Associative:
( ka )·b = (kb )·a = ka·b
•
Distributive:
a·(b + c) = a·b + a·c
Other basic properties:
0
If a and b are parallel vectors in the same direction <=>
a·b = lallbl cos0° = lallbl
6
If a and b are parallel vectors in the opposite direction <=>
a·b = jal lbl cos 180° = -lal lbl
@
If a and b are perpendicular vectors <=>
a·b = lallbl cos90° =0.
0
If a and b are equal vectors, a = b
<=>
a·a = jaj2
!j
2 2 2
Mathematically, if a= x 1i + y 1j + z1k and b = x i + y j + z k, then
a·b = ( x1i + y 1j + z1k )·( x) + y 2 j + z2k) (vector form)
or [
~:}[ ~:)
(column vector form)
I Examples I
CD
a·b = !al lbl cos 40° = 6.f2. (3 sf)
;;{ ..
45°
b·c = jbj lei cos 120° =-6
c=4
3.2
a=4
the scalar and vector products of vectors
3-15
(l)
The scalar product of the two vectors [
=
[~]{:J
~J
and [:]
= (1)(4)+ (2)(5)+ (3)(6) = 32
Angle between two vectors
O The angle (), between two non-zero vectors a and b,
can be calculated from its scalar product,
a·b = lal lblcos() .
L
a
O Thus the angle () between the two vectors is given by
a·b
cos() = lallbl
Example
CD
or
() =cos
-1(
a·b)
la
l bl
I
Find the angle between a = 2i + 4 j and b = -i + 2j , giving your
answers to the nearest degree.
(!)·(;1)
a.b
cos()=--1
--;=====--r======= = -2 + 8 = 0 6
J(2)2 +(4)2J(-1)2+(2)2
J20J5 .
a II b I
:. () = cos- 1 0.6 = 53.1° (3 sf)
~
Find the angle between a = 2i + 4j-k and b = -i + 2j + 6k,
giving your answers to the nearest degree.
a.b
cos()=---
!a II b I
J(2)2 +(4)2 +(-1)2 J(-1)2 +(2) 2 +(-6)2
- -2+8-6 - 0
- ££ 1
() = cos- 0
3-16
AL eGuide Mathematics
=90° (perpendicular to each other)
C cosmic
Geometrical interpretation of the scalar product
u a·b = \al \bl case = \b\ \a\cose = \blxOM
A
= lblxp
= jbj x the length of projection of a on b
Thus, length of projection (p) of a on b,
~ = Ja·bl (where b is a
p=
1'
p =a cos e
M
b
unit vector)
•
In another words, the projection of vector a onto vector b is like a
"shadow" of a being cast onto b, i.e., p = a·b where pis the length
of the "shadow" cast and 6 is a unit vector in the direction of b.
•
It may also be called the component of a in the direction of b.
Example
I
CD Given two vectors a= 2i - j + 2k and b = 3i - 4 j- 2k , find the
length of projection of a on b.
Length of projection of a on b
=
= a·b = ja.bl
Jbj
[a[~J m
6
-t9+16+4
-
3.2
the scalar and vector products of vectors
3-17
Vector product
The vector (or cross) product of two non-zero vectors a and b (in that
order), written as ax b is defined as the vector that has a direction, as
given by the right-hand rule, that is oeroendicular to both a and b and a
magnitude given by Ila xbj = lallbl sine !, where
e is the angle between a
and b.
D
......-------~
'
''
h
' ',,
'
A
axb
C
''
a
'
''
'
bxa
O Given two vectors a " a,i + a,j + a,k or [ ::
•
O
Jand b b,i + b,j + b,k or
0
Do note that each component of the vector product does not
contain the corresponding components of the operands (the
vectors being operated upon, i.e., i: (a 2b3 - a3b2 ) does not have a 1
nor b1.
If a, b and c are non-zero vectors, the laws of vector product are:
•
Anti-Commutative:
a xb = - bxa
•
Distributive:
ax (b + c) = a x b +a x c
(a+b)xc=axc+bxc
3-18
AL eGuide Mathematics
@cosmic
0
Other basic properties:
0
A, (a x b) = A.a x b = a x A.b
6
If a and b are parallel vectors <=>
axb = 0
@
If a and b are perpendicular vectors <=>
laxbl =lal lbl
0
If a and b are equal vectors, a = b
<=>
axa = 0
I Example I
CD
Find the vector product between the vectors a = i + 2j + 3k
and b=2i-3j+k.
j
axb = 1 2
2 -3
k
3 =(11)i+(5)j+(-7)k
1
(ans)
O Some general results:
i x j=k
jxk=i
kxi=j
j x i = -k
k x j = -i
ix k = - j
ixi=O
jxj = O
kxk=O
Area of parallelogram
z
O
J
y
~-------.
If vectors a and b represent two adjacent
sides WX and WZ of a parallelogram
WXYZ, then
h
Area of parallelogram = lax bl I
W
a
Proof:
Area of parallelogram = base x height= (WX)h
=/a/lb/ sin()= la xbl =
Example
CD
lwx x wzl
I
A parallelogram has two adjacent sides a = i + 2j + 3k and
b = 2i - 3j+k.
Area= laxbl =l(i+2j+3k)(2i-3j+k)I
=ji(11) + j(5) + k(-7)j = J195 unit2
3.2
the scalar and vector products of vectors
3-19
Area of triangle
y
O If vectors a and b represent two sides WX
and WY of a triangle WXY, then
J
Area of triangle = ! [a x b[
Example
CD
J
I
w
a
x
A triangle has two sides a= i + 2j + 3k and b = 2i - 3j + k.
Area= ! [a x bJ = -tj(i+2j+3k)x(2i-3j +k)j
=! Ii(11) + H5) + k ( - 7)I =../195I2 unit2
Worked Examples
[Example i
I
Consider a triangle as shown and the scalar
product c·c = (a - b )-(a - b) , deduce the cosine
rule.
Solution:
a
c·c = [c[2 = c 2 = (a - b )·(a - b) = a·a + b·b - a·b -b·a (distributive)
2
2
2
2
2
2
= [a[ + [b[ - 2a·b = a + b -2Ja[[b[cose = a + b -2ab cose
=>
[Example
c = a + b - 2ab cos e
2
J
2
2
J
21
Given that A(2 , 3, 4 ), B(-2, 1, 0) and C(4, 0, 2), find the exact value of
sin LBAC and hence the area of the triangle ABC.
Solution:
Position vectors A, Band C
a = 2i + 3j + 4k, b = -2i + j, c = 4i + 2k
~
AB = b - a = -4i - 2j - 4k
~
AC= c - a = 2i - 3j - 2k
->
->
cos BAG= AB·AC = --;====-=8=+=6-;:+
= 8= = =
IAal lAcl ~42 + 22 + 42 ~22 + 32 + 22
3-20
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C cosmic
1
J17
4
sinBAC =
4
4
~1-cos 2 BAC = ~1- 171 = -J17
- = J17
17
Area of MBC = ±IABllAClsinBAC =
=12units
2
±J36m ~
(ans)
IExample 31
Find a unit vector perpendicular to both a = 2i - j + 2k and b = 3i - 4j - 2k.
Solution:
l
Vector ax b is J_ to both a and b
2
[3 ] [(-1)(-2)-(-4)(2)]
[10:
[2 ]
a x b = -1 x -4 = 2(3)-2(-2)
= 10 = 5 2
[
-2
(2)(-4)-(3)(-1)
-5
-1
2
A unit vector perpendicular to both a and b
2i + 2j - k
---=====
-- - (2"I + 2'J- k)
1
~22+22+12
3
3.2
the scalar and vector products of vectors
3-21
Worked Problems
·~ ample 1
I
\
Referred to the origin 0, the position vectors of the points A and Bare
i-j+2k
and
2i+4j+k
respectively.
(a) Show that OA is perpendicular to OB. [2]
(b) Find the position vector of the point M ( m =OM) on the line segment
AB such that AM : MB = 1 : 2 . [3]
(c) The point Chas position vector - 4i+2j+2k . Use a vector product to
find the exact area of triangle OAC. [4]
Solution:
(a) Taking the dot product of vectors OA and OB:
OA·OB = (i- j + 2k)·(2i+ 4j+k) = (1)(2)+ (-1)( 4)+ (2)(1) = 0
=> Vectors OA and OB are perpendicular. (QED)
(b) Using ratio theorem,
2
m = a+b = .1(2a+b)
3
2+1
=
+(2 (i - j + 2k) + ( 2i + 4j + k))
=+(4i+2j+5k)
(ans)
0
(c) Area of triangle OAC= -tlOAxOCI
J[Tl
=±i(i- j+2k)x(-4i+ 2j+2kJ! =, [
= -tli(-2- 4) + j(-8-2) + k(2- 4)1 = -tl-6i-10j- 2kl
=-t.J36+100+ 4 =-!35 units
2
(ans)
I Example 21
z
The figure shows a right
triangular prism with its
bottom base lying in the xyplane . The point P divides
DE in the ratio 2: 1. Give the
co-ordinates of A, B, C, D, E
and find the angle between
the lines OP and AB.
Solution:
3-22
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D
3
A
x
C cosmic
A= (2, 0, 0) (ans)
B = (0, 3, 0) (ans)
c = (0, 0, 3) (ans)
0 = (2, 0, 3) (ans)
E = (0, 3, 3) (ans)
Given that DP : PE
=2 : 1,
=
oP woE+oD)
=tl[:J+[~]] =+[:]
:. P=(f,2,3)
Let () be the angle between OP and AB.
-->
-->
-->
AB= OB-OA =
[-2]
~
14
11v'13
g = 69.3 °
(ans)
3.2
the scalar and vector products of vectors
3-23
Three-dimensional geometry
Content
3.3.1
Three-dimensional geometry
Learning Outcomes
Include:
(a)
Vector and Cartesian equations of lines and planes
(b)
Finding the distance from a point to a line or to a plane
(c)
Finding the angle between two lines, between a line and a plane, or between two planes
(d)
Relationships between
(i)
(ii)
(iii)
(iv)
(e)
two lines (coplanar or skew)
a line and a plane
two planes
three planes
Finding the intersection of lines and planes
Exclude
3-24
•
finding the shortest distance between two skew lines
•
finding an equation for the common perpendicular to two skew lines
AL eGuide Mathematics
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Three-dimensional geometry
Vector equation of a line
B
O A straight line is uniquely located in space
if it passes through
•
a known fixed point and has a known
direction, or
L
0
•
two known fixed points.
0
Vector equation of line L passing through the point with position
vector a and is parallel to d is
L : r = a + Jcd , Jc E JR
where
r is the position vector of any point on the line L,
a is the position vector of a fixed point on the line L, and
dis the direction vector which is parallel to the line L.
•
•
For each value of A, , the equation gives the position vector of
one point on the line.
•
The vector equation is not unique as it depends on the values
of a and d.
Vector equation of line L passing through two fixed points A and B
with position vectors a and b respectively is
L : r = OA + JcAB , Jc E JR
where
r is the position vector of any point on the line L,
OA is the position vector of a fixed point on the line L, &
AB is the direction vector which is parallel to the line L.
•
I Examples I
CD
Find the vector equation of the straight line through the point
A [-:] and parallel to the vector 2i + 3j- k .
3.3
three-dimensional geometry
3-25
~
Find the vector equation of the straight line through the points
{fland{J
2
L
r=OA+AAB
=[~}µl[~J[~ Jl=[~} µ[~J
Cartesian and parametric equation of a
line
[j
The Parametric and Cartesian form of an equation of line can be
attained by rearranging its Vector form.
•
J
Consider the line with equation r = a + Ad . If a = ( :: and
d = (:+hen we have
Vector form r = ( ::
J+ A (
:J ~
AE
x = a1 + A-d1
Parametric form : y = a2 + A-d2
z = a3 + A-d3
•
If we arrange the three equations by making }., the subject, we get
x-a
y-a
z-a
Cartesian form : (;., =) - -1 = __2 = _ _3
d1
dz
d3
3-26
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•
If any one of the numbers d1 , d2 and d3 is zero, the cartesian form
becomes:
y-a
z-a
_
x- a
z-a
_
x -a y-a
d,
d2
0
2
3
If d 1 = 0 x = a1 - = -or
'
'
d2
d3 '
6
1
3
If d2 - 0 y = a2 - =- and
'
' d1
d3 '
@
lfd3-0, - - 1, - - 2, Z=83.
Example
CD
I
Express r = (2i-3j+k)+A.(i-k), A. E IR in parametric and
Cartesian form .
Vector form r
Let r =
[;IzJ
~ ~3)+ •[:]
[
=> Parametric form : ; :
~;A.
Z=1-4A.
Rearranging the parametric form making A. the subject,
1-z
Cartesian form: x- 2 = - - , y = -3
4
3.3
three-dimens ional geometry
3-27
Equation of a plane
n
oc
O A plane can be uniquely specified in
A
several ways:
0
By using two distinct concurrent
lines (such as AB and BC),
············.•
0
,' ~...... ~./~
··.,,/a:f
a·•
·'l; •..
I
-
_-....-r
6
By using three non-collinear points
in space (such as A, B, C),
@
By a direction that perpendicular to the plane (such as normal n)
and the perpendicular distance from the origin (such as d) or a
known reference point, or
0
By using a given point on the plane (such as A) and a direction
that is normal to the same plane (such as n).
Vector equations of a plane
p)
0: A plane n, which contains two distinct concurrent lines AB (AB=
and BC (BC=
q).
v._.
A
C
B__.,.- ,'r
0
II :
r = a + ,1,p + µq ,
,1,, µ E JR
where
r is the position vector of any point on the plane,
a is also a position vector of a point on the plane, and
p, q are two distinct direction vectors on the plane.
Example
I
CD Given two concurrent lines with equations
'-i:
r=(2i+j)+s(i - j+3k)
~ :
r=(i+j)+t(2i - j+2k) where s,tElR
The two planar direction vectors parallel to the plane are [
3-28
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C co•mic
A point on the plane is
m.
The vector equation of the plane:
@:
A plane II, which contains three distinct non-collinear points A, Band
C, whose position vectors are a, b and c respectively.
A c-a C
.. ······.:t-
•:.:···
b-a\ ../
~-
,,'ff'
,,•
0
n: r =a+ A. (b - a)+µ ( c - a),
A.,µ E JR
where
r is the position vector of any point on the plane,
a is also a position vector of a point on the plane, and
(b - a), ( c - a) are two distinct direction vectors on the
plane.
Example
I
CD Given three non-collinear points A(O, 1, 1), 8(2, 1, 2), and
C(1, 1, 2).
Taking a as
..
[rJ,
the two planar direction vectors are:
The vector equation of the plane:
3.3
three-dimensional geometry
3 -29
n
@+O : A plane II, which contains the
point A with position vector a and
which is normal to a non-zero vector
n, can be specified by a vector
equation relating the position vector r
of any point on the plane to n and a:
n
0
r n = a·n =D
ll :
0
where
r is the position vector of any point on the plane
a is also a position vector of a point on the plane
n is the vector perpendicular to the plane
D is given by a·n
=lnl x length of projection of a onto n in
the direction of n.
•
The vector equation of the plane is not unique as n can be any
vector normal to II.
•
A plane II with a vector equation
n : r·n = D has a unique
In : Ml d I
D
where n = R and d = R
standard form
n
=
A
•
dis the length of projection of r onto n in the direction of n
•
ldl is the perpendicular distance of the plane from the origin 0 .
Example
I
CD Given two concurrent lines with equations
~:
r=(2i+j) +s(i- j+3k)
~:
r =(i+j)+t(2i - j+2k) where s,tElR
The two planar direction vectors parallel to the plane are
A position vector that is on the plane, (
3-30
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~) ~
( a)
Cl cosmic
Deduce the perpendicular vector using n = p x q = [
..
The vector equation of the plane:
~1] ~1]
x[
r·(i - j + k) == 1
Cartesian equation of a plane
O The Cartesian equation of the plane is given by:
kn1x + kn2 y + kn3 z == D
where kn, , kn, and kn, are the direction ratios of n = [
~:]
• the
normal to the plane
•
The cartesian form of the equation of plane can be derived from its
vector form, i.e.,
n
r.n
=D, given that r =[~] and n =[~J
~ =[~JG]= n,x+n,y+n,z =D
r·n
(QED)
I Examples I
<D Find a vector and Cartesian equation of the two planes which
are at right angles to the vector 4i- 4j + 7k and which are at
unit distance from the origin.
Let the normal, n == 4i- 4 j + 7k
=>
~ == ~4 + (~4 ) +
2
2
72
(
4i - 4 j + ?k) =
H4i - 4j + ?k)
Vector equation of the 2 required planes are
r·n == ±1
Cartesian equation: 4x- 4y + 7z == ±9
3.3
three-dimensional geometry
3-31
~
Find the equations of plane for x = 0 , y = 0 and z = 0.
Deduce all the planes' distances to origin are zero.
Deduce the unit normals are i, j and k respectively.
=> r·i = 0, r·j = 0, r·k = 0 respectively.
Distance of a point from a line
CJ Consider a point P (position vector p) and a line
with vector equation L : r = a+ A-b , A- E JR .
p
\~
~
L
CJ Deduce that the shortest distanced from the point
P to the line L is PQ, where Q is the foot of
perpendicular drawn ("drop") from P to L.
CJ Some common methods in finding the perpendicular distance, d:
O
Deduce that since PQ is perpendicular to the line L, then
PQ·b = 0. The coordinates of Qis then found. The perpendicular
distance d = IPQI.
6
Select a (convenient) point A on the line L.
p
Find the angle e between AP and b, using
-··~·\_.e·_ _·~ --~--
IAPllblcose.
The perpendicular distance d = IAPI sine.
the relationship, AP·b =
@
. . _. . ....! _I _d
A
Q
L
b
Select a (convenient) point A on the line L. Find the er endicular
_
distance d using the relationship, d = IAPj sine =
Example
IAP xbl
lbl .
I
L
Let Q be the foot of perpendicular from P to L.
3-32
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Method 0:
Since
OQ =[ ~:Al
1
Q
is on the line L,
Since PQ1-L
=>
for some A E JR .
PQ{~l=O,
"" (oo-oPJW=
""
rPtifJ1Hl=
=0
,,,
0
, , w~n-uJirn
0
16A-4+9A-2h 0
=>A-=1
.·. oo =m
Hence.
Po=[~J
Shortest distance from P to L is
IPQI = ~(3) + (-4 ) + (3)
2
2
2
= ,,/34 units.
Method 6:
Ta king A as 0, a (convenient) point on L will be A [
AP
l
~
=OP-OA=[~J-m=[ _:J.
Let e be the angle between AP and L.
Consider
AP·b = IAPIJbJ case,
cosB=
sin B
=~1-
mm
~1 + 7 + (- 3) • ,J4 + 3
2
2
cos B
2
2
2
r
4+21
5../59
5
=--=-
.J59
=~1 -( }sg = ~: =~
Perpendicular distance, d
J1 -
= jA"Pj sing= ~(1 2 + 7 2 + 3 2 )~
= ,,/34 units
3.3
three-dimensional geometry
3-33
Method@:
Ta king A as O, a (convenient) point on L will be A [
AP = OP - OA =
~J.
l~J-m l
= [ _:
Perpendicular distance from P to L:
IAP xbl
~
d =
IAPI sin() = lbl
=
UJl~J
~42 +32
_19 + (-12)+ (-25)1 _ .Jg'+ 12' + 25'
-
5
-
5
.Jaso
r;::;;
.
=- = v34
units
5
Distance of a point from a plane
O Consider a point P (position vector
p) and a plane with vector equation
n : r·n = a·n D => dn =
(where dn is the perpendicular
=
n
r----
r·n
,/
p
t
----------~
·-·---··········· llp ,'/
I
I
I
I
I
I
distance of the plane n from the
origin).
I
d
I
I
I
O Deduce that the shortest distance d
from the point p to the plane n is
PQ, where Q is the foot of
perpendicular drawn ("drop") from P
ton.
O
Some common methods in finding the perpendicular distance, d:
0
Deduce that the line PQ is parallel to n, i.e., PQ = ,1,n . Since the
point Q is on the plane, it must also satisfies the planar equation,
i.e., OQ·n = D. The coordinates of Q is then found. The
perpendicular distance d =
@
IPQI.
Construct a similar plane that is parallel to the plane fl and
= dP ( dP is the perpendicular
contains the point P, i.e., nP :
p·n
distance of the plane nP from the origin). The required
perpendicular distance d = \dn
3-34
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-dP\.
C cosmic
@
p
Choose a (convenient) point Ron
the plane . The projection of the line
PR on the unit normal
of the
plane II gives the perpendicular
n
distanced= IPR·nl (=PR cos¢).
0
Example
CD
I
Find the distance d from P( 1, 1, 3) to the plane
3x+2y+6z = 6 .
The vector equation of the plane is fl : r{
normal n=
l~I =+[~),and
~)
= 6 , the unit
let Q be the foot of perpendicular
on the plane from the point P, with position vector OQ = [
~:J
.
Method 0:
Deduce that the line PQ is parallel ton, i.e., PQ = A-n
~
PQ = OQ-OP =
00-m
dn = A [
~) m)
=
Since the point Q is on the plane , it must also satisfies the
planar equation, i.e., OQ·n = 0.
The perpendicular distance, d =
IPQI = - li [ ~) = I- :i x 7j = ¥
3.3
three-dimensional geometry
J-35
Method 6:
Construct a similar plane that is parallel to the plane II and
contains the point P, i.e., IIP : p·n = dP.
===>
n p ·.
[~ ]·.i[~] = n. =
7
3
n
7
dp
6
•HJ=6 ~ n
r·t[~]=f=d,,
The perpendicular distance, d
=ldn - dP I= It- /I =-¥2
Method@:
Choose a point Ron the plane to be (0, 0, 1) (take x =0, y =
0, then z = 1).
The perpendicular distance d
=[PR·ri[ =n+[~] =1-¥-1 =!f
= [
~
Angle between two lines
{j
The angle Ba (or ()0 ) between two lines L,
and ~ is determined by the angle
between their direction vectors, d1 and d 2
respectively.
O Consider two lines whose equations are:
L, : r = a 1 + A-d 1
Li : r = a2 + A,d2
Let B be the angle between the two lines
d1·d2
cos () -- ldJjCiJ
•
If d1 ·d 2 = 0 , the 2 lines are perpendicular,
•
If d 1 ·d 2 =
•
Depending on the directions of the direction vectors, the angle (),
between the two lines may be acute ea or obtuse e0 • The angles
or
() -- cos -1 ( 1dJjCiJ
d1·d2
J
:::::>
ldl d I or d = kd the 2 lines are perpendicular, and
1
2
1
2 ,
are related by IBa + B0 = 180° I·
3-36
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I Examples l
CD
Find the angle between the lines ~ :
and ~ :
x-4
2
z-7
2
- - = y- 3 = - -
x + 1 = y = z + 1.
3
2
6
The direction vectors are ( 2i + j + 2k) and ( 3i + 2j + 6k)
respectively.
Let () be the angle between the lines,
cosB=
mm
.J22 +f +22.J32 +22 + 52
=~
21
1
()=cos - (20)
21' =17.8°(3sf) (ans)
~
Showthatthelines ~:
~ :
r=(i+j)+p(j+k) and
r = (j + k) + q (i + k), where p and q are parameters, do
not intersect. Determine the obtuse angle between the two
lines.
By inspection, the direction vectors, (j + k)
* c (i +k) , where c
is a multiple, i.e., not parallel.
If L1 and L2 intersect, then [ 1:
p] ~
= [
1
J
for some p, q e IR
=> then, q = 1 and p = 0, but p * 1+q, i.e., not intersecting.
The lines L1 and L2 are not parallel and do not intersect.
They are skew lines . (QED)
Let () be the angle between the two lines,
The obtuse angle between the two lines is 120°. (ans)
3.3
three-dimensional geometry
3-37
Angle between a line and a plane
(j
r = a+ A,d ,
r·n = D is given
The angle between a line L:
A, E JR and a plane II:
by
Isine = d·n I or [e = sin- d·n) [
1
(
· .
d·n d·n d·n
ldlfrlf
= 1"I = ldf .
•
Also, d·n =
•
The line and the plane are parallel if g = 0, i.e.,
d·n = 0 or
d·n = 0.
•
If only acute angle is considered, then sineacute =
Example
ld·nl .
I
CD The angle g between a line L:
a plane II :
r=(2i+4j)+A-(2i-j+k) and
r·(i + 2j + 2k) = 4 is given by
[~ ] m =~
1
sinB=
•
·h +1+1 .J1+2 +2
2
2
2
2(3)
g = 19.5° (3 sf)
t'J For a line to lie on a plane, it is suffice to show any one of the following
criteria:
0
(a+ A.d)·n = D for all A, E JR
6
Any two distinct points on the line lie on the plane
@
The line is parallel to the plane and a point on the line is on the
plane
©
Similarly, if any one criterion is not met, then the line does not lie on the
plane.
Example
I
CD Consideraline L:
II :
0
r =(i+2j+2k)+A-(2i-j-k) and a plane
r·(i + j + k) = 5 .
For (a+ A-d)·n = D (in its usual notations),
LHS = ((i + 2j + 2k) + A, ( 2i - j- k) )·(i + j + k) = 1+2 + 2 = 5
RHS =5
(a+ A.d)·n = D is true, the line lies on the plane.
3-38
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6
Choose a convenient point on the line, A (i + 2j + 2k),
when A.= 0. Then, (i+2j+2k)·(i+ j+k) = 1+2+2 = 5 .
Thus, point A lies on the plane .
Choose another convenient point on the line, B(3i + j + k),
when A.=1. Then, (3i+j+k)·(i+j+k)=3+1+1=5 .
Thus, point B lies on the plane.
Since two distinct points on the line lie on the plane, the
line lies on the plane.
@
Consider d·n = ( 2i - j - k)·(i + j + k) = 2 -1- 2 = 0 .
Thus, the line and the plane are parallel.
Choose a convenient point on the line, A (i + 2j + 2k),
when A.= 0. Then, (i+2j+2k)·(i + j+k) = 1+2+2 = 5.
Thus, point A lies on the plane.
Since the line is parallel to the plane and a point on the
line is on the plane, the line lies on the plane.
Angle between two planes
6
The angle between two planes, fli : r•n 1 = 0 1 and II2 :
given by
icos t9 =1\•n 2 I or le=cos- (n 1°n 2
1
r·n 2 = 0 2 , is
)i
•
•
The two planes are parallel if t9 = 0 .
•
If only acute angle is considered, then t9acute = cos- (Jn 1 •1\J) .
1
Example
I
CD The angle e between the planes fli :
II2 :
r·(i + j + k) = 3 and
r·(i + 2j + 2k) = 4 is given by
case~ rn . m ~-5
.J1+1+1 .J1+2 2 + 22
3../3
t9 = 15.8° (3 sf)
3.3
three-dimensional geometry
3-39
Relationships between two lines
(J
In general, two lines in space (3-dimensional) can
be classified into four groups:
Line 1
fa
0
The lines are not parallel and intersect
(intersecting lines)
f)
The lines are parallel and intersecting at every
point (identical lines)
@
The lines are distinct and parallel (non-intersecting lines)
0
The lines do not intersect and are not parallel (skew lines)
2
identical lines
14-__:_:_.:..:..c:..-=.=-"----:__o_-..i intersecting
nonintersecting
intersecting
lines
lines
skew lines
non-intersecting.._____ _ _ __., non-parallel
©
In a 2-dimensional plane, unless the lines are parallel to each other, a pair of
lines will a lways intersect.
o
When two distinct lines intersect, the point of intersection must have a
unique position vector that satisfies both the equations of the line .
Consider two lines whose equations are:
•
Li : r = a + A.d 1 , A. E IR
1
L;z : r = a 2 + µd 2 , µ E IR
Let: a1 + A.d 1 = a 2 + µd 2 .
CD
If a solution can be found for equation CD, the two lines intersect.
o Skew lines and distinct parallel lines do not intersect while identical
lines intersect at every point of both lines.
o For two lines to be parallel, their direction vectors must be multiples of
each other. i.e., d1 = kd 2 .
I Examples I
<D Find the intersection point of the two lines L, : r = (
AE
3-40
AL eGuide Mathematics
~] + ~
A(
J
Rand~: r=[~]+µm µER.
C cosmic
•
For the lines to intersect, we try to equate
[~)+A[:]=(~}µ[!}
1+4A- = 5
CD
2 + 5,1, = 6 + µ
@
3 + 6,1, = 7 + 2µ
@
CD: A-= 1
@:
µ =1
)
To confirm,@:
3+6(1)=7+2(1)
=> The lines intersect at (5, 7, 9).
@
Determine whether the lines L,:
[~]+ µ[~), µ
r =[
~1)+ ~1l A
A[
e R and
L,:
r=
•
For the lines to be parallel, their direction vectors must be
e R are parallel, intersecting or skew.
~J ~
a multiple of each other, i.e., [ 1 = k [ )- Si nee no value
for k can be found, the two lines are not parallel.
•
For the lines to intersect, we try to equate
[J{+[~J+µm
1+A-=2 + 2µ
CD
-1-A-=4+µ
@
3+.-1, = 3µ
@
CD+@:
0=6+3µ
=>
µ=-2
Substitute µ = -2 into CD:
A-= -3
Substitute µ = -2 and A-= -3 into@:
0
* -6
There is no unique value for A- and µ , thus the lines do
not intersect.
•
Since the lines are not parallel and they do not intersect,
they are skew lines.
3.3
three-dim ensional geometry
3-41
Relationships between a line and a plane
Intersection
O If a line and a plane are not parallel, then they will intersect at a point.
The point of intersection is given by
I Place the Equation of line i!J.!Q the Equation of plane
Example
CD
I
Find the point of intersection of a line
L: r = (1+2,t)i + (2-3,.t )j + (-3-5,.t )k and a plane
r = (j + k) +a (i - 2k) + ,B (i + 2j + k).
17:
l[
At the point of intersection between line and plane,
a + ,8
1+2/3
[
1+ 2,.1, ]
2-3,.1,
=
1- 2a + /3
-3 - 5,.1,
Solving,
we have a = 4, f3 = -1, ,t = 1
:.
Common point is (3, -1, -8).
Projection
O The length of projection p of vector AB or a
(or line segment AB) onto a plane
17 : r·n = 0 is given by
OR
Alternatively, first find the height of the projection,
h = jA"Bj cos¢ =
1H1 = la·nl
Using Pythagoras' theorem,
p2 = lal2 - h2
3-42
AL eGu/de Mathematics
=>
Ip +~jaj2 - h2 I fl
=
C co• mic
I Example I
<J)
Find the length of projection p of a
line segment AB or a on a plane
II: r·(2i+ j+k) = 0 and the points
A(0,0,0) and B(1,2,1).
AB " OB-OA " a " [
l
~J-m "[~
and lal=~
""J.m
Method 0:
--+
The length of projection p of AB on II is given by
Method@:
The height of the projection is given by
The length of projection p of AB on II is given by
p
=+~lal 2 -h =+~( ~)2 -(Js)2 =1: units
2
3.3
three-dimensional geometry
3-43
Relationships between two planes
Intersection
D Consider two planes, IL, : r•n 1 = 0 1 and II2 :
r•n 2 = 0 2 .
D If two planes are not parallel, they will intersect at a line
L:
•
r = a+A-d = (8)+82j+83 k)+ A- (d1i+d2j+d3k), A- E JR
As the line of intersection of two planes is contained in both
and
planes, it is perpendicular to both
n1
n2.
D The equation of the line of intersection can be found by:
•
Converting the equations of
plane, from vector form to
Cartesian form,
•
II1
line of
Combining the two formed
Cartesian equations in
terms of only two of its
variables, such as x = f(y)
--+.--4-.------'C.:::...L.....J.<--- intersection,
L
and x = f(z) .
.
y-82
and
1.e., x = - -
z-8
0
X=--3
d2
d3
Thus, a Cartesian form of the equation of line of intersection L is
formed:
x-0
y-8
z-8
1
d2
d3
- - = - -2 = - -3
•
(provided d 2 , d 3
* 0 ), and
Its vector form is then given by
L:
r = (82j + 83k) +A- (i + d 2j + d 3 k) , A- E JR
OR
D Alternatively, the equation of the line of intersection can be fou'nd by:
Id= n x n I.
•
First finding the direction vector of the intersecting line,
•
Choose a convenient point that lie on both planes, such as A- = 0
and let 8 1 = 0. From the two equations of plane, form the pair of
1
2
simultaneous equations in terms of its remaining unknowns,
8 2 , 8 3 and solving them:
i.e., l82n1.2 +83n1,3 = 011 and l82n2.2 +83n2,3 = 021
•
6
The vector equation of line of intersection L is then given by
L : r = (82j + 83k) +A- ( d1i + d 2j + d 3k) , A- E JR
3-44
AL eGuide Mathematics
e cosmic
I
Example
CD
Find the equation of the line of intersection of the two planes,
II,: r·(i+ j - 3k) = 6 and Il2 : r·(2i-j+k) = 4 .
II, : r·n 1 =r·(i + j- 3k) = 6
=>
x+ y-3z = 6
0
II,_ : r·n 2 = r·( 2i - j + k) = 4
=> 2x- y+z = 4
6
Method 0:
0+6:
3x-2z = 10 =>
X=
2z+10
3
0+6x 3:
7x - 2y = 18 =>
X=
2y+18
(eliminate z)
7
:. L:
(eliminate y)
x = 2y + 18 = 2z + 10
7
3
:. The equation of line for the intersection of the planes is
L:
r=(-9j-5k)+,t(i+fj+%k), AElR
Method 6:
If two planes are not parallel, they will intersect at a line
L:
r = a+ ,td = ( a1i + a2 j + a3 k) + ,t ( d1i + d2 j + d3 k), ,t E JR
•
Direction vector of line L:
d = n1 x n2
•
Position vector of line L:
a = (Oi + a2 j + a3k), then:
0: (a 2 j+a3 k)·(i+j-3k)=6
6: ( a2 j + a3 k )·( 2i - j + k) = 4
:. The equation of line for the intersection of the planes is
L:
r=(- 9j-5k)+,t(-2i - 7j-3k), AElR
3.3
three-dimensional geometry
3-45
Relationships between three planes
O Three planes in space may intersect as follows:
•
The planes may coincide with each
other, as a single plane or double
planes, or are distinctly separated,
with no intersection.
~--~---=­
~--~-~
2 planes are parallel and the 3rd is not.
f)
@
parallel
planes
3 planes are parallel to each other.
0
•
The 2 parallel planes may coincide
with each other forming a single
plane or are distinctly separated .
•
The 3rd plane intersects each of
the parallel planes in a line. The
intersection lines are parallel.
intersection
lines
No planes are parallel.
•
The planes should normally intersect in 3 separate parallel
lines. However, these intersection lines may also coincide as a
single line as shown.
intersection
line
intersection
lines
•
If none of the intersection lines are parallel and the lines of
intersection will coincide at one point.
point of
'-\..:::-1-__:~c--~t\::::::--- intersection
Example
<D
I
Find the intersection point p = (xi+ yj + zk) of the 3 planes:
IIi:
r·(i+j+k)=4, ll2 :
ll3 :
r·(i-j+k) = 0
r·(2i+j-k)=3 and
(xi+ yj + zk )·(i + j + k) = 4
=>
x+y +z = 4
(xi + yj + zk )-( 2i + j - k) = 3
=>
2x + y - z = 3
(xi+ yj + zk )·(i - j + k) = 0
=>
x- y +z= 0
Solving,weget x=1 , y=2, z=1
3-46
AL eGuide Mathematics
=> p=(i+2j+k)
Cl cos mic
Worked Examples
I Example 11
A line has equation x 3
5
= 1 - y = 2z + 4 .
(a) Find a vector equation of the line.
(b) Determine whether the point (-1, 3, -3) lies on the line.
Solution:
x-5
x-5 y-1 z+2
(a) - - = 1 -y=2z+4 <=:>--=--=--
3
3
-1
1
Direction vector: 3i - j + 1 k
The line passes through the point (5, 1, - 2)
Vector equation of the line is
r=5i+j-2k+ A,(3i-j+1k), AER
(ans)
(b) Consider the points (-1, 3, -3),
Substitute x = -1, y = 3, z = -3
x-5
:. - - = -2 , 1 - y = -2 , 2z + 4 = -2
3
Therefore the equation is satisfied, which makes the point lie on the
line. (ans)
IExample 21
Find the foot of the perpendicular from the origin to the line L with equation
r = i - 4j - 3k + A, (i + 4j + 2k), A, E R
Deduce the perpendicular distance from the origin to the line.
Solution:
Let N be the foot of .l from 0 to L.
Nlies on Lo;
Now
oN {~]+Am for some A e R
oN ~Lo>oNGJ~o
:.[-~:~,i]·[:] =O =>
-3+2,1,
1+A,-16+16A,-6+4A=0
=>
A=1
2
3.3
three-dimensional geometry
3-47
:.l. distance from origin to L =ON= ~2
2
+1 2 =JS units
(ans)
IExample 3j
Given that
oA =[~1]. oB =[:]. oC =[~+how
that the vector
[:;a]
is perpendicular to the plane ABC.
Solution:
[
~ ~o}AB= [~ ~oJG) =
(-7j1) + (-1 D)(2) + (9)(3)
=o
[~~+AB
Hence
3-48
(:;a]
AL eGuide Mathematics
is perpendicular to the plane ABC. (QED) (ans)
C cosmic
I Example 41
Find an equation of the plane passing through the points A(3, 1, -1) and
8(4, 2, 0) and containing the line with equation
r =4i + 2j + A. (2i - j + 3k)
Find also a unit vector normal to the plane.
Solution:
The two vectors parallel to plane are
AB=[~] and [~ 1]
Therefore the vector equation of the plane is
r
=[~}Am+ µ[~1}
A,µ ER
(ans)
~H~ 1] =[~ ~]
A vector normal to the plane =[
:. Unit vector normal to plane =
1 [~11
~42 + (-1)2 +(3)2
1
= - -(4i -j- 3k)
ffe
-3
(ans)
3.3
three-dimensional geometry
3-49
Worked Problems
IExample i I
The line R passes through the points A and B with coordinates (1, 2, 4)
and (-2, 3, 1) respectively. The plane p has equation 3x - y + 2z = 17 .
Find
(a) the coordinates of the point of intersection of R and p,
(b) the acute angle between f and p, [3]
[5]
(c) the perpendicular distance from Atop. [3]
Solution:
(a) R:
p:
r=(i+2j+4k)+A,(AB)
=> r=(i+2j+4k)+A,(-3i+j-3k)
r·( 3i - j + 2k) = 17
fl
0
For the line and plane to intersect (put 0 into @),
p:
( ( 1- 3A,) i + ( 2 +A,) j + ( 4 - 3A,) k )·( 3i - j + 2k) = 17
3 ( 1- 3A, )- ( 2 +A,)+ 2 ( 4 - 3A,) = 17
=>
9 -16A, = 17
The position vector for the point of intersection is
r = (i +2j+ 4k)-·t(-3i + j-3k) =(t i+! j+11k)
(ans)
(b) The acute angle between f and p is given by:
-16
=~~=
J19.J14
= -78.8° (3sf) (ans)
(c) Choose a convenient point Con the plane to be (0, 0, ¥) (take x =0,
y= 0, then z = 1).
The perpendicular distance from A top is
3-50
AL eGuide Mathematics
Cl cosmic
I Example 21
The variable point U which is in the plane fl has position vector
(-p + 2q + 1)i + (11p-4q + 1)j + (5p- 2q + 2)k, where p,q E IR
Find an equation of fl in the form r =a + a b + j3 c, where b and c are
perpendicular vectors.
Solution:
fl: r =
l
[~~: !~: ~i [~ p[~1 ] q[~ 4]
=
5p - 2q + 2
Leta=
m
and b =
1
+
2
+
5
p, q ER
- 2
~[~~] {~;
[~ ~} ~1 ] ~ [~
1
A normal vector of the plane is n =
[
4:
A vector parallel to the plane and perpendicular to b is:
-~bx -(~H~+ffJ
n=
=c
Since fl is the plane which passes through the point with the position
vector a and is parallel to b and c, its equation is:
r=a+ ab+ jJc
3.3
three~dimensiona l geometry
3-51
IExample 3j
The line L has equation r
=[~}Am
(a) The point A on Lis such that OA is perpendicular to L, where 0 is the
origin. Find the position vector of A.
(b) Find the position vectors of the two points on L such that
I\
cosPOA =
1
J7 .
Solution:
(a) A lies on L
=> OA [ -5 +1+2A]
4A
for some A R
=
E
3+ A
=> [ ~::A .~ j = 0
- 1+2A][21
OA _L L
Solving,
A= -1
Therefore
(b) PliesonL
=>OP=[~::;]
for some µER
3 +µ
I\
cosPOA =
--+
=>
1
J7
--+
OP.OA
[~ ;]·[~ 3:
: :
-
3+µ
2
1
- ~
!o--P!! oAI - ~(-1+2µ)2 + (5 + 4µ)2 + (3 + µ)2 ffe - J7
=> µ =- 3 or 1
Therefore the position vectors of the required points are:
[!]or[~~] (ans)
3-52
AL eGulde Mathematics
C cosmic
I Example 41
(a) Given A(1, 1, 1), B(1, -1, 3), C(2, -1, 0), find the perpendicular
distance of A from the line through Band C.
(b) Find the vector equation of the line through A(2, -1, 5) which is
perpendicular to and intersects the line I whose equation is
1
1
x - 3 = Y - = z + . Deduce the shortest distance of A from the line
2
2
I.
Solution:
(a) Vector equation of the straight line OA:
'"[}{]
aC
µER
"[~ 3]
Vector equation of the straight line BC:
Let X be the point on the line BC such that AX .l BC
=> OX =
1+A.l
-1
for some A. R
E
[3-34
1
l
xA =oA-ox = [~]-[ ~1 =[ -2 ]
1
Since
4
3 - 34
- 2 + 34
XA _!_BC , XA ·BC = 0
4
i.e, [
4
-2
- 2 + 34
][~]-o
===>
-4(1)+2(0)+(-2+34)(-3)=0
3 -
Distance between A and BC= l~
xAI = (- 3) + 22 + ( - 51)
2
2
5
fi10
. (ans)
=- units
5
3.3
three-dimensional geometry
3-53
(b) L : r = 3i + j - k + A, (i + 2j + 2k)
for A, ER
Let the foot of perpendicular of the line through A to L be N
oN{}{~J
for some AER
AN is perpendicular to L: ANHJ~ 0
+ A, - +21
(ON-OA} [ 12J= 0 ~ [31+2A,
. [1]
2 =0
-1 + 2A, - 5 2
2
i.e.,
l
AL eGuide Mathematics
l
~ [A,2A,++1 2 . [12] = 0
2A, - 6
2
[.lj- ]
A,++12 = 32
AN= 2A9
[
2A- - 6
- ~o
The shortest distance is
3-54
l
/AN/ =8 ~ units (ans)
Cl cosmic