Vectors JS 31 Solutions
Exercise Page 611
1.
Sum of two unit vectors is a unit vector is possible
in an equilateral triangle ABC of side unity
Where side BC , CA and AB
are represented by vectors a , b and c respectively .
We have aˆ  1, bˆ  1, aˆ  bˆ  1
we have to find a  b .
Produce AC to a point A such that
CA   bˆ then BA  aˆ  (bˆ)  aˆ  bˆ
Also BCA  120
cos120 

12  12  BA2
2  1 1
BA  3
Thus aˆ  bˆ  3
2.
Use the result that if a and b are non collinear vectors such that  a   b  0
Then   0,   0
3.
3λ
AC  AB
λAE  AB 4
BP λ
 . AP 

Let
PE 1
λ 1
λ 1
 2 AC  AB 
 . We easily get  = 8 : 3.
 v 1 
But AP  vAD  v 
4.
Let A be origin AB  a , AD  b . Let DE : EL   :1
Now AE 
 AL  AD
2

a
b
 2
.
 1

Also AE   AC   a  b

1
On equating two representations of AE we get
/2
1
 ,

 1
 1
on dividing we get

2
1
   2 whence  
1
3
which proves our assertion .
5.
The equations of two planes can be written as
r  (1  1 )a  1b  1 c ,
r  (2  2 )a  (22  2)b  2 c
If the two planes intersect then for the line of intersection we must have
1  1  2  2
, 1  22  2 ,
1  2
Now 1  1 , 2  1 , 1  21  2
2  1  1  21  2  1  1  21  2
Thus position vector of points on the line of intersection of two planes
satisfy ( on putting all parameters in terms of 1 in the first equation )
r   31  2  a  (21  2)b  1c
For a particular point on the line set 1  0
Thus one of the points on both planes or line of intersection has position vector 2a  2b
To write the equation of the line in the parametric form r  a  b we need to find a vector to
which the line is parallel to . For which we can subtract two position vectors on line
Indeed subtract r values for 1  l and 1  m we get a vector parallel to required line
as 3(l  m)a  2(l  m)b  (l  m)c or line is parallel to the vector 3a  2b  c .

Hence the equation of required line is r  2a  2b  k 3a  2b  c

( In the above working it has been assumed that a , b , c are non coplanar )
6.
It is clear that X lies on produced OA while Y lies internally on OB .
Let O origin . OA  a , OB  b
Let
AP
XP
k ,
m
PB
PY
then OP 
kb  a
k 1
( from OAB )
2
Also OP 
mOY  OX
m 1
2
m b  2a
 3
m 1
( from OAY )
2
m
k
Since a and b are non collinear we easily get
 3
,
k 1 m 1
On dividing last two relations we get k 
On putting k 
Whence
m
3
m
1
2

in second equation we get
m
3
1 m 1
3
 m3
1
2 1
   k 1
k 1 4 2
 P is mid of AB with
8.
1
2

k 1 m 1
XP
 3 :1
PY
Give PQ  AP  PB  PC
 PC  CQ  AB  BP  PB  PC
 CQ  AB

(*)
ABQC is a parallelogram and Q
therefore a fixed point since relation (*) does not depend up to position of P .
9.
In the figure F is mid-point of BD
and E is the mid-point of AC
We have AB  AD  2 AF ,
And CB  CD  2 CF ;
 AB  AD  CD  2 ( AF  CF )
  2 ( FA  FC )  2 (2 FE )
  4 FE  4 EF .
10.
We may take three face diagonals as OR , OT and OS ( see figure )
Take iˆ , ˆj , kˆ along
OA , OB and OC in the figure
Now unit vector along OR 
jk
.
2
3
 vector of magnitude a along OR 
a ˆ ˆ
( j  k)
2
Similarly other two vectors described in the
are
question
2a ˆ ˆ
3a ˆ ˆ
(k  i ) and
(i  j )
2
2
 Vector sum 
 R
5a ˆ 4a ˆ 3a ˆ
i
j
k  R (say)
2
2
2
25a 2 16a 2 9a 2


 5a
2
2
2
5a 4a 3a
1
4
3
The direction cosines of vector R are 2 , 2 , 2 or are
.
,
,
5a 5a 5a
2 5 2 5 2
Finally it is easy to observe that OR  OS  OT  2(iˆ  ˆj  kˆ)  2OP
11.
We have OC 
OM
  n  1 OC  OC  CM
n 1
OC 1

CM n
 nOC  CM 
Let A be origin . AO  a , AM  b then AC 
Also OB 
1b  na
n 1
1
1
b  AB  a  b
n
n
 CB  position vector of B - position vector of C
1
b  na
 a b
n
n 1

b  na
 AC
n 1
n 
1 

1
 1 
a  
b
 n 1 
 n n 1 

a
b

n  1 n(n  1)
 A, B, C lie on a line.
Exercise Page 621
1.
aˆ  bˆ  aˆ  bˆ  1
(Given )
2
Now aˆ  bˆ  1
 (aˆ  bˆ).(aˆ  bˆ)  1
 1  1  2aˆ. bˆ  1
4
 aˆ.b  
1
2
2


Now aˆ  bˆ  aˆ  bˆ . aˆ  bˆ
3.

 1  1  2aˆ.bˆ  3  aˆ  bˆ  3.
2
A B C  A B C  A B C  A B C
2
 A. A  B.B  C.C  2 A.B  2 B.C  2 A.C  A. A  B.B  C.C  2 A.B  2 B.C  2 A.C
From this equation B.C can be determined since all other terms are known.
and AC
. gets cancelled after which angle  between B and C is given by
cos  
4.
B.C
B C
etc.
Correction : C  B  A  B
B  1 , A B  5 , C  B  A  B
(i)
Taking dot product of equation (ii) with A
We get AC
.   A.B
(ii)

A.( B  A)  0

whence (i) is proved
Again taking dot product of (i) with B we get B.C  B.B  1
(ii)
 Angle between B and C is obtuse.
2
2
Finally C  B  A  B  2B.( B  A)
5.
6 .

Let A(-1, 3), B(2, 5) and C(, ) be the vertices and H(1, 2) be the orthocenter of the triangle
ABC then.
5
7
as  ,
7.
C 
Take the required vector r as λ a  b then determine  by using
 λa  b    2iˆ  ˆj  kˆ   8  0 .
6.

 5  1 +zero  6 
β2
3

α 1
2
( HC  AB) and
β 3
1

α 1
3
( BH  AC ) . We get C(, )
17 
.
7 
Take a  a1iˆ  a1 ˆj  a3kˆ then b  4i  k  (a1iˆ  a2 ˆj  a3kˆ)  (4  a1 )iˆ  a2 ˆj  (1  a3 )kˆ
Now a.b  1 , a  b
 2iˆ  ˆj  8kˆ will imply
a1 (4  a1 )  a2 2  a3 (1  a3 )  1 (i) and
5
i
j
k
a1
a2
a3  2iˆ  ˆj  8kˆ
4  a1 a2 1  a3
(ii)
The relation (ii), on equating the coeffs of non coplanar unit vectors iˆ, ˆj
a2 (1  a3 )  a2 a3  2
and k̂ we get
(iv)
a1 (1  a3 )  a3 (4  a1 )  1
(v)
a1a2  a2 (4  a1 )  8
(vi)
From (iv) a2  2 from (v) a1  4a3  1
Putting a1  4a3  1 , a2  2 , a3  a3 in (i) we get
 4a3  1 4  (4a3  1)  4  a3 (1  a3 )  1

17 a32  9a3  0
But a3 
 a3  0 , a3 
9
17
9
is rejected as it is given that a has integer components  a3  0
17
whence a1  1
Thus a1  1, a2  2, a3  0  a  iˆ  2 ˆj
b is easily found as 3iˆ  2 ˆj  kˆ by using b  4iˆ  kˆ  a .
8.
Correction :- Prove that OA. AB  6
x0  1 , y0  2  OA  iˆ  2 ˆj .
Now
dy
1

 1 at (1, 2)
dx
x
 Equation of tangent at (1, 2) is y  2  1( x  1)
It meets x-axis at B ( 1, 0)  OB  i  AB  2iˆ  2 ˆj
 OA . AB  2  4  6
10.
Take A as origin
AB BM

AC MC

Position vector of M is determined. Again BC = 10 and
position vector of N is determined.
11.
12.

 a b

c a

 a   a  c 
 b   b  a 
 c  which is zero.
 2

 2

 2

( AB CD) and b  a  c  b  0
We have b  a  c  d


b c

LHS = c  b  




Take the dot product of first equation with b and eliminate b  b .
6
13.
( correction (r 2  b 2 ) instead of (r 2  b 2 ) )
r  b  r  b  2k (Sum of distances from two fixed points is constant )

(r  b).(r  b )  (r  b).(r  b )  2k

r 2  2b .r  b 2  r 2  2b .r  b 2  2k

  2b . r    2b .r  2k


(
r 2  r . r , b 2  b . b etc)
(   r 2  b2 )
  2b . r    2b . r  2  2  4(b . r )2  4k 2
 2  4(b .r )2  2k 2   
 2  4(b .r )2  4k 4   2  4k 2
On dividing by 4 and transposing all terms to LHS we get
k 4  k 2 (r 2  b 2 )  (b . r ) 2  0 . Put k  a .
14.
Area of triangle 
1
BC  BA
2

1
1
(c  b )  ( a  b )  a  b  b  c  c  a
2
2
Since points are collinear Area  0 etc.
15.
If d , d  , d  are points where exterior bisectors meet opposite sides then
d 
bb  cc
cc  aa
, d 
,
bc
ca
d  
aa  bb
a b
 (b  c)d  (c  a)d   (a  b)d   0 etc.
16.
Equation of principal diagonal originating from origin is
r   (aiˆ  bjˆ  ckˆ)
(i)
An edge not intersecting it may be taken as.
an edge joining (a, b, 0) and (0, b, 0) whose equation is
r  bjˆ   (aiˆ)  bjˆ  iˆ
(ii)
( taking the point (0, b, 0) )
We can easily find SD between lines (i) and (ii).
to obtain one of the given expressions .
18.
Let A be origin AB  b
then AM 
, AC  c
b c
2
( AB  c , AC  b , BC  a according to triangle notations )
By geometry BD : DC  c : b
7
AD 

 cc  bb  c  b
cc  bb
 AD  AM  

cb
2
 cb 
c(c  b )  b(b  c )
2(c  b)
 ( Area ABC )
19.

(b  c ) (b  c)
2 (b  c)
bc

bc
Area  ADM
bc

etc .
Area  ABC
2(b  c)
LHS  (b  a ).(d  c )  (c  b ).(d  a )  (a  c ).( d  b )
(i)
 b . d  b .c  a .d  a .c  c .d  c .a  b .d  b . a  a .d  a . b  c . d  b . c  0
LHS  (c  a ).(c  a )  (d  b ).(d  b )  c 2  d 2  a 2  b 2  2a . c  2b .d
(ii)
RHS  (b  a ) 2  (d  c ) 2  2(c  b ).(d  a )  a 2  b 2  c 2  d 2  2a. c  2b . d
20.
In the figure we have a triangle PQR
RP  B , PQ  C , RQ  A
Area 

1
1
( B  A) 
B  (B  C)
2
2

i j k
1
1
 (B  C)  d 3 4
2
2
3 1 2

1 
(10iˆ  ˆj (2d  12)  kˆ(d  9)   5 6 (given )


2
 100  (2d  12)2  (d  9)2  600 . This will give d  5 or d  11
Also , on comparing iˆ , ˆj , kˆ coeffs in A and B  C  a  d  3 , b  4, c  2 etc
21.
iˆ  (iˆ  ˆj )  kˆ
(iˆ  ˆj )  ( ˆj  kˆ)  iˆ  ˆj  iˆ  kˆ  ˆj  kˆ  kˆ  ˆj  iˆ
 a is along kˆ  (kˆ  ˆj  iˆ)
o r along ˆj  kˆ  kˆ  iˆ
or along iˆ  ˆj
 angle  between (iˆ  ˆj ) i.e a and iˆ  2 ˆj  2kˆ is given by
Case 
(iˆ  ˆj ).(iˆ  2 ˆj  2kˆ)

2. iˆ  2 ˆj  2kˆ
3
1

   45
2.3
2
Page 629-630
1.
After following hints given in the question take dot product of both sides
8
with b  c , c  a , a  b in succession.
2.
We can assume r  xa  yb  z a  b ( Since a , b and a  b are non coplanar )
The equations becomes
3.
 xa  yb  z a  b   b  a

xa  b  zero  z b  (a  b )  a

xa  b  z (b .b ) a  (b . a )b  a

x a b

x 0 , z 


 1  (b . b ) z  a  0
1
b.b
(
a and a  b are non collinear )
The equation cab be written as (r  c )  b  0  r  c and b are parallel .

r c  b
Taking dot product with a we get
(
4.
a.b  0)
(
 
c .a
b.b
r . a  0) etc .
If O be origin in tetrahedron OABC
OA  a , OB  b , OC  c
Let N1 , N2 , N3 , N4 be out word normals
of faces OBC , OCA , OAB and ABC ,
then N1 
1
1
1
1
(b  c ) , N 2  (c  a ) , N 3  (a  b ) , N 4  (a  b )  (c  b )
2
2
2
2
1
N 4 is simplified to N 4  (a  c  a  b  b  c )
2

5.
N1  N2  N3  N4  0 .
Since a , b , c are non coplanar  a , b , c   0
Now  a , a  b , a  b  c 


 a . (a  b )  (a  b  c )  a b c   0
 set {a , a  b , a  b  c} also forms a basis
Therefore x a  yb  zc  la  m(a  b )  n (a  b  c )
Since a , b , c are non coplanar we can compare coeffs . of a , b , c
9
to get
l  x  y , m  y  z ,n  z
thus co-ordinates of vector with respect to new basis are x  y , y  z , z
6.
The equation of plane containing the line of intersection of planes
r . n1  p1  0 , r . n2  p2  0
must be of the form r .(n1   n2 )  p1   p2  0
(i)
If the plane r . n3  p3 contains the same line then they must exist  such that
n1   n2   n3
(ii)
p1   p2   p3
(iii)
Taking cross product of equation (ii) with by n1 , n2 and n3 in succession we get
 n1  n2   (n1  n3 )
n2  n1   n2  n3 , n3  n1   n3  n2  0
Let us express all cross products in terms of n1  n2 .
Indeed n2  n3  
n1  n2

n3  n1   n3  n2 
,

(n  n )
 1 2
 LHS  p1 (n1  n3 )  p2 (n3  n1 )  p3 (n1  n2 )
 p p

 n1  n2   1  2  p3   0 by (iii)

 

7.
Apply the formula (a  b ).(c  d ) 
a .c
a .d
for all the three terms on LHS .
b .c b .d
8.
9.
a  (b  c )  (a  b )  c



(a . c ) b  (a . b )c   (c .b )a  (c .a )b

(c . b )a  (a .b )c  0 Or  b  ( c  a )  0 etc .
The equation of a line through any point a  bt of the given line normal to the given plane
must be r  a  bt  pn . Since the projection of the point a  bt on the given plane lies on
the plane we must have (a  bt  pn ). n  q .

p
q  (a  bt ).n
. Thus the projection of a  bt on the plane is
n2
10
a  bt 
q  (a  bt ).n
n
n2
q  a.n  
b .n 

 a 
n   t b  2 n .
2
n
n 

 
q  a.n  
b .n 

n   t  b  2 n .
 required projection is the line r   a 
2
n
n 

 
Miscellaneous Exercise page -639-641
1.
Consider A  B  C
2
as A  B  C
2


 A B C . A B C

 A . A  A .( B  C )  B .B  B ( A  C )  C.C  C .( A  B )
 9  16  25  50 etc .
2.
Since A is perpendicular to both B and C it must be in the direction of B  C . Let

A  λ BC
A  λ BC

3.

 1  λ sin 30
 λ  2.
If possible let  ABC   0 , then A, B, C must be non-coplanar. Now X  A  X  B  X  C  0

X is perpendicular to A, B and C which is not possible since A, B, C are non-
coplanar. Thus  A B C   0 .
4.
Transpose RHS term to the left and equate coeffs of iˆ , ˆj , kˆ to zero since iˆ , ˆj , kˆ
are non-coplanar. For non trivial solution put D  0 to determine  .
5.
(a  b )  (a  b )  2b
(a  b )  (a  kb )  (1  k ) b
 Points whose position vectors are a  b , a  b and a  kb are collinear since the
two vectors formed by them are collinear for all k .
8.
9.
We must have  DA , DB , DC   0


Let OA  a , OB  b . Draw AN  OB
Now ON  ( unit vector along ON) ON

b
a cos 
b

a .b
b
2

b
b
a
a. b
a b
b
Also NA  position vector of A -position vector of N
11
 a
a .b
b
11.
b.
2
This was IIT-96 incorrect question . The representation of a in this form is possible
if the angle between b and c is 90 .
In this case a  (a . b ) b  (a . b ) c   a b c  (b  c )
Which following by assuming a  b   c  b  c
12.
a  b  1 , c  2 , a  (a  c )  b  0

(a . c ) a  (a.a ) c  b  0  (a .c ) a  c   b

2 cos  a  c   b

 (2cos )a  c  .  2cos a  c 

4cos2  a  4cos  a. c  c  b

2
b
2
a .c
a c



2
2
4cos 2   8cos 2   4 1
cos   

Let
 

Area  6 a  b etc.
(Same sign area)
14.
cos  
3
 acute  is 30 .
2
1
1
Area of quadrilateral = a  10a  2b  10a  2b  b
2
2

13



AM  a , AN  b . Then AP  ka , AQ  lb .
a b
ka  lb
and AC 
.
2
2
λ
1
a  b  ka  lb
Since B, A, C are collinear
2
2
PQ  lb  ka  k b  a

=k=l

Now AM 





15.

PQ  kMN etc.
(a)
LHS  u v cos 2   u v sin 2 
(b)
RHS  (1  u .v )2   u  v  u  v  . u  v  u  v 
2
2
2

2

u v  RHS
2
2
 1  (u.v ) 2  2u.v  u  u .v  zero
2
 v . u  v  zero  zero  zero  u  v
2
 1  u v  (u .v ) 2  (u  v ) 2
2
2
2

 1 u
2
2
 1 v  .
 1  u  v  u v from (a) part .
2
12
2
2
2
16.
This was IIT -2001 incorrect question .If we assume v  2iˆ
v2  x2iˆ  y2 ˆj  z2 kˆ
x2  1 , x3  3

, v3  x3iˆ  y3 ˆj  z3kˆ then the given relations become
v1.v2   2 , v1.v3  6
x2 2  y2 2  z2 2  2 , x2 x3  y2 y3  z2 z3  5
x32  y32  z32  29
On putting the values of x2 , x3 in this system we get
y2 2  z2 2  1 , y32  z32  20, y2 y3  z2 z3  2
Again the system is insolvable since there are four unknown and three equations .
Again taking z2  1 we get y2 2  0  y2  0 from first equation
Whence y32  z32  20 , z3  2
 y32  16
 x32  9
x32  y32  z32  29
Thus one set of vectors satisfying is v1  2iˆ , v2   iˆ  kˆ , v3  3iˆ  ˆj  kˆ
17.
18.
Define a function h(t) = f1(t)g2(t) – f2(t)g1(t)
Which is also continuous for t  [0, 1]. Now h(0) and h(1) are of opposite signs

h(t)
= 0 for some t  (0, 1) 
A(t) and B(t) are parallel for some t.
i j k
V  W  2 1 1  3iˆ  7 ˆj  kˆ, V  W  59
1 0 3


Now UVW   U V  W  U V  W  V  W  59
19.
Given that u , v , w are three non coplanar unit vectors . Angle between u and v is  ,
between v and w is  and between w and u it is  .In figure OA and OB represent
u and v . Let P be a point on angle bisector of AOB such that OAPB is parallelogram.
Also PAO  BOP 
APO  BOP 

2
,

2
In OAP , OA  AP,


OP  OA  AP  u  v
u v
u v
OP 
i.e. x 
u v
u v

2
But u  v  (u  v ).(u  v )  1  1  2u.v
 2  2cos   4cos2

2
13
u  v  1

u  v  2cos
1

 x  sec (u  v )
2
2

2
1

1

Similarly , y  sec (v  w), z  sec ( w  u )
2
2
2
2
Now  x  y y  z z  x 
 x y z 
2

1

1
Now  x y z    sec (u  v ), sec (v  w),
2
2
2
2
1


sec (w  u )
2
2

1



 sec sec sec u  v v  w w  u 
8
2
2
2

1



sec sec sec u v w
4
2
2
2
x y z
2

v  w w  u   2 u v w )
1



2
u v w sec2 sec2 sec2
16
2
2
2

From (*) and (**),  x  y y  z z  x 
20.
u  v
(

(**)
1



2
u v w sec2 sec2 sec2 .
16
2
2
2
Given that incident ray is along v , reflected ray is along w and normal is along
(See figure )
Since incident ray reflected ray and normal lie in a plane ,
and angle of incident = angle of reflection .
 a will be along the angle bisector of w and v , i.e
a
w  ( v )
wv
(*)
 bisector will along a vector dividing in same
[
ratio as the ratio of sides forming that angle ]
w  v  OC  2OP  w cos  2cos
But a is a unit vector where
Substituting this value in equation (*), we get a 
Hence w  v  (2cos )a  v  2(a . v )a

wv
2cos
a.v   cos  .
Objective Exercise Page 645-647
1.
2
2
We have u  v  a  b  0
2
2
2
2
2
u v
 u  v  u v sin 2   u v
2
2
2
2
u  a  b  8, v  a  b  8  u  v  64  u  v  8
Only choice (a) is equal to 8.
14
2
2
a, outwards.
3.
i j k
1
Area = AB  AC 1 (given ), Now AB  AC  r 1 0  (r  s)kˆ
2
s 1 0
 AB  AC 1 (r  s)
4.
 r  s  2.
We have c. a  c .b  cos , a .b  0 , Now c   a   b   (a  b) (Fundamental law)
Taking the dot product of both sides with a we get , c .a   cos
(
2
a 1, a .b  0)
  cos
Similarly
2
Now taking the dot product with a  b we get [c a b ]   a  b  
Now  2  [c ab]2  [ab c ]2
a. a a.b a. c
1
 b. a b. b b. c  0
c. a c.b c. c cos
cos 
cos  1  cos 2   cos  ( cos  )   cos 2
1
0
1
cos
Thus     cos ,  2   cos 2
5.

The equality
( r . c)  b  0  r  c   b
Find  by taking dot product of both sides with a .
6.
From the relation a  2b  3c  0, We easily get c  a 
2
1
a  b , b c  a  b
3
3
Where a  b  b  c  c  a  2(a  b).
8.
9.
2
a b c 0
2
2
2
 a  b  c  2a  b  2b  c  2c  a  a  b  b  c  c  a  25
Let the unit vector be xiˆ  yjˆ then x 2  y 2  1 and from the given information
cos 45 
x y
2 x2  y 2
,cos60 
3x  4 y
 x + y = 1, 3x – 4y = 5/2
5 x2  y 2
 x = 13/14, y = 13/ 14, which is not possible since x2 + y2 = 1
Thus none of the choices (a), (b) , (c) are correct.
10.

  
 B  C  A   A  C  B  2 A  B C  0




The given relation  A  C B  A  B C  B  A C  B  C A



Since A, B, C are non coplanar B  C  0, A  C  0, 2 A  B  0

 
 

whence A  B  C  A  C B  A  B C  0
12.
a  b  3 a  b  a b sin   3 a b cos  tan   3    60
15
14.
AB is normal to the plane  Equation of plane must be of the form
r . AB  d Or r .(5i  3 j  11k )  d
3 7
2 2
9
2
Now plane passes through M , the mid point of AB or passes through  , ,  

7
9 
3
 d   iˆ  ˆj  kˆ  . 5iˆ  3 ˆj  11kˆ
2
2 
2


135
2
 ( a ) is correct answers .
16.
Take AB  a , BC  b then AD  2b (why)?
AC  a  b , AE  AD  DE  2b  a
EB  a  (2b  a )  2a  2b
FC  2a .
The given relation becomes 2b  2a  2b  2a   a
 4a   a    4 .
19.


a  b  a b sin  nˆ (whence  is the angle between a and b )
a  b   c   a

b sin  nˆ  c  a b c  sin   cos 
where  is the angle between a  b and c


Now a  b  c
a b
c if and only if sin  = 1, cos  = 1
 a and b are perpendicular and c is parallel to a  b
20.
 c is perpendicular to both a and b
If the points are designated by A, B, C respectively then
AB  20iˆ  11 ˆj, BC   a  40  iˆ  44 ˆj, if A, B, C are collinear AB   BC
 20iˆ  11 ˆj    a  40  iˆ  44 ˆj 
22.
    a  40  20 iˆ  11  44  ˆj  0
Since iˆ and ˆj are non collinear vectors (a – 40) + 20 = 0, 11 - 44 = 0
  = ¼, a = -40
a  (2 p)iˆ  ˆj , a  ( p  1) iˆ ' ˆj 
(i , j  are unit vectors in new system )
Since both system are rectangular a  (2 p)2  1  ( p  1) 2  1 etc.
23.
Let c  xiˆ  yjˆ , then 4x+3y = 0 since c  b. Let piˆ  q ˆj be a vector whose projections along
( piˆ  qjˆ).(4iˆ  3 ˆj )
( piˆ  qjˆ) .( xiˆ  yjˆ)
 1,
2
b and c are 1 and 2 respectively , then
4iˆ  3 ˆj
xiˆ  yjˆ

4 p  3q  5, px  qy  2( x 2  y 2 )1/ 2
16
1/ 2
10 x
4x
4q 

 16 
On putting y  
the second relation becomes x  p 
  2 x 1   
3 
9 
3
3


If x = 0, then y = 0 and c becomes a zero vector which is not the case
4q 10
If x > 0, then p 
or 3 p  4q  10

3
3
Solving it with 4 p  3q  5, we get p  2, q  1. Thus 2iˆ  ˆj is one such vector .
Again for x < 0 the equation becomes 3 p  4q  10
2
11
2 11
whence we get p   , q 
and the vector is  iˆ  ˆj.
5
5
5
5
2 11
Thus the vectors are 2iˆ  ˆj and  iˆ  ˆj. the first one appears in choice (b)
5
5
24.
SEE IIT Mains section Question 28 .
26.
It is better to proceed from choices. The projection of the vector given in choice
(a) on vector a 
 2iˆ  3 ˆj  3kˆ    2iˆ  ˆj  kˆ   2  
2iˆ  ˆj  kˆ
6
2

3
2
(neglecting sign)
3
All other choices are wrong. The subjective solution is as follows.
Any vector d coplanar with b and c  xb  yc

 

 x iˆ  2 ˆj  kˆ  y iˆ  ˆj  2kˆ   x  y  iˆ   2 x  y  ˆj    x  2 y  kˆ
Now projection of this vector on a  2iˆ  ˆj  kˆ

2 x  y    2x  y   x  2 y x  y
x  y
2

. As given

3
6
6
2iˆ  ˆj  kˆ
 x + y = -2  y = -2 – x
 d  2iˆ   2 x  2  x  ˆj    x  4  2x  kˆ
 2iˆ   x  2  ˆj   x  4  kˆ   iˆ   2  x  ˆj   x  4  kˆ 


For x = -1, d becomes the vector given in choice (a)
27.
OR 
2q  3 p
2q  3q
 1
, OS 
, OR .OS  ( 4q 2  9 p 2 )    .
5
1
 5
Since OR and OS are perpendicular 9 p 2  4q 2
28.
The magnitude of difference of any two vectors 
(   ) 2  (    ) 2  (   ) 2
 The three points are forming an equilateral triangle .
17
IIT Exercise 648-653
2.
Let d  d1iˆ  d 2 ˆj  d 3 kˆ then a . d  0  d1  d 2  0
0 1 1
Now [b c d ]  1 0 1  0  d1  d 2  d3  0
d1 d 2 d3
(1)
(2)
Also d12  d22  d32  1
(3)
From (1) d1  d 2 and from (2) d3  2d 2 , On putting in (3) we get
1
d 2 2  d 2 2  4d 2 2  1  d 2  
etc.
6
NOTE:7.
It is a divisible to proceed from choices.
p , q , r may be taken as iˆ, ˆj , kˆ respectively , Now first term on L H S
 (iˆ .iˆ) x  ( iˆ .x ) iˆ  ˆj
= iˆ  (( x  ˆj )  iˆ)  iˆ  ( x  iˆ  kˆ)
= iˆ  ( x  iˆ)  iˆ  kˆ
 x  (iˆ.x ) i  j
 
 3x   iˆ  x  iˆ   ˆj  x  ˆj   kˆ  x  kˆ    iˆ  ˆj  kˆ   0


= x  iˆ  x iˆ  ˆj , We can similarly sort out second and third terms Thus LHS = 0




1 ˆ ˆ ˆ
i  jk
2
(NOTE :- Any vector x  iˆ  x iˆ  ˆj  x ˆj  kˆ  x kˆ )
 3x  x  iˆ  ˆj  kˆ  0
x
 
  
8.
11..
If three vectors are linearly dependent then they are coplanar
i j k
a  b  2 1 2  2iˆ  2 ˆj  kˆ  a  b  3
1 1 0
Now
 a  b   c  a  b c sin 30  32 c
Now c  a  2 2
 c 2  a 2  2a  c  8
( a  c  c given)

Take c  xa  yb  x  2iˆ  ˆj  kˆ   y  iˆ  2 ˆj  kˆ 
 c2 + 9 – 2c = 8
12.


 c  1 . Whence a  b  c 
=
3
2
 2x  y  iˆ   x  2 y  ˆj   x  y  kˆ
Since c is perpendicular to a we must have 2(2x + y) + (x + 2y) + x – y = 0
Again, as c is a unit vector (2x + y)2 + (x + 2y)2 + (x – y)2 = 1 etc.
13.
2
u  a  ( a .b )2 b  2( a .b )2  1  cos2   2cos2   sin 2 
2

2
v
2
 a b
2
 sin 2 
Again u .b  a .b  (a .b )  0
 Choice (c) is also correct.
 v u
( b .b  1)
18

a .b  cos

14.
 a  b   c  d   0
 a  b and c  d are parallel
 Normal to planes P1 and P2 are parallel
 planes P1 and P2 are parallel.
17.

Expression = 6 - 2 a  b  b  c  c  a

(1)
Now a  b  c  0


 3 + 2 a b  b c  c  a  0
19.
21.
(2)
From (1) and (2) the expression ≤ 9.
dV
d 2V
1
V = 1+ a3 – a,
 3a 2  1, 2  6a  V has a minimum at a 
da
da
3
Proceed from choices OR else let the required unit vector be d then
d  x 2iˆ  ˆj  kˆ  y iˆ  ˆj  kˆ   2 x  y  iˆ   x  y  ˆj   x  y  kˆ

 

Since d is orthogonal to 5iˆ  2 ˆj  6kˆ we have 5(2x + y) + 2(x – y) + 6(x + y) = 0
22.
Since d is a unit vector (2x + y)2 + (x – y)2 + (x + y)2 = 1, etc
(2)
ˆ
A vector d in the plane of a and b is d  xa  yb   x  y  i   2x  y  ˆj   x  y  kˆ
Since projection of d on c is

23.
(1)
 x  y    2x  y    x  y 
3
1
3

we have
1
d c
1

c
3
 2x  y  1
3
Since in d components of iˆ and kˆ are equal only choice (a) is correct (2x - y = 1 is also
satisfied)
The line of intersection of P1 is along  x 2iˆ  3kˆ  4 ˆj  3kˆ and that of P2 is along

 

 ˆj  kˆ   3iˆ  3 ˆj 
 A is along  2 ˆj  3kˆ    4 ˆj  3kˆ    ˆj  kˆ    2iˆ  3 ˆj 

 

ˆ
ˆ
Or along 54  ˆj  k     ˆj  k   Angle  between A and given vector 2iˆ  ˆj  2kˆ is given by
cos   
2  0  1  1   2  1
2  1   2 
2
24.
2
0  1  1
2
2
2

1
2
 

4
or
3
4
Since  2 i  j  k , i   2 j  k , i  j   2 k are coplanar
(C):

2
 2
1
1

1
1
1
2
 2   4  1  1  2  1  1   2   0
1 0
 2

6   2   2  1  1   2  0   6  3 2  2  6 - 32 – 2 = 0  (2 + 1)(4 - 2 – 2) = 0
19
 (2 + 1)2(2 – 3) = 0  Two real solutions.
25.
(a).
aˆ.aˆ aˆ.bˆ aˆ.cˆ
1 1/ 2 1/ 2
ˆ
ˆ
ˆ
ˆ
b.aˆ b.b b.cˆ  1/ 2 1 1/ 2
1/ 2 1/ 2 1
cˆ.aˆ cˆ.bˆ cˆ.cˆ
OP  aˆ cos t  bˆ sin t
26. (a) :

= 1  sin 2taˆ.bˆ
uˆ 
aˆ  bˆ
aˆ  bˆ
2
27. (d) :

1
2
1
2
 V

 cos 2 t  sin 2 t  2 cos t sin taˆ.bˆ
 OP
 uˆ 
V2

max

 1  aˆ.bˆ

1
2
when, t 
1
2
 
1
2
ˆˆ
= 1  2 cos t sin tab

4
aˆ  bˆ
aˆ  bˆ
2
The direction cosines of each of the lines L1 , L2 , L3 are proportional to(0,1,1).
28.
(b) :
i j k
i  7 j  5k
3 1 2   i  7 j  5k . Hence unit vector will be
5 3
1 2 3
29.
(d) :
S .D 
30.
(c) :
Plane is given by –(x+1)-7(y+2)+5(z+1)=0  x  7 y  5 z  10  0
 distance =
31.
(1  2)(1)  (2  2)(7)  (1  3)(5)
5 3

17
5 3
.
1  7  5  10
13
.

75
75
 (1  3 μ)iˆ  (1  μ) ˆj  (2  5 μ) kˆ
For a particular μ OQ  r


1
2
PQ  position vector of Q  position vector of P
 OQ  (3iˆ  2 ˆj  6kˆ)
 (2  3 μ)iˆ  ( μ  3) ˆj  (5 μ  4)kˆ
PQ must be perpendicular to normal to the plane

32.
1(2  3 μ)  4( μ  3)  3(5 μ  4)  0

μ  1/ 4 
(a) is correct.
Let the angles between a and b be θ between c and d be φ and between a  b and c  d
be ψ . Now (a  b).(c  d )  1

sin θ sin φ cos ψ  1

θ  90, φ  90, ψ  0

a  b, c  d , (a  b) (c  d )
since all the three are proper fractions we must have

20
[abc]  0 and [abd ]  0
33.

Options A and B are incorrect. We can easily rule out option D by drawing a figure.
Hence option C is correct.
(A):
PQ  6iˆ  ˆj
QR  3iˆ  3 ˆj  4iˆ  iˆ  3 ˆj
PS  3iˆ  2 ˆj  2iˆ  ˆj  iˆ  3 ˆj
SR  3iˆ  3 ˆj  3iˆ  2 ˆj  3iˆ  ˆj



Now
PQ.PS  6iˆ  ˆj . iˆ  3 ˆj  3  0

Adjacent sides are not perpendicular
 Square and rectangle are ruled out.
Here PQ SR and PS QR  (A) is correct.
34.
Since AB. AD  0  θ is acute.
(B):
Now cos θ 
AB. AD

AB AD
8
9
Also cos  α  θ   0  cos α.cos θ  sin α.sin θ  0
 8cos α  17.sin α  0
64 cos 2 α  17 sin 2 α  cos α 
35.
17
 (b) is correct
9
(5): Note that a and b are unit vectors and a .b  0

 
  

b   a.b  a  2  a.b  b  2 b a.  b  2a
 2a  b .l   2a  b . 2a  b   5

m  a  b  a  2b  a  b  a  2 a  b  b
a

36.
2
 x(iˆ  ˆj  kˆ)  y (iˆ  ˆj  kˆ)
1
v  ( x  y )iˆ  ( x  y ) ˆj  ( x  y )kˆ since projection of v on c is
3
Sol:- (C)
we must have
Let v  xa  yb
v.c
c


1
3
x  y  ( x  y)  ( x  y) 1

 y  x 1
3
3
21
 v  ( x  1  x)iˆ  ˆj  ( x  x  1)kˆ  v  (2 x  1)iˆ  ˆj  (2 x  1)kˆ
for x  1 it is same as the vector given in choice ‘C’
37.
Sol:- (A & D)
A vector c is coplanar with a & b is given by
c   a  b
 (   )iˆ  (  2 ) ˆj  (2   ) kˆ
If c is perpendicular to iˆ  ˆj  kˆ ,
38.
  0
for   1 then c   ˆj  kˆ & ˆj  kˆ
Thus c   ( ˆj  kˆ)

      2  2    0
( A) & ( D) are correct.
r  b  c  b (given).
SOL:- (9)
a  ( r  b )  a  (c  b ) 

r  0  a  (c  b )
iˆ
Now c  b  1
On taking cross product of both sides with a we get
(a.b)r  (a.r )b  a  (c  b)
r  a  (c  b )
ˆj kˆ
2 3  iˆ(0  3)  ˆj (3  0)  kˆ(1  2)  3iˆ  3 ˆj  3kˆ
1 1 0
iˆ

r  a  (c  b)  1
ˆj
kˆ
0
1  iˆ(0  3)  ˆj (3  3)  kˆ(3  0)  3iˆ  6 ˆj  3kˆ
3 3

39.
3
r.b  (3iˆ  6 ˆj  3kˆ).(iˆ  ˆj )  3  6  9
If a , b , c are unit vectors then a .b  b .c  c .a 
2
1
3
a  b  c  3  



2
2
It will be equal to 3 / 2 if a  b  c  0
 unit vectors a , b , c are along sides of an
equilateral triangle is the style shown
in the figure
 Angle between any two of them is 120 .
2
2
Now a  b  b  c  c  a  9
 aˆ. bˆ  b .c  c . a   3 / 2 .
 a , b , c satisfy the situation shown the figure  a . b  b  c  c .a  
22
1
2
Now 2a  5b  5c
 1
 1
 1
 4  25  25  20     30     20     9
 2
 2
 2
2a  5b  5c  3.

40.
2
a  (2iˆ  3 ˆj  4kˆ)  b  (2iˆ  3 ˆj  4kˆ)
 (a  b )  (2iˆ  3 ˆj  4kˆ)  0

a  b and 2iˆ  3 ˆj  4kˆ are parallel

a  b   (2iˆ  3 ˆj  4kˆ)
But
a  b  29    1

  1(2iˆ  3 ˆj  4kˆ). (7iˆ  2 ˆj  3kˆ)
(a  b ). (7iˆ  2 ˆj  3kˆ)
 4 or  4
among the given choices 4 matches.
41.
(c)
a  b  3iˆ  ˆj  2kˆ
c  iˆ  2 ˆj  3kˆ
a  b  iˆ  3 ˆj  4kˆ

 

= a  b  a  b  1  10iˆ  10 ˆj  10kˆ
2 a  b  10iˆ  10 ˆj  10kˆ
= a  b  5 iˆ  ˆj  kˆ
Now volume of parallelepiped  cab   c . a  b  51  2  3 = 10
42.
There
are
four
sets
S1  iˆ  ˆj  kˆ, iˆ  ˆj  kˆ , S2  iˆ  ˆj  kˆ, iˆ  ˆj  kˆ ,
S3
of

 

 iˆ  ˆj  kˆ, iˆ  ˆj  kˆ, S  iˆ  ˆj  kˆ, iˆ  ˆj  kˆ
parallel
vectors
4
If two vectors from the same set S i  i  1, 2,3, 4  are taken and one vector from remaining then
the
8
43.
three
vectors
will
be
coplanar
 Number
C3  4  6  32  2  p  5
P
5
2
 a b c   2 We know that  a  b, b  c, c  a    abc   4



 

23
of
non
coplanar
vectors
  2  a  b  , 3  b  c  , c  a   6 a  b, b  c, c  a   6  4  24
P  (3)
Here  a b c   5
Q




3 a  b , b  c,2 c  a   6 a  b, b  c, c  a 




 6.2  a  b, b  c, c  a   6  2  5  30
Q  (4)
1
a  b  20  a  b  40
2
1
1
5
Now 2a  3b  a  b  5a  b   40  100
2
2
2
 R  (1)
R

 

a  b  30
S


Now a  b  a  b  a  30  S   2 
Thus (C) is the answer.
44.
Given x  y  y  z  z  x  2  2 cos
Also a  x  0, a  ( y  z )  0
π
 1 (They are equally inclined)
3
b  x  0, b  ( z  x)  0
Using vector triple product formula
  

 xz y  x y z
x  ( y  z)


 yz
Since a is  to x and y  z , a is in the direction of x  y  z or a is in the direction of y  z


aλ yz

Taking dot product of both sides with y


a  y  λ y  y  z  y  λ (2  1)  λ

a  (a  y)( y  z )
Similarly b  (b  z )( z  x)
Also a  b  (a  y)(b  z ) ( y  z  y  x  z  z  z  x)
 (a  y)(b  z )(1  1  2  1)
 (a  y)(b  z )

(A), (B), (C) are correct
24
45.
a b  bc  c a 
1
2
Now given relation is
a  b  b  c  pa  qb  rc
Taking dot product of both sides with a
a b c   p  q  r


2 2
(i)
Taking dot product of both side with b
p
r
q 0
2
2
(ii)
p q
Taking dot product with c we get  a b c     r
2 2
From (i) and (iii) p 
(iii)
q r p q
   r
2 2 2 2

p  r whence from (ii) q  r

p 2  2 q 2  r 2 r 2  2r 2  r 2

4
q2
r2
AIEEE/IIT MAINS Page 653-657
2.
If D is the mid point of BC then AD 
6.
We are given [abc]  0
AB  AC
 4iˆ  ˆj  4kˆ  AD  33
2
Now [a  2b  c ,  b  4c ,(2  1)c]
 (a  2b  3c) .
 (b  4c)  (2  1)c 
 (a  2b  3c) .  (2  1) (b  c)   (2  1) [a b c]
Which vanishes only for   0 ,  
7.
u v w
As given
2
since [a b c ]  0.
 u 2  v 2  w2  2u . v  2v . w  2u . w
v.u
u
10.
1
2

w. u
, v .w0  u v  w
14
etc.
u
i
j
Let a  a1i  a2 ˆj  a3kˆ then a  iˆ  a1 a2
1 0

2
k
a3  a3 ˆj  a2 kˆ
0
(a  iˆ)2  (a  iˆ).(a  iˆ)  a32  a22
Similarly (a  ˆj )2  a12  a32 ,(a  kˆ)2  a12  a22
25
 LHS  2(a12  a2 2  a32 )  2a 2 .
11.
12.
a a c
We must have 1 0 1  0
c c b

a(0  c)  a(b  c)  c(c  0)  0

c 2  ab  c is GM of a and b
 (a  b )  2 a

Also  a

c   . 2 .  a  b
b  c b 
b c 
  4  a b c 
 a . ((b  c )  b )  a . (c  b )    a b
On equating we get  4  1

a b c   0


c 

 No real  are possible.
13.
Correction : b  xiˆ  ˆj  (1  x)kˆ
1 0
1
a b c   x 1
1 x


y x 1 x  y
  1( x  1  x)  1
0
0
1
 1
1
1 x
1 x x 1 x  y
C1  C1  C3 
  a b c  is independent of x and y .
 [(a  b)iˆ  2 ˆj ] .[(a  1)iˆ  6kˆ]  0  (a  2)(a  1)  0  a  2,1
14.
AC.BC  0
14A.
(a  b )  c  a  (b  c )
  c  (a  b )  a  (b  c )

15.
c
a.c
a  c and a are parallel .
a.b
uˆ  vˆ  sin 

  (c . b )a  (c .a )b  (a.c )b  (a.b )c .

uˆ  vˆ  1
2uˆ  2vˆ  4sin  1 
16.
 a b c   0 etc .


19.
By cutting unit vectors
sin  
1
which has only one solution.
4
ˆj  kˆ
iˆ  ˆj
along b and
along c
2
2
26
we get a vector d along bisector of b and c as d 
iˆ  ˆj ˆj  kˆ
iˆ  2 ˆj  kˆ


2
2
2 2
Another vector along d may be taken as iˆ  2 ˆj  kˆ
Thus  iˆ  2 ˆj   kˆ   (iˆ  2 ˆj  kˆ) for some 
    , 2  2 ,   
20.
a  8b , c  7b
     1  (b) is correct.
8
 a  c
7
 a and c are anti parallel
 Angle between them is 
21.
3u pv pw  3 p2 u v w
 pv w qu    pq u v w
2w qv qu    2q2 u v w
The given relation becomes
(3 p  pq  2q2 )
u
v w  0
But u v w  0  2q 2  pq  3 p 2  0
Which is possible if and only if p  0 , q  0
( Since as a quadratic in q , D  p 2  24 p 2  0 for all p )
iˆ
22.
(a):
ˆj
kˆ
1  2iˆ  ˆj  kˆ given aˆ  bˆ  cˆ  0
1 1 1
aˆ  cˆ  0
1


 aˆ.bˆ aˆ   aˆ.aˆ  bˆ  aˆ  cˆ  0
3aˆ  2bˆ   2iˆ  ˆj  kˆ   0ˆ
2bˆ  3  ˆj  kˆ    2iˆ  ˆj  kˆ 

aˆ  aˆ  bˆ  aˆ  cˆ  0ˆ 
23.

bˆ  iˆ  ˆj  2kˆ
(a):
aˆ  iˆ  ˆj  2kˆ, bˆ  2iˆ  4 ˆj  kˆ and ĉ  λiˆ  ˆj  μkˆ
Given , aˆ , bˆ, cˆ are mutually orthogonal, aˆ.cˆ  0 , bˆ.cˆ  0

λ 1 2μ  0 ,
2λ  4  μ  0
solving (i) and (ii) we get λ  3 & μ  2
24.
Note that a.a  b . b  1 , a.b  0
(2a  b ).
 (a  b )  (a  2b ) 
27


 (2a  b ).  a  (a  b )  2b  (a  b ) 
 (2a  b ).  (a.b )a  (a . a )b  2(b .b )a  (2b.a )b 
= (2a  b ).(2a  b )
 a.b  0, a.a  b.b  1
 (2a  b ) . (a  b )  a  (a  b )  2b
 4a.a  b .b  5
25.
b  c  b  d  b  (c  d )  0  c  d and b are parallel  c  d  b
Taking dot product with a we get c .a  d .a   b .a


26.

c .a
b .a
( a.d  0 , b .a  0)
d c
a.c
b
a.b
Correction : The three vectors are piˆ  ˆj  kˆ , iˆ  qjˆ  kˆ and iˆ  ˆj  rkˆ
p 1 1
we must have 1
1
27.
q 1 0
1 r

pqr  ( p  q  r )  2  (a) is correct.
Since a  3b is collinear with c , a  3b  c for some scalar  (i)
Similarly b  2c   a for some scalar  .
Substituting a from first equation in to second equation we get
b  2c   ( c  3b )  (1  3 )b  (2   )c  0
Since b and c are non collinear 1  3  0 , 2    0

28.

1
,   6 whence a  3b  6c  0 from (i).
3
Indeed r  q  AE
(i)
Now AE  projection q on AD 
 AE 
q. p
p
q. p p
( q. p ) p
.

2
p p
p
28
Thus from (i) r  q 
( p.q ) p
p

29.
( a ) is correct.
c.d  0 will give 5  8  6aˆ.bˆ  0
 aˆ.bˆ  
30.
(c)
(d)
1
 angle between â and b̂ is 60 .
2
Let A  0,0,0 then B  3,0, 4  and C  5, 2, 4 therefore length of median
through A =
31.
2
16  1  16  33
    
  a  b   u   c  a    a  b    u  a  c   u  c  a   a  b   b c a  c  b c c  a
a  b b  c c  a 



Let u  b  c  a  b  b  c  c  a
2
 b c a   a b c    a b c   λ  1
29