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VECTOR ALGEBRA

VECTOR ALGEBRA
1
...
If c be a vector such that the


Let a  iˆ  ˆj  kˆ

7
...


 
6 3  3 

(a) 6 3  3

(b) 3  3

(c)

(d) 6

and bˆ

Let aˆ



 





3 1

between aˆ

and bˆ,

 S1 : 2 aˆ  bˆ

 aˆ  bˆ

If    0,  

is the angle





aˆ on aˆ  bˆ is

(a) Only (S1) is true
(b) Only (S2) is true
(c) Both (S1) and (S2) are true
(d) Both (S1) and (S2) are false

Let a  a1iˆ  a2 ˆj  a3 kˆ ai  0, i  1, 2,3







5
...



4
5
(c)
6

Let a   iˆ  2 ˆj  kˆ


4
3
(d)
4

and b  2iˆ   ˆj  kˆ,



(b) 

where

  R
...
Let v be a vector in the plane of a


2

and b whose projection on c is

...
Also, let the projection of a on the

vector 3iˆ  4 ˆj be 7
...
If a , b and x-axis are coplanar,

then projection of a vector b on 3iˆ  4 ˆj is equal to

4
...
let the angle between

   


a and b be acute
...
If



 

c  2 2 a  b  2b , then the angle between b and c

be two unit vectors such that

aˆ  bˆ  2 aˆ  bˆ  2
...





and c  2iˆ  3iˆ  2k
...
,10 is :

6
...


equal to
(a) 10
(b) 7
(c) 9
(d) 14

Let a be a vector which is perpendicular to the vector






1ˆ
j  2kˆ
...
Let
and
a   iˆ  3 ˆj  kˆ,
b  3iˆ   ˆj  4k
 ˆ
c  i  2 ˆj  2k ,  , R , be three vectors
...
Let A, B, C be three points whose position vectors
respectively are :

VECTOR ALGEBRA

a  iˆ  4 ˆj  3kˆ

b  2iˆ   ˆj  4kˆ,   R

c  3iˆ  2 ˆj  5kˆ

4
5

16
...
Let
be
a
triangle
such
that
ABC
      

BC  a , CA  b , AB  c , a  6 2, b  6 3, and
 
b  c  12
consider
the
statements
:
 
 

 S1 : a  b  c  b  c  6 2 2  1

(c)



 







 2
 S 2  : ABC  cos  
...
Let a  iˆ  ˆj  2kˆ and b be a vector such that

 
 
a  b  2iˆ  kˆ and a  b  3
...
Let a   iˆ  ˆj  k and b  2iˆ  ˆj   kˆ, and   0
...
A vector a is parallel to the line of intersection of the
plane determined by the vectors iˆ, iˆ  ˆj and the plane

(a)

determined by the vectors iˆ  ˆj , iˆ  kˆ
...
Then the

 

projection of b  2a on b  a is equal to

(c)

(d)

(a) 2

(b)

39
5

(c) 9

(d)

46
5


and b   iˆ   ˆj  2kˆ
...
Let


a  2iˆ  ˆj  5kˆ



If



(d) 17


18
...
Let a vector b be
(c)

21

such that for every  x, y   R  R   0, 0  , the vector




xa  yb is perpendicular to the vector 6 ya  18 xb
...
Let S be the set of all a  R for which the angle

between the vectors u  a  log e b  iˆ  6 j  3kˆ and

v   log e b  iˆ  2 j  2a  log e b  kˆ,  b  1 is acute
...
Let a  3iˆ  ˆj and b  iˆ  2 ˆj  kˆ
...
If b c are non-parallel,





then the value of  is:
(a) 5
(b) 5
(c) 1
(d) –1
21
...
If  is the angle between the
4
and aˆ  2bˆ  22 aˆ  bˆ , then the







value of 164 cos 2  is equal to :
(a) 90  27 2

(b) 45  18 2

(c) 90  3 2

(d) 54  90 2

VECTOR ALGEBRA
  
22
...
If the product
14
of
their
magnitudes
is
and
   
   
   
a  b  b  c  b  c   c  a    c  a   a  b  168
  
then a  b  c is equal to :





 







(a) 10
(b) 14
(c) 16
(d) 18


ˆ
23
...
If a is a vector such that
 
 
and a  b  21  0,
then
a  b  13iˆ  ˆj  4kˆ
   
  
b  a  k  j  b  a  i  kˆ is equal to


24
...
Given a  b  1 and a  b  3
...
Let a, b , c be three non-coplanar vectors such that

 
  
 

a  b  4c , b  c  9a and c  a   b ,   0
...

36


27
...
Then
2
a is equal to ………
...
(c)

2
...
(b)

4
...
(d)

6
...
(d)

8
...
(c)

10
...
(a)

12
...
(a)

14
...
(a)

16
...
(b)

18
...
(b)

20
...
(a)

22
...
14

24
...
00 25
...
00

26
...
00

27
...
00

Solutions
1
...


2



 aˆ  bˆ   2  aˆ  bˆ   2,   0,  
 aˆ  bˆ  2  aˆ  bˆ   aˆ  bˆ  2  aˆ  bˆ  4
aˆ  bˆ  4  aˆ  bˆ   0  4
2

2

Let the angle be  between aˆ and bˆ
2  2 cos   4 sin 2   4
2  2 cos  – 4 cos 2   0
Let cos   t then
2t 2 – t – 1  0
2t 2 – 2t  t – 1  0
1
or t  1
2
1
cos   –
2
not possible as    0,  
2

3

t–

Now,
 
 2 
S1 2 a  b  2sin 

 3 
 
 2 
a – b  1  1 – 2 cos 

 3 
 1
 2 – 2  – 
 2

VECTOR ALGEBRA
 3
S1 is correct
...




aˆ  aˆ  bˆ
aˆ  bˆ



 2 

3 
2
2  2 cos
3

  1  cos 

1
2
1

1–

1
2
(c) option is true
...



a  a1iˆ  a2 ˆj  a3 kˆ

 1 ˆ 1 ˆ 1 ˆ  ˆ ˆ ˆ

a  
i
j
k 
i  jk
3
3 
3
 3


Now projection of a on b  7
 
a b
  7
b







ˆ ˆ ˆ
ˆ
ˆ
 i  j  k   3i  4 j 
7
5
3

 5 3

a  5 iˆ  ˆj  kˆ

Now b  5 iˆ  ˆj  kˆ    i 
 
a b  0









 25  3  5  0
 15    0    –15

b  5 –2iˆ  ˆj  kˆ

b 5 3





1

  

2


5
b 
–2iˆ  ˆj  kˆ
2

Projection of b on 3iˆ  4 ˆj is

b  3iˆ  4 ˆj
5  –6  4 


 2
5
2 5 





4
...


   

 


 














v   a  b

v   1,1, 2     2, –3,1

v    2  ,  – 3 , 2   

2
 ˆ
 c
v  j  7v   
c
3
 
 – 3  7 v  c  2

  2  –   3  2    2
2  6   2
  3  1
 – 3  7
2  8
4
  –1

We get v   2, 7, 7 
6
...



a  iˆ  ˆj – kˆ

c  2iˆ – 3 ˆj  2k
  
b c  a

b  1, 2
...

 
Now a  c  2 – 3 – 2  –3  0
  
This b  c  a is not possible
...
of vectors b  0

VECTOR ALGEBRA

1  
 
ab  2 2  ab  4 2
2
   

a  1 and a  b  a  b

Area 

 cos   sin 

 
4



 
 a  b  4 2  a b sin  4 2
4

 b 8


 
Now, c  2 2 a  b – 2b




c 

2 2 

2



 


 2
a b  2 b

2

 16 2

 
2
Now, b  c  –2 b
 8  16 2  cos   –2
...


2

 

3
4



a   iˆ  2 ˆj – kˆ, b  –2iˆ   ˆj  kˆ,

Area of parallelogram  aˆ  bˆ
aˆ  bˆ 

  2 

2

  – 2    2  4 
2

Given aˆ  bˆ  15  2  4 

2  2  4    2  4   15  2  4 
2



2

 4   13  2  4 
2

  2  4  13   2  9
2
  2
2 a  a b b





2
a   2  4 1   2 5
2
b  4   2 1   2  5

a  b  –2  2 –1  –1
2
  2
2 a  ab b





2  2  5  – 1 2  5    2  5  14

2

VECTOR ALGEBRA

9
...


 
a  c 10
 
c
3

  6  2 10

  2
1 4  4 3
ˆj
iˆ
kˆ
And 3 –  4  –6iˆ  ˆj  kˆ
1 2 –2


 2  – 8  –6    1
  3

11
...


9
9
82
 16  
4
4
2

  
a b c  0
  
b c  a
2 2
  2
b  c  2b  c  a
2
c  36

c 6
    
S1: a  b  c  b – c
   
a  c  b – c

VECTOR ALGEBRA
  
–b  b – c
0 – 6  –6
  
S2 : a  b  c  0
 

b  c  –a
2 2
 
2
a  b – 2 a b cos  ACB   c
cos  ACB  

13
...


21

 
a  b  1 –   iˆ   2 – 2  ˆj   – 2  kˆ
 
Projection of a  b on iˆ  2 ˆj  2kˆ
 
a  b  iˆ  2 ˆj  2kˆ

 30
3







 2 2    91  0
   7, 

15
...
R of a  iˆ  kˆ

D
...




2 3

3
4



Let a   iˆ  ˆj   kˆ, b  3iˆ  5 ˆj  4kˆ
 
a  b  iˆ  9 ˆj  12kˆ
iˆ

3

ˆj

kˆ


–5 4
1

  4  5  iˆ   3 – 4  ˆj   5  3 kˆ
 iˆ  9 ˆj  12kˆ
 4  5  1,3  4  9, 5  3  12

  1,   3


 a  3iˆ  ˆj  kˆ, b  3iˆ  5 ˆj  4kˆ
 
 a  b   4 jˆ  3kˆ
2
2
a  11, b  50
 
a  b  9   5  4 �

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