Vector Algebra Questions and Answers

JEE ADVANCED - VOL - I VECTOR ALGEBRA
 ^ ^  ^ ^ ^
60. OA  4 i  3 k ; OB  14 i  2 j  5 i
MULTI ANSWER QUESTIONS ^ ^ ^ ^ ^
^ 4 i  3 k ^ 14 i  2 j  5 k
a ;b 
  5 15
55. Let a  2iˆ  4 ˆj  5kˆ and b  iˆ  2 ˆj  3kˆ.
   ^ ^ ^ ^ ^

Then the diagonals of the parallelogram are r   12 i  9 j  14 i  2 j  5 k 
 ^ ^ ^  ^ ^ ^
15  
p  3i 6 j 2 k, q   i 2 j 8 k    ^ ^ ^

So, unit vectors along the diagonals are r  2 i  2 j  5 k 
15  
1 ^ ^ ^
 1  ^ ^ ^

 3 i  6 j  2 k  and   i  2 j  8 k   2  ^ ^ ^

7  69   r i  j  2 k

15  
56. We have ,
 ^ ^ ^  ^ ^
1 2 3
AB   i  j  4 k , BC  3 i  3 j and
0   0
 ^ ^ ^ 61. For coplanar vectors,
CA  4 i  2 j  4k 0 0 2  1
  
Therefore. AB  BC  3 2 and CA  6 1
  2  1   0    0,
 2  2  2 2
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Clearly , AB + BC = AC  ^ ^ ^  ^ ^ ^
62. Let   i  x j  3 k ,   4 i   4 x  2  j  2 k
Hence , the triangle is right -angled isosceles
triangle  
Given, 2   
57. use  a b c   0
2
 2 10  x 2  20  4  2 x  1
x 2 5
58.    k  say 
1 y z  10  x 2  5   4 x 2  4 x  1
where k  R
        3x 2  4 x  4  0
59. a  2 i  2 j  k,b  i  2 j  2k
  2
 | a | 3, | b | 3  x  2, 
3
 a vector along bisectors
 2 2 2 

a b
      
2 i  2 j  k i  2 j  2k
63. We have, a  b  a  b  2 a.b  
     
|a| |b| 3 3  2 2 2  
 a  b  a  b  2 a b cos 2
 1 1 4 
 i  k, i  j  k
 2  
3 3 3
 the required vector = 2

 a  b  2  2 cos 2  a  b  1 
 1  2
1 4   a  b  4sin 2 
i k i  j k
3 3 3
.
2
,2.
2 2  
2  1  1  4 2  a  b  2 sin 
1     3    3  1
 3    
 
Now, a  b  1

 2 sin   1
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, VECTOR ALGEBRA JEE ADVANCED - VOL - I

1        
 sin  
2
Hence, a  b  b  c  c  a  2 a  b  
   
   5   
or 6 b  c or 3  c  a 
   0,  or    ,       
 6  6  67. a  b  c,b  c  a
    
64. a  b  a  2b Taking cross with b in first equation, we get
     
2  
b  a b  b c  a
  b 2      
 a.b  
2  
 b a  a.b b  a  b  1 and a.b  0
2       
  1 b 2 1 Also a  b  c  a b sin  c
a .b    2 1 2
Also 2  
b 2 2 b 2 a c

 2  1 Using A.M.  G.M. 1
68. x.y  y.z  z.x  2  2   1.
  2
65. Consider V1.V2  0  A  900
       
Let a   x  y  z   x.z y  x.y z
A

 
   
 yz 
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V1  3 a  b  
V2  b  a.b a


a.y  

2
 
 a  a.y y  z 

C
Similarly b   b.z  z  x 
B


  
 
b  aˆ.b a 3 aˆ  b a.b    a.y  b  z 
Using the sine law, 
sin  cos  69. The given vector equation can be written as
^ ^
      1  a  x  3 y  4 z  i   x  (3  a ) y  5 z  j
1 b  a .b

a 1  
a b a   
 tan       ^
3 a b 3 a b  3x  y  az  k  0
 (1-a)x+3y-4z=0
  

1 a  b a sin 90
0
1
x-(3+a)y+5z=0 and 3x+y-az =0
    The above system of equation has a non-trivial
3 a b 3 solution

 1 a 3 4
 
6  1  3  a  5  0
 
66. We know that a  b  c  0 , then 3 1 a
     
a b  b c  c a  a  0, 1
 
Given a  2b  3c  0  ^ ^
      70. We have a  2 p i  j
 2a  b  6b  c  3c  a
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