Vector Algebra Questions and Answers pt4

JEE ADVANCED - VOL - I VECTOR ALGEBRA

3    6.................  iii 
MULTI ANSWER QUESTIONS
on solving the equations (i), (ii) & (iii) , we have
49. As forces are in equilibrium we have   1,   2,   3
    
F1  F2  F3  F4  0 aˆ  ^ ^  bˆ  ^ ^ 
53.  . a b   . a b 
^ ^ ^ a   b  
equating the coefficients of i , j , k and solve the
    
equations a b
50. Let point A be taken as origin. Then the position
    

or  a b

. a  b  0
   
vectors of B,C and D are a, a  b and b  a  
 
respectively which is possible when a  b
     
aba  b a b
P.V.of M a
2 2  
If angle between a and b is  , then
 DN : NM  4 :1   

 b   a.a  a a cos 
4 a    b  a
P.V. of N   2 3   cos   1 ,   00
5
 ab
5
 
 3  54. cos 2   sin 2   sin 2   1
AN  AC
1
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5
 3sin 2   1  sin 2  
51. Given that 3
   
 
3 4q  3r  2q  3r
 sin  
1
  3
 10q  12r
 6 tan   2
 q   r.......  i         
5 2) a.b  c  0 , b c 
 a  
0 , c 
ab  0 
   5     
(i) p =4 q + 3    q  by u sin g (i )   a.b  b.c  c.a  0
 6  2 2 2 
 3  a  b  a  b  2a.b
 p q  
2  9  16  2 a b sin 
   
 p  q and directions of p and q are same  2
a  b  25  2  3 4  sin 
  6   
(ii) p  4  r   3r = 25  24sin 
 5   
 9  a  b  25  24sin 
p r  
5
52. We have, a  b  25  24  7
        
    
 i  3k   2i  j  r i  j  k  a  b  25  24  1
    
 2i  5 j  6k range of a  b is 1,7   25  24  1
On comparing, we get  
  2    2............  i  3) similarly we have calculate range of b  c
    5 .................(ii)
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, VECTOR ALGEBRA JEE ADVANCED - VOL - I
 57. According to the given conditions a.b  0 and
55. Let c  c1iˆ  c2 ˆj  c3kˆ
2 b.c  0 , where c   0,1,0  . Thus
 c12  c22  c32  iˆ  ˆj  2
2x 2  3x  1  0 for x  0 . But
Also,
2x 2  3x  1 is greater than zero for all x  0
58. Let b  xi  yj . Since a is perpendicular to b so
 c iˆ  c ˆj  c kˆ  .iˆ  ˆj    c iˆ  c ˆj  c kˆ   1
1 2 3 1 2 3
 4 
2 2 2 4x  3y  0 . Thus b  x  i  j  . Let
 3 
 c1  c2  c2  c3  1  c1  c3 and
c  ui  vj be the required vector. According
c2  1  c3 to the given condition
2 1 c.a
Hence from (1), 3c3  2c3  1  0  c3   1  4u  3v  5 .
3 a
1
or 1 when c3   c.b ux    vx
3 2  2
Also b x 2 1    

c   iˆ  4 ˆj  kˆ when c3  1, c  iˆ  kˆ
1
3
   3u  4v  10
Solving these equations we have
D C
u  2 and v  1 or u   ,v  11/ 5
59. The diagonals are given by
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 6, 1
E AB  BC  4i  2j  4k, AB  BC  2i  2j
56. These vectors have magnitudes 6 and 2 2 ,
respectively, and their dot product is 12.
B
A 1, 2  Therefore the angle between them is
E is the mid-point of A and C 12 1  3
 cos1  cos1  or
Position vector of C is 11i  6  2 2  2 4 4

AEB  DEA  60.
2 ADE is right angled at E
AB 2  AD 2  2 AE 2    h  Area ABC 
1
If B or D is  x, y  3
  2 2 1  
BE. AE  0 h   AB  AC
  3 2
and DE. AE  0
 5 x  y  29 D
2 2
and  x  1   y  2   52
4 h
 x 2  12 x  35  0
A1,1,1
 x  5, y  4
2
x  7, y  6 E
      B D1 C
OB  7i  6 j OD  5i  4 j and area = 1, 0, 0   2, 0, 0   3, 0, 0 
52 sq.units

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