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JEE Main Paper

Vector
1. If a = î + ˆj + k̂ & a . b = 1 & a × b = ˆj – k̂ then b is equal to -

(1) 2 î (2) î – ˆj + k̂ (3) î (4) 2 ˆj – k̂

              
2. a , b , c are three non-coplanar vector such that [a  b  c abc 2a  b  c] = k [a b c] then k
is-
(1) 3 (2) –3 (3) 2 (4) 6

       
3. If a  î  ˆj  k̂ and b  î  2ˆj  k̂ , then the vector c such that a . c = 2 and a  c = b is-

1 1
(1) (î  2 ĵ  k̂) (2) (î  2 ĵ  5k̂ )
3 3
1 1
(3) (î  2ˆj  5k̂) (4) (î  2 ĵ  5k̂)
3 3

4. Let a, b, c be distinct non-negative numbers and the vectors a î + a ˆj + c k̂ , î + k̂ , c î + c ˆj + b k̂ lie in a

plane, then the quadratic equation ax2 + 2cx + b = 0 has -
(1) real and equal roots (2) real unequal roots
(3) unreal roots (4) both roots real and positive

^ ^ ^ ^ ^ ^ ^ ^ ^
5. A unit vector is orthogonal to 5 i + 2 j + 6 k and is coplanar to 2 i + j + k and i – j + k then the vector is -

3 ĵ  k̂ 2î  5 ĵ 6î  5k̂ 2î  2 ĵ  k̂

(1) (2) (3) (4)
10 29 61 3

         
6. If three non-zero vectors a , b , c are such that a × b = 2 a × c , | a | = | c | = 1, | b | = 4 and the angle
  1   
between b and c is cos–1   then b – 2 c =  a , where  is equal to :
 4

1 1
(1) ± 2 (2) ± 4 (3) (4)
2 4

 
7. The lines r = î – ˆj + (2 î + k̂ ) and r = 2 î + ( î + ˆj – k̂ ) :

(1) are parallel (2) do not intersect

(3) intersect r = 3 î – ˆj + k̂ (4) none of these

Page No. : 1
  
8. The shortest distance between the lines whose equations are r = t( î + ˆj + k̂ ) and r = k̂ + s ( î – 2 ˆj + 3

k̂ ) is :

3 3 2
(1) 3 (2) (3) (4)
38 14 13

9. Vector equation of the line 6x – 2 = 3y + 1 = 2z – 2 is :

 
1 1 
(1) r = î – ˆj + k̂ +  ( î – ˆj + k̂ ) (2) r = î + 2 ˆj + k̂ +   î – ˆj  k̂ 
3 3 
  1
(3) r = î – ˆj + k̂ + ( î – ˆj + k̂ ) (4) r = ( î – ˆj + 3 k̂ ) + ( î + 2 ˆj + 3 k̂ )
3

   
10. Let the unit vectors a , b , c be the position vectors of the vertices of a triangle ABC. If F is the position
   
vector of the mid point of the line segment joining its orthocentre and centroid then ( a – F )2 + ( b – F )2 + (
  2
c –F) =
(1) 1 (2) 2 (3) 3 (4) None of these

 
11. If the shortest distance between the line r  (î  2 ĵ  3k̂ ) +1 ( 2 î  3 ĵ  4 k̂ ) and r  (2î  4ˆj  5k̂ ) + 2

(3î  4 ĵ  k̂ ) is x, then cos–1(cos 6 x) is equal to -

(1) 1/2 (2) 0 (3) 1 (4) 2


12. The image of the point having the position vector î  3ˆj  4k̂ in the plane r . ( 2 î – ĵ  k̂ ) + 3 = 0 is

(1) 3î  2 ĵ  k̂ (2) 3î  5ˆj  2 k̂

(3) – 3î  5ˆj  2 k̂ (4) 3î  2ˆj – 5k̂

              
13. If a + b + c =  d , b + c + d =  a and [ a b c ]  0 then a + b + c + d equals -
  
(1)  a (2)  b (3) 0 (4) (+) c

14. If the position vector of the points A, B, C, D are 3î  2ˆj  k̂ , 2î  3 ĵ  4k̂ , – î  ĵ  2k̂ , 4î  5ˆj  k̂

respectively. If the points A, B, C, D lie on a plane then the value of  equals-

146 146
(1) (2)
17 17
17 17
(3) (4)
146 146

Page No. : 2
            
15. If a , b , c are non-coplanar vectors such that b × c = a , a × b = c and c × a = b then -
     
(1) [ a b c ] = 1 (2) [ a b c ]  1
  
(3) | a | + | b | + | c | = 3 (4) None of these

 
16. If the vector a = (c log2 x) î – 6 ˆj + 3 k̂ and b = (log2 x) î + 2 ˆj + (2c log2 x) k̂ make an obtuse angle for

any x  (0, ) then the interval to which 'c' belongs is -

(1) (0, ) (2) (– , 0)
(3) (– 4/3, 0) (4) (–1, 0)  (0, 2/3)

     
17. If a , b are non-zero vectors such that | a + b | = | a – 2 b |, then -
   
(1) a.b  2 | b | 2 (2) a.b  | b | 2

 4  4
(3) Least value of a.b   2 is 2 2 (4) Least value of a.b   2 is 2 2 – 1
| b | 2 | b | 2

        
18. If | a | = | b | = | c | = 1 and a . b = b . c = c . a = cos  then maximum value of  is-
 
(1) (2)
2 5
2 
(3) (4)
3 9

   
19. Let a = 1 î + 2 ˆj + 3 k̂ , b = 1 î + 2 ˆj + 3 k̂ and c = 1 î + 2 ˆj + 3 k̂ and | a | = 2 2 makes the angle

1  2 3
   
/3 with the plane of b and c and angle between b & c is /6 then the value of 1  2 3 equals -
1  2 3

  n   n/2
 | a || b |   | c || b | 
(1)   (2)  
 2 3   2 3 
   
  n
 3 (| b || c |)   3 (| a | | b |) 
n

(3)   (4)  
 2   2 
   

         
20. a = î + 2 k̂ , b = î + ˆj + k̂ , c = 7 î – 3 ˆj + 4 k̂ , then the vector d satisfies the relation d × b = c × b , a . d

= 0 is given by –

(1) 2 î + 4 ˆj + k̂ (2) 2 î – 8 ˆj – k̂

(3) 2 î + 5 ˆj +2 k̂ (4) None of these

Page No. : 3
  
   
   3 
21. If the vectors b =  tan ,1, 2 tan  , c =  tan , tan , are orthogonal and

 2   
 sin 
 2 

vector a = (1, 3, sin ) makes an obtuse angle with z- axis then  equals:
(1) (2n +1)+ tan–1 2 (2) (2n + 1)– tan–12
(3) n – tan–1 2 (4) None of these

22. ( a + 3 b – c ). {( a – b ) × ( a – b – c )} equals
(1) 4 [ a b c ] (2) 2 [ a b c ] (3) – 4 [ a b c ] (4) –2 [ a b c ]

         
23. If a  2b  3c  0 , then a  b  b  c  c  a is equal to
     
(1) 6(b  c) (2) 6(c  a ) (3) 6(a  b) (4) none of these

            2    1
24. If a , b, c are such that [ a , b, c ] = 1, c  (a  b), a b  , and | a | = 2 , |b|= 3 , | c | , then the
3 3
 
angle between a and b is
   
(1) (2) (3) (4) 
6 4 3 2

       
25. A unit vector a in the plane b  2î  ĵ and c  î  ˆj  k̂ is such that a  b  a  d where d  ˆj  2 k̂ is

î  ĵ  k̂ î  ˆj  k̂ 2î  ˆj 2î  ˆj
(1) (2) (3) (4)
3 3 3 5

      
26. If a1 , a 2 , a 3 are non-coplanar vectors and (x + y– 3) a 1 + (2x – y + 2) a 2 + (2x + y + ) a 3  0 holds for some
‘x’ and ‘y’ then  is
7 10 5
(1) (2) 2 (3)  (4)
3 3 3

  
27. Let a , b and c be non-coplanar unit vectors equally inclined to one another at an acute angle . Then
  
[a , b , c] in terms of  is equal to

(1) (1 + cos ) cos 2 (2) (1 + cos) 1  2 cos 2

(3) (1 – cos ) 1  2 cos 2 (4) none of these

            
28. Let r , a , b and c be four non-zero vectors such that r . a = 0, | r  b | = | r | | b | , | r  c | = | r | | c | , then [a, b, c] =
(1) | a | | b | | c | (2) – | a | | b | | c | (3) 0 (4) None of these

Page No. : 4
          
29. If b and c are two non-collinear vectors such that a . (b  c) = 4& a × (b  c) = (x2 – 2x + 6) b + (siny) c ,
then the point (x, y) lies on
(1) x = 1 (2) y = 1 (3) y =  (4) x + y = 0

   
30. vectors a , b and c with magnitude 2, 3 & 4 respectively are coplanar. A unit vector d is perpendicular to

    î ĵ k̂     
all of them. If (a  b)  (c  d)    and the angle between a and b is 30°, then c. î  c. ĵ  c. k̂ is
6 3 3
equal to
5 5 5 5
(1) (2) (3) (4)
3 9 12 18

 
31. If the vector a = (c log2x) î  6ˆj  3k̂ and b = (log2x) î  2ˆj + (2c log2x) k̂ make on obtuse angle for any

x(0, ) than the interval to which ‘c’ belongs is

(1) (0, ) (2) (– , 0) (3) (– 4/3, 0) (4) (–1,0)  (0, 2/3)

      
32. If a , b are non zero vectors and a is perpendicular to b then a non zero vector r satisfying r . a = , for
  
some scalar  a × r = b is:
           
 a  ( a  b) a  a  b  a  (a  b )  a  (a  b )
(1)  (2)  (3)  (4) 
| a |2 | b |2 | a |2 | b |2

     
33.  
If (a  b)  (c  d) . ( a × d ) = 0, then which of the following is always true-
       
(1) a , b , c , d are necessarily coplanar (2) either a or d must lie in the plane of b and c
       
(3) either b or c must lie in plane of a and d (4) either a or b must lie in plane of c and d

      
34. a , b, c are three vectors of which every pair is non-collinear. If the vector a  b and b  c are collinear
    
with c and a respectively then a  b  c is -
(1) a unit vector (2) the null vector
  
(3) equally inclined to a , b, c (4) none of these

35. Let , ,  be distinct real numbers. The points with position vectors  î   ĵ  k̂ ,  î  ˆj  k̂ ,

 î  ˆj   k̂ -

(1) are collinear (2) form an equilateral triangle

(3) form a scalene triangle (4) form a right angled triangle

Page No. : 5
 JEE MAIN PAPER (VECTOR)

ANSWER KEY
1. (3) 2. (4) 3. (2) 4. (1) 5. (1) 6. (2) 7. (2)
8. (2) 9. (4) 10. (3) 11. (3) 12. (3) 13. (3) 14. (1)
15. (1) 16. (3) 17. (4) 18. (3) 19. (3) 20. (2) 21. (2)
22. (1) 23. (1) 24. (2) 25. (2) 26. (3) 27. (3) 28. (3)
29. (1) 30. (4) 31. (3) 32. (3) 33. (3) 34. (2) 35. (2)
 Solution for

JEE Main
Practice
Questions Set
28
 HINT & SOLUTION
1.

Sol. Let b =  î +  ˆj +  k̂
 
a × b = ˆj – k̂

î ˆj k̂
 1 1 1 = ˆj – k̂
  

  –  = 0,  = 1,  = 1
  =   = 1 + = 1 + 
 
a . b = 1  = 1
  + 1 +  = 1  = 0
  = 1,  = 0

 b = î .

2.
1 1 1

Sol. STP = 1  1 1 [a b c]
2 1 1

3.
           
Sol. a  b  a  (a  c)  (a . c) a  (a . a) c = 2a  3c

i j k
 
ab  1 11 = 3î  3k̂
1 2 1

 1    1
 c  (2a  a  b) = (  î  2ˆj  5k̂ )
3 3

Page No. : 1
4.
a a c
Sol. 1 0 1 = 0  c2 – ab = 0
c c b

0 = 4c2 – 4ab = 4 (c2 – ab) = 0

2c c
roots are = – =– 0
2a a
So roots are real and equal.

5.

Sol. Let a be the unit vector
 ^ ^ ^ ^ ^ ^
 a = (2 i + j + k ) + ( i – j + k )

 | a | = 1 (2+ )2 + (– )2 + (+ )2 = 1
 62 + 4 + 32 = 1
 ^ ^ ^
a is orthogonal to 5 i + 2 j + 6 k so,

5(2+ ) + 2(– ) + 6(+ ) = 0

18+ 9 = 0  2 = – 
  62– 82 + 122 = 1
 102 = 1
1 2
=±  = 
10 10

 3 ĵ  k̂
 a =
10

6.
   
Sol. a × b =2a × c
  
 a × ( b – 2 c ) = 0
  
b –2 c =a
  
Taking modulus on both sides and then squaring we get, | b – 2 c |2 = 2 | a |2
    
 | b |2 + 4| c |2 – 4| b | | c | cos = 2| a |2
1
 16 + 4 – 4 . 4 . 1. = 2.1
4
 2 = 16
 =±4

Page No. : 2
7.
         
Sol. Line r = a1 +  b1 and r = a 2 +  b 2 do not intersect if ( a 2 – a1 ). ( b1 × b 2 )  0 and are not parallel if
b1  b2.

8.
Sol. Shortest distance between two lines is given by

(î  ˆj  k̂ )  (î – 2 ĵ  3k̂)

= k̂
(î  ˆj  k̂ )  (î – 2 ĵ  3k̂)

(5î – 2 ĵ – 3k̂ )
= k̂
25  4  9

3
=
38

9.
Sol. Here,
x  1/ 3 y  1/ 3 z 1
= = …(i)
1/ 6 1/ 3 1/ 2
  
Line r = a +  b can be written as
x  a1 y  a2 z  a3
= =
b1 b2 b3

Where a = a1 î + a2 ˆj + a3 k̂

and b = b1 î + b2 ˆj + a3 k̂
so the equation (i) can be written as :

1 1 
r  î – ĵ  k̂  + ( î + 2 ˆj + 3 k̂ ).
3 3 

10.
Sol. Let the circumcentre of the triangle be the origin.
  
   a bc
 orthocentre is a + b + c and the centroid is
3
2   
 F = (a + b+ c )
3
     
 ( a – F )2 + ( b – F )2 + ( c – F )2
1         
= [( a – 2( b + c )2 ) + ( b – 2( a + c )2) + ( c –2 ( a + b )2)]
9
1
= (27) = 3
9

Page No. : 3
11.
Sol. Line of shortest distance will be along
î ˆj k̂
2 3 4 = – î  2ˆj – k̂
3 4 5

(– î  2 ĵ – k̂) 1
 x = ( î  2 ĵ  2 k̂ ) . =
6 6

 cos–1 (cos 6 x) = cos–1(cos1) = 1

12.
Sol. Equation of line passing through given point P and normal to given plane is

r = ( î  3ˆj  4 k̂ ) +  ( 2 î – ĵ  k̂ )
Image point Q
 ((1  2 ) î  (3 –  ) ˆj  ( 4   ) k̂ )
Now mid point of
PQ  [(  1) î  (3 –  / 2)ˆj  ( 4   / 2) k̂ ]
satisfies given plane that gives  = –2
So position vector of image is – 3î  5ˆj  2k̂

13.
     
Sol. a + b + c + d = ( +1) d = ( +1) a
  1 
 d= a
 1
      1  
So a + b + c =  d =    a
  1
   1   
 a 1   + b + c = 0
  1 
  
 [ a b c ]  0  = –1
   
 a + b+ c + d=0

14.
         
Sol. If the points are coplanar then their P.V. a , b , c , d must satisfy the condition [ b – a , c – a , d – a ]=0
1 5  3
146
 4 3 3 = 0 = –
17
1 7  1

Page No. : 4
15.
        
Sol. b × c = a  [a b c ] = a . a = |a | 2
      
similarly a × b = c  [ a b c ] = | c |2
      
c × a = b  [a b c ] = | b |2
           
Also, [ a b c ] = [ b × c c × a a × b ] = [ a b c ]2
  
 [a b c ] = 1

16.
 
Sol. For vectors a and b to be inclined at an obtuse angle
 
a . b < 0 , x  (0, )
c(log2 x)2 – 12 + 6 c log2 x < 0  x  (0, )
cy2 + 6 < cy – 12 < 0 y  R y = log2 x
c < 0 and 36 c2 + 48 c < 0
c<0 c (3c + 4) < 0
c  (– 4/3, 0)

17.
       
Sol. 2 a .b = – 4 a .b + 3 b2  2 a .b = b2
  4
a.b +  2
| b | 2
  4
= (1 + a . b ) +   –12 2 –1
2(a . b  1)

18.
     
a .a a . b a . c

       
Sol. [ a b c ]2 = b . a b . b b . c
     
c .a c .b c .c

1 cos  cos 
= cos  1 cos 
cos  cos  1

= 2 cos3  – 3 cos2  + 1
= (2 cos  +1) (1 – cos )2
  
[ a b c ]2 0  1 + 2 cos  0
1 2
 cos  – 
2 3

Page No. : 5
19.
n
1 2 3
  
Sol. 1 2 3 = (| a . ( b × c )|)n
1 2 3

    
= (| a | | b | | c | cos sin )n
6 6
  n
 3 | b || c | 
= 
 2 
 

20.
   
Sol. d × b= c × b
  
 (d – c ) ×b = 0
  
 d = c + b

 d = (7+ ) î + (–3 + ) ˆj + (4 + ) k̂
 
 a .d = 0
 7 + + 2(4 + ) = 0  = – 5

So d = 2 î – 8 ˆj – k̂

21.
   
Sol. b perpendicular c  b . c = 0
 tan2  – tan  – 6 = 0
 tan  = 3, –2

also a make an obtuse angle with z- axis

therefore a . k̂ < 0  sin 2 < 0
2 tan  4
if tan  = 3 then sin 2 = = >0
1  tan 2  3
Now tan 2> 0, sin 2 < 0 third quadrant and tan  = –2
tan ( – ) = 2  = (2n +1)  – tan–1 2, n  I

22.
Sol. ( a + 3 b – c ). {( a – b ) × (– c )}

= – (a + 3b) . {a ×c – b× c }

= a .( b × c ) – 3 b . ( a × c )

= [ a b c ] + 3[ a b c ]

= 4 [a b c ]

Page No. : 6
23.
   
Sol. a  2b  3c  0
       
 a  b  3(c  b) = 0 i.e. a  b  3 (b  c)
        
a  c  2(b  c)  0 i.e. 2(b  c)  c  a
           
 a  b  b  c  c  a  3(b  c)  b  c  2(b  c)
 
 6(b  c)

24.
  
Sol. c  (a  b)
 
   c.c | c | 2 1 1
1 = (a  b). c    
 2   3 3
2 2 
c   (a  b)
1 1 1
 = (a2 b2 sin2) = × 2 × 3 sin2
3 9 9
1 
sin2 =  =
2 4

25.
  
Sol. Let a   b  c
 
a.b a.d
then 
ab ad
     
(b  c).b (b  c).d
i.e. 
b d
[(2î  ĵ)  (î  ˆj  k̂ )].(2î  ˆj)
i.e.
5
[(2î  ˆj)  (î  ˆj  k̂ )].(ˆj  2k̂ )
=
5

i.e.  (4 + 1) + (2 – 1) = (1) + (–1 + 2)

i.e. 4= 0 i.e. =0
 î  ĵ  k̂
 a
3

26.
  
Sol. Since a 1 , a 2 , a 3 are non-coplanar vectors
 x+y–3=0 .....(i)
2x – y + 2 = 0 .....(ii)
2x + y +  = 0 .....(iii)
From (i) and (ii) x = 1/3, y = 8/3
10
 from (iii)  = –
3

Page No. : 7
27.
  
a.a a.b a.c 1 cos  cos 
2
  
Sol. [abc] = b.a b.b b.c  cos  1 cos 
  
c.a c.b c.c cos  cos  1

28.
         
Sol. r . a = 0, | r  b | = | r || b | and | r  c | = | r | | c |
   
 r  a , b, c
  
 a , b, c are coplanar
  
 [a, b, c] = 0

29.
   
Sol. a . b  a . c = 4 ……..(i)
      
& (a.c)b  (a . b)c = (x2 – 2x + 6) b + (siny) c
 
 b & c are non-collinear vectors
 
 a.c = x2 – 2x + 6 & a.b = – siny
putting in (i) we get
x2 – 2x + 6 – siny = 4
 (x – 1)2 + (1 – siny) = 0
 x = 1 & siny = 1

30.
   
Sol. a  b = | a | | b | sin 30° n̂ = 3n̂
  
since d is perpendicular to a & b

 n̂ = ± d
  
 a  b = ± 3 d
      
Now (a  b)  (c  d) = ± 3d  (c  d)
    î ĵ k̂  î  2ˆj  2k̂
 ± 3 {(d.d)c  (d.c)d}    or ± 3c 
6 3 3 6
 1  1  1
 | c. î |  , | c. ĵ |  , | c.k̂ | 
18 9 9

31.
 
Sol. For vectors a & b to be make an obtuse angle
 
a . b < 0, x(0, )
c(log2x)2 – 12 + 6c log2x < 0 for all x (0,)
cy2 + 6cy – 12 < 0 yR y = log2x
c < 0 and 36c2 + 48c < 0
c < 0 c(3c + 4) < 0
 4 
c   , 0 
 3 

Page No. : 8
32.
  
Sol. a× r=b
    
 a × (a × r ) = a ×b
       
or ( a . r ) a – ( a . a ) r = a × b

33.
Sol. (a  b )  (c  d ) . a  d  = 0
     
 [a c d] b  [b c d] a  . a  d  = 0
    

 [a c d ] [ b a d ] = 0
   
   either c or b must lie in the plane of a and d .

34.
     
Sol. Here, a  b = t c , b  c = sa . Subtracting,
   
a  c = t c – sa
 
or (1 + s) a = (1 + t) c .
 
But a , c are non-collinear.

  
 1+ s = 0, 1 + t = 0. hence, a + b = – c .

35.
Sol. Let P, Q and R be points having position vectors

 î   ĵ  k̂ ,  î  ˆj  k̂ and  î  ˆj   k̂ respectively.

  
Then | PQ | = | QR | = | RP | = (  ) 2  (  ) 2 (  ) 2

Hence, PQR is an equilateral triangle.

Page No. : 9