Test-1-: Vector Algebra Time-:1hr.

Multi choice single correct ( +3,-1)

 
Q1. The component of vector P  6iˆ  2jˆ  5kˆ along vector Q  3iˆ  4ˆj is
(A) 2 (B) 3 (C) 4 (D) 4

Q2. The dot product of the vectors

 
A = 2 î  3 ĵ + k̂ and B = 3 ĵ + 2 k̂
(A) 7 (B) 0 (C) 12 (D) 13
 
Q3. For the resultant of two non-zero vectors A and B to be maximum, then
       
(A) A  B = 0 (B) A  B = 0 (C) | A  B | AB (D) A  B = 0
 
Q4. Find the component of A  2iˆ  3ˆj along the vector B  3iˆ  4ˆj .
(A) 18/5 (B) 18/7 (C) 18 (D) 15
     2 2 2
Q5. Two vectors A and B are such that A  B  C and A + B = C . Which of the
following is correct?
   
(A) B and A are parallel (B) B is perpendicular to A
 
(C) A and B are equal in magnitude (D) none of the above

Q6. The angle between ˆi  ˆj  kˆ and ˆi  ˆj is

(A) /6 (B) /4 (C) /3 (D) /2

 
    
Q7. Two vectors A and B are  i  j  and  2 i  12 j  respectively. Find the angle
   
 
between A and B
 3  1  3  1
(C) sin 1  (D) cos 1  
0 0
(A) 105 (B) 165
 2 2   2 2 
   
  
Q8. Let a  2iˆ  ˆj  kˆ , b  ˆi  2jˆ  kˆ and c  ˆi  ˆj  2kˆ be three vectors. A vector in the
   2
plane of b and c whose projection on a is of magnitude is
3
(A) 2iˆ  3ˆj  3kˆ , – 2iˆ  ˆj  5kˆ (B) 2iˆ  3ˆj  3kˆ , 2iˆ  ˆj  5kˆ
(C) – 2iˆ  ˆj  5kˆ , 2iˆ  3ˆj  3kˆ (D) 2iˆ  ˆj  5kˆ , 2iˆ  3ˆj  3kˆ
 
Q9. If vectors P  3iˆ  ajˆ  kˆ and Q  2iˆ  ˆj  2kˆ are mutually perpendicular, then the
value of ‘a’ is
(A) 1 (B) 2 (C) 3 (D) 4
   
Q10. If P  ˆi  2jˆ  3kˆ and Q  2iˆ  ˆj  kˆ then the angle between the vectors (2P  Q) and
 
(P  2Q) is
(A) 30 (B) 37 (C) 45 (D) 60
       
Q11.. If P  3 i  4 j and Q  2 i  3 j, what is the value of P  2 Q ?


(A) 5  2 13  
(B) 5  2 13
101  (D) 99
( C)
Q12. Six forces are acting on a particle. Angle between two adjacent force is 60. Five of
the forces have magnitude F1 and the sixth has magnitude F2(<F1). The resultant of
all the forces will have magnitude of:

1
 (A) zero (B) F1 + F2 (C) F1  F2 (D) F2 
     c
Q13. If a, b, c and d such that they make a closed quadrilateral now c is
reversed remaining all other vector are unchanged. The new
 
resultant will be
     d b
(A) 2c (B) 2c (C) d  a  b 
a
Q14. At what angle the two vectors of magnitudes (A + B) and (A  B) must act, so that the
resultant is A 2  B2 ?
A 2  B2 A 2  B2 A 2  B2 A 2  B2
(A) cos1 (B) cos1 (C) cos1 (D) cos1
A 2  B2 B2  A 2 2(A 2  B2 )
2(B2  A 2 )
Q15. If  and B̂ is a unit vector and its resultant is also unit vector. Then Aˆ  3Bˆ is equal
to
(A) 13 (B) 3 3 (C) 3 (D) 2 3
 
Q16. A vector C having magnitude equal to that of A  3iˆ  5ˆj  2kˆ and directed along

vector B  2iˆ  ˆj  2kˆ is given by
(A) 4iˆ  2jˆ  4kˆ (B) 6iˆ  3ˆj  6kˆ (C) 3iˆ  2jˆ  3kˆ (D) 5iˆ  3ˆj  5kˆ
        
Q17. If A  B  C and A + B = C , where A , B and C are any three vectors. What is the

 
angle between B and C ?
0 0 0
(A) 0 (B) 90 (C) 180 (D) Date insufficient
Q18. The resultant of two equal forces is less than either of force. If angle between the
vector is , then
(A) 120° <  < 180 (B) 90° <  < 120 (C) 30° <  < 90 (D) None
Q19. Forces 3N , 4N and 12N act at a point in mutually perpendicular directions. The
magnitude of resultant force in Newton is
(A) 12 (B) 13 (C) 7 (D) 19
 
Q20. If P  miˆ  2ˆj  2kˆ and Q  2iˆ  njˆ  kˆ are parallel to each other, then
(A) m = 2, n = 3 (B) m = 4, n = 1 (C) m = 3, n = 2 (D) m = 1, n = 4
Q21. The minimum and maximum magnitude which is possible by adding four forces of
magnitudes 1N, 3N, 9N and 10N is
(A) 0 and 23 N (B) 1N and 23 N
(C) 2N and 23 N (D) 3N and 23 N
  
Q22. a, b and c are three orthogonal vectors with magnitudes 3, 4 and 12 respectively.
  
The value of a  b  c will be
(A) 19 (B) 96
(C) 13 (D) none

Q23. The angle made by the vector V  3iˆ  4jˆ  12kˆ with the Z axis is
 5 4
(A) sin 1   (B) tan 1  
 13  3
13
  4
(C) cos1   (D) cot 1  
 12  3
Multi choice multi correct ( +4,-2,0)
   
Q24. If A  B  C  0, which of the following can be correct?
(A) A + B > C (B) A – B < C (C) A + B = C (D) A – B = C
    
Q25. The two vectors A and B are drawn from a common point and C  A  B , then angle
 
between A and B is
0 2 2 2 0 2 2 2
(A) 90 if C = A + B (B) greater than 90 if C < A +B
0 2 2 2 0 2 2 2
(C) greater than 90 if C > A + B (D) less than 90 if C > A + B

2
   
Q26. Vector b with magnitude 5 units is added to a  5iˆ  5 3 ˆj . If b can assume any
 
direction then maximum and minimum inclination of a  b from x-axis is
(A) max = 60 (B) max = 90
(C) min = 30 (B) min = 45
   

Q27. If the vectors i  j, j  k and a form a triangle, then a may be
    
(A)  i  k (B) i  2 j k
   
(C) 2iˆ  j  k (D) i  k

 dA
Q28. A can be:
dt
(A) 1 (B) 0 (C) negative (D) infinite

  
Q29. Vectors A and B are perpendicular to each other while C is any vector coplanar
 
with A and B . Therefore
   
(A) A can be expressed as A  xB  yC where x and y are scalars.
  
(B) A  (B  C) = 0
     
(C) magnitude of (A  B)  C is simply the product of magnitudes of A , B and C .
 
(D) A  B = 0

Q30. Which of the following operation(s) is/ are not obeying the commutative Law
( A) Addition of Two vector ( B) Subtraction of Two vector
( C) Dot product of Two vector ( D) Cross product of Two vector
Integer type ( +3,-1,0)
    y
Q31. A and B are shown in the figure. where | A | = | B | = 5 units.
     B
C is a vector such that A  B  C = 0. If the magnitude of C is
x
k a , where k and a are positive integer, ranging between 0 to A
9, then the value of k + a is……….

Q32. If the vector that points from the origin towards a position half way between the tips of
   aiˆ  bjˆ  ckˆ
two position vectors r1  4iˆ  2kˆ and r2  ˆi  3ˆj  2kˆ is r  where a, b,c
d
abc
and d are positive integer, ranging between 0 to 9, then find
d
Paragraph for Q. No. Q33 to Q35

In parallelogram ABCD, two points X and Y Y 4

D 1 C
are taken on the side BC and CD
respectively, such that BX : XC = 4 : 1 and  Z 1
d
CY : YD = 4 : 1 as shown in the figure. Let X
 
b and d be the position vectors of B and D 4
points w.r.t. point A as origin. The line XY  
A B
cuts the diagonal AC at point Z. O b
Q33. The ratio XZ : ZY =
(A) 1 : 4 (B) 21 : 4
(C) 1 : 3 (D) None
Q34. The ratio AZ : ZC =
(A) 1 : 4 (B) 21 : 4
(C) 1 : 3 (D) None

3
Q35. The P.V. of Z is given by
   
bd 4(b  d)
(A) (B)
25 25
21  
(C) (b  d) (D) none
25
Paragraph for Q. No. Q36 to 38
A mosquito net over a 7 ft  4 ft bed is 3 ft high. The net has a hole at one corner of the bed
through which a mosquito enters the net. Take the hole as origin, the length of the bed as the
x-axis, its width as the y-axis and vertically up as the z-axis. Now answer the following
question based on the paragraph.

Q36. What is the unit vector along the plane of the bed, which is directed away from the
hole along the diagonal of the bed?
(A) 7iˆ  4ˆj (B) 4iˆ  7ˆj
7iˆ  4jˆ 4iˆ  7jˆ
(C) (D)
65 65

Q37. What is the magnitude of displacement of the mosquito, if it comes out of the opposite
diagonal of the net?
(A) 14 ft (B) 65 ft
(C) 74 ft (D) can’t be determined.
  
Q38. If the length of the bed is  , width is b and height is taken as h . Then, find the
  
value of (   b)  h
3 3
(A) 28 ft (B) 84 ft
3 3
(C) 21 ft (D) 12 ft