Title 01. Vectors Easy Ans.pdf ✅

1. (d) As the multiple of j ˆ in the given vector is zero therefore this vector lies in XZ plane and projection of this vector on y- axis is zero. 2. (b) If a point have coordinate (x, y, z) then its position vector . ˆ ˆ ˆ = xi + yj + zk 3. (c) Displacement vector r xi yj zk ˆ ˆ ˆ =  +  +  i j k ˆ (5 5) ˆ (4 3) ˆ = (3 − 2) + − + − i j ˆ ˆ = + 4. (d) The component of force in vertical direction = F cos = F cos 60 2.5 N 2 1 = 5  = 5. (d) 2 2 | B| = 7 + (24) = 625 = 25 Unit vector in the direction of A will be 5 ˆ 4 ˆ 3 ˆ i j A + = So required vector =         + 5 ˆ 4 ˆ 3 25 i j i j ˆ 20 ˆ = 15 + 6. (a) Let the components of A makes angles ,  and  with x, y and z axis respectively then  =  =  cos cos cos 1 2 2 2  +  +  =  3 cos 1 2  =  3 1 cos = 3 cos A  Ax = Ay = Az = A  = 7. (a) A i j k ˆ 5 ˆ 4 ˆ = 2 + −   | | (2) (4) ( 5) 45 2 2 2 A = + + − =  45 5 , cos 45 4 , cos 45 2 cos −  =  =  = 8. (b)Unit vector along y axis j ˆ = so the required vector )] ˆ 7 ˆ 6 ˆ ) (3 ˆ 2 ˆ 3 ˆ [( ˆ = j − i − j + k + i + j − k i j k ˆ 5 ˆ 2 ˆ = − 4 − + 9. (b) F3 F1 F2    = + There should be minimum three coplaner vectors having different magnitude which should be added to give zero resultant 10. (d)Diagonal of the hall = 2 2 2 l + b + h 2 2 2 = 10 + 12 + 14 = 100 +144 +196 = 400 = 20m 11. (d) Total angle =   2 50 100  = So all the force will pass through one point and all forces will be balanced. i.e. their resultant will be zero. 12. (d) Total angle =   2 50 100  = So all the force will pass through one point and all forces will be balanced. i.e. their resultant will be zero. 13. (a) P i j ˆ 2 1 ˆ 2 1 = +   2 2 2 1 2 1 | |         +         P =  = 1  It is a unit vector. 14. (b) 15. (c) R ˆ i j i j R R ˆ 2 1 ˆ 2 1 1 1 ˆ ˆ | | 2 2 = + + + = =  16. (c) R i j k ˆ 2 ˆ ˆ = 3 + +   Length in XY plane = 2 2 2 2 Rx + Ry = 3 + 1 = 10 17. (a) If the angle between all forces which are equal and lying in one plane are equal then resultant force will be zero. 18. (b) A i j ˆ ˆ = +   | | 1 1 2 2 2 A = + = = = = cos 45 2 1 | | cos A Ax    = 45 19. (c) 20. (c) 21. (d) All quantities are tensors.
22. (d) P Q PP ˆ QQ ˆ + = +   23. (b) r a t i a t j ˆ ( sin ) ˆ = ( cos ) +   a t i a t j dt dr v ˆ cos ˆ = = −  sin +     As r.v = 0   therefore velocity of the particle is perpendicular to the position vector. 24. (d) Displacement, electrical and acceleration are vector quantities. 25. (b) Magnitude of unit vector = 1  (0.5) (0.8) 1 2 2 2 + + c = By solving we get c = 0.11 26. (b) Displacement AC = AB + BC AC (AB) (BC) (400 ) (300 ) 500 m 2 2 2 2 = + = + = Distance = AB + BC = 400 + 300 = 700 m 27. (a) Resultant of vectors A and B R A B i j k i j k ˆ 8 ˆ 3 ˆ ˆ 6 ˆ 3 ˆ = + = 4 + + − + − R i j k ˆ 2 ˆ 6 ˆ = 3 + − 7 ˆ 2 ˆ 6 ˆ 3 3 6 ( 2) ˆ 2 ˆ 6 ˆ 3 | | ˆ 2 2 2 i j k i j k R R R + − = + + − + − =  = 28. (a)  = B.A . In this formula A is a area vector. 29. (a) r a b c     = + + i j i j k ˆ ˆ 2 ˆ 3 ˆ ˆ = 4 − − + − i j k ˆ ˆ ˆ = + − 3 ˆ ˆ ˆ 1 1 ( 1) ˆ ˆ ˆ | | ˆ 2 2 2 i j k i j k r r r + − = + + − + − = =  30. (d) 9 16 25 9 16 25 9 16 25 | || | . cos + + + + + + = = A B A B  = 1 50 50 =  cos = 1  cos (1) −1  = 31. (a) r t i t j k ˆ 7 ˆ 4 ˆ 3 2 2 = + +  t = 0 , r k ˆ 1 = 7  at t = 10 sec , r i j k ˆ 7 ˆ 400 ˆ 2 = 300 + +  , r r r i j ˆ 400 ˆ  = 2 − 1 = 300 +   | r| | r r | (300 ) (400 ) 500 m 2 2  = 2 − 1 = + =   32. (b) Resultant of vectors A and B R A B i j i j ˆ 8 ˆ 8 ˆ 3 ˆ = + = 4 − + + i j ˆ 5 ˆ = 12 + 2 2 (12) (5) ˆ 5 ˆ 12 | | ˆ + + = = i j R R R 13 ˆ 5 ˆ 12i + j = 33. (a) 2 ) ˆ ˆ )( ˆ 3 ˆ (2 | | . i j i j i j A B + + = +   2 5 2 2 3 = + = 34. (a) 9 16 25 9 16 25 ) ˆ 5 ˆ 4 ˆ )(3 ˆ 5 ˆ 4 ˆ (3 | || | . cos + + + + + + + − = = i j k i j k A B A B  0 50 9 16 25 = + − =  cos = 0 ,   = 90 35. (a) For 17 N both the vector should be parallel i.e. angle between them should be zero. For 7 N both the vectors should be antiparallel i.e. angle between them should be 180° For 13 N both the vectors should be perpendicular to each other i.e. angle between them should be 90° 36. (b) A B i j i j i j ˆ 5 ˆ 10 ˆ 8 ˆ 6 ˆ 3 ˆ + = 4 − + + = + 2 2 | A + B| = (10) + (5) = 5 5 2 1 10 5 tan = =        = − 2 1 tan 1  37. (d) From figure v j ˆ 1 = 20 and v i ˆ 2 = −20) ˆ ˆ v = v 2 − v1 = −20(i + j | v | = 20 2 and direction = =  − tan (1) 45 1  i.e. S–W 38. (b) Let 1 n ˆ and 2 n ˆ are the two unit vectors, then the sum is 1 2 n n ˆ n ˆ s = + or 2 1 2 cos 2 2 2 1 2 ns = n + n + n n = 1 + 1 + 2cos
Since it is given that s n is also a unit vector, therefore 1 = 1 + 1 + 2cos  2 1 cos = −   = 120  Now the difference vector is 1 2 n ˆ n ˆ n ˆ d = − or 2 1 2 cos 2 2 2 1 2 nd = n + n − n n = 1 + 1 − 2 cos(120 )  2 2( 1 / 2) 2 1 3 2 nd = − − = + =  nd = 3 39. (b) ) ˆ 2 ˆ ) 3(6 ˆ ˆ ) 2(3 ˆ ˆ A − 2B + 3C = (2i + j − j − k + i − k i j j k i k ˆ 6 ˆ 18 ˆ 2 ˆ 6 ˆ ˆ = 2 + − + + − = i j k ˆ 4 ˆ 5 ˆ 20 − − 40. (a) P mv i mv j ˆ cos ˆ 1 = sin −  and P mv i mv j ˆ cos ˆ 2 = sin +  So change in momentum , ˆ P = P2 − P1 = 2m v cos j | P| = 2m v cos 41. (b) 2 cos 2 2 R = A + B + AB By substituting, A = F, B = F and R = F we get =  = 120  2 1 cos  42. (a) 43. (d) If two vectors A  and B  are given then the resultant Rmax = A + B = 7N and Rmin = 4 − 3 = 1N i.e. net force on the particle is between 1 N and 7 N. 44. (b) If C  lies outside the plane then resultant force can not be zero. 45. (d) 46. (c) = + + 2 cos 90  1 2 2 2 2 F F1 F F F 2 2 2 = F1 + F 47. (a) 48. (c) 49. (c) 2 2 C = A + B The angle between A and B is 2  50. (c) R A B    = + = i j i j ˆ 4 ˆ 3 ˆ 7 ˆ 6 + + + = i j ˆ 11 ˆ 9 +  | | 9 11 81 121 202 2 2 R = + = + =  51. (c) R 12 5 6 144 25 36 205 14.31 m 2 2 2 = + + = + + = = 52. (c) A i j k ˆ ˆ 2 ˆ = 3 − +  , B i j k ˆ 5 ˆ 3 ˆ = − +  , C i j k ˆ 4 ˆ ˆ = 2 − +  | | 3 ( 2) 1 9 4 1 14 2 2 2 A = + − + = + + =  | | 1 ( 3) 5 1 9 25 35 2 2 2 B = + − + = + + =  | | 2 1 ( 4) 4 1 16 21 2 2 2 A = + + − = + + =  As 2 2 B = A + C therefore ABC will be right angled triangle. 53. (c) 54. (b) C A B    + = . The value of C lies between A − B and A + B  | C| | A| or | C| | B|       55. (a) 56. (d) 57. (d) Here all the three force will not keep the particle in equilibrium so the net force will not be zero and the particle will move with an acceleration. 58. (a) A + B = 16 (given) ...(i) =  + = tan 90 cos sin tan    A B B  A + Bcos = 0  B −A cos = ...(ii) 8 2 cos 2 2 = A + B + AB ...(iii) By solving eq. (i), (ii) and (iii) we get A = 6N, B = 10 N 59. (c) | P| = 5  , | Q| = 12  and | R| = 13  13 12 cos = = R Q         = − 13 12 cos 1  60. (b) 2 cos 2 2 2 A B AB B = + + ...(i)  cos 0 cos sin tan 90  + = +  =    A B A B B  B A cos = − Hence, from (i) 2 2 3 4 2 2 2 2 B A B A A B = + −  =
 2 3 cos = − = − B A    = 150  61. (b) i − j + k + i + j −k + R = i  ) ˆ ˆ ˆ ) (2 ˆ 2 ˆ 2 ˆ (  Required vector R  = i j k ˆ ˆ ˆ − 2 + − 62. (a) Resultant R P Q P Q P       = + + − = 2 , The angle between P  and P  2 is zero. 63. (b)    cos sin tan 90 P Q Q +  =  P + Q cos  = 0 Q −P cos =          − = − Q 1 P  cos 64. (a) According to problem P + Q = 3 and P − Q = 1 By solving we get P = 2 and Q = 1  = 2 Q P  P = 2Q 65. (c) 66. (c) 67. (c) 68. (d) F1 + F2 + F3 = 0  0 ˆ 6 ˆ 4i + j + F3 =  F i j ˆ 6 ˆ 3 = −4 −  69. (a)  =          = 2 sin90 2 v 2v sin v  = 2100 = 200 km/hr 70. (c) 71. (d) Resultant velocity 2 2 = 20 + 15 = 400 + 225 = 625 = 25 k m/hr 72. (a) 2 2 C = A + B = 3 4 5 2 2 + =  Angle between A and B is 2  73. (c) 74. (d) If the magnitude of vector remains same, only direction change by  then 2 1 v = v − v , ( ) 2 1 v = v + −v Magnitude of change in vector        = 2 | | 2 sin  v v         =   2 90 | v| 2 10 sin = 10 2 = 14.14 m / s Direction is south-west as shown in figure. 75. (a) AC = AB + BC2 2 AC = (AB) + (BC) 2 2 = (10) + (20) = 100 + 400 = 500 = 22.36 km 76. (b) | || | . cos 1 2 1 2 F F F F  = 25 100 400 100 25 225 ) ˆ 15 ˆ 5 ˆ ).(10 ˆ 20 ˆ 10 ˆ (5 + + + + + − − − = i j k i j k 525 350 50 − 50 + 300 =  2 1 cos =   = 45 77. (d) If two vectors A and B are given then Range of their resultant can be written as (A − B)  R  (A + B). i.e. Rmax = A + B and Rmin = A − B If B = 1 and A = 4 then their resultant will lies in between 3N and 5N. It can never be 2N. 78. (d) A = 3N , B = 2N then 2 cos 2 2 R = A + B + AB R = 9 + 4 +12 cos ...(i) Now A = 6N , B = 2N then 2R = 36 + 4 + 24 cos ...(ii) from (i) and (ii) we get 2 1 cos = −   = 120  79. (a)In N forces of equal magnitude works on a single point and their resultant is zero then angle between any two forces is given N 360  = = = 120  3 360 If these three vectors are represented by three sides of triangle then they form equilateral triangle –v1 → v1 → v → v2 →  E N S W