USS Physics Vectors PDF

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 Vectors 1 y ˆj x k̂ î z Fig. 0.1 Chapter 0 Vectors Introduction of Vector (7) Orthogonal unit vectors ˆi , ˆj and kˆ are called orthogonal unit Physical quantities having magnitude, direction and obeying laws of vectors. These vectors must form a Right Handed Triad (It is a coordinate vector algebra are called vectors. system such that when we Curl the fingers of right hand from x to y then Example : Displacement, velocity, acceleration, momentum, force, we must get the direction of z along thumb). The impulse, weight, thrust, torque, angular momentum, angular velocity etc. If a physical quantity has magnitude and direction both, then it does ˆi  x , ˆj  y , kˆ  z not always imply that it is a vector. For it to be a vector the third condition x y z of obeying laws of vector algebra has to be satisfied.  x  xˆi , y  yˆj , z  zkˆ Example : The physical quantity current has both magnitude and (8) Polar vectors : These have starting point or point of application. direction but is still a scalar as it disobeys the laws of vector algebra. Example displacement and force etc. Types of Vector (9) Axial Vectors : These represent rotational effects and are always along the axis of rotation in accordance with right hand screw rule. Angular (1) Equal vectors : Two vectors A and B are said to be equal when they velocity, torque and angular momentum, etc., are example of physical have equal magnitudes and same direction. quantities of this type. (2) Parallel vector : Two vectors A and B are said to be parallel Axial vector Axis of rotation when (i) Both have same direction. (ii) One vector is scalar (positive) non-zero multiple of another vector. Anticlock wise rotation Clock wise rotation (3) Anti-parallel vectors : Two vectors A and B are said to be Axis of rotation Axial vector Fig. 0.2 anti-parallel when (10) Coplanar vector : Three (or more) vectors are called (i) Both have opposite direction. coplanar vector if they lie in the same plane. Two (free) vectors are always (ii) One vector is scalar non-zero negative multiple of another coplanar. vector. (4) Collinear vectors : When the vectors under consideration can Triangle Law of Vector Addition of Two Vectors share the same support or have a common support then the considered If two non zero vectors are represented by the two sides of a vectors are collinear. triangle taken in same order then B the resultant is given by the (5) Zero vector (0 ) : A vector having zero magnitude and arbitrary closing side of triangle in opposite R  AB direction (not known to us) is a zero vector. order. i.e. R  A  B (6) Unit vector : A vector divided by its magnitude is a unit vector. Unit B vector for A is A ˆ (read as A cap or A hat).  OB  OA  AB O A A Fig. 0.3 ˆ  A  A AA Since, A ˆ. (1) Magnitude of resultant A vector Thus, we can say that unit vector gives us the direction.  2 Vectors AN (2) Direction In  ABN , cos   AN  B cos B CN B sin tan    BN ON A  B cos sin   BN  B sin B Polygon Law of Vector Addition In OBN , we have OB  ON  BN 2 2 2 If a number of non zero vectors are represented by the (n – 1) B sides of an n-sided polygon then the resultant is given by the closing side or the n side of the polygon taken in opposite order. So, th R B B sin R  ABCD E   O A A N OA  AB  BC  CD  DE  OE B cos D D C 2Fig. 0.4  R  ( A  B cos  )  (B sin ) 2 2 E C  R 2  A 2  B 2 cos 2   2 AB cos   B 2 sin2   R 2  A 2  B 2 (cos 2   sin2  )  2 AB cos  E B  R 2  A 2  B 2  2 AB cos  R B  R A 2  B 2  2 AB cos O A (2) Direction of resultant vectors : If  is angle between A and A Note :  Resultant Fig.of0.6 two unequal vectors can not be zero. B, then  Resultant of three co-planar vectors may or may not be | A  B|  A 2  B 2  2 AB cos zero  Resultant of three non co- planar vectors can not be If R makes an angle  with A, then in OBN , zero. BN BN tan    Subtraction of vectors ON OA  AN B sin Since, A  B  A  ( B) and tan   A  B cos Parallelogram Law of Vector Addition | A  B |  A 2  B 2  2 AB cos If two non zero vectors are represented by the two adjacent sides of  | A  B|  A 2  B 2  2 AB cos (180 o   ) a parallelogram then the resultant is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors. Since, cos (180   )   cos (1) Magnitude Since, R 2  ON 2  CN 2  | A  B |  A 2  B 2  2 AB cos   R 2  (OA  AN )2  CN 2 R sum  A  B  R 2  A 2  B 2  2 AB cos  B  R | R | | A  B |  A 2  B 2  2 AB cos  B C  1 2 A 180 –  R  AB B B B sin B    R diff  A  ( B ) O A N A B cos Fig. 0.7 Fig. 0.5 B sin tan 1  Special cases : R  A  B when  = 0 o A  B cos R  A  B when  = 180 o B sin(180   ) and tan  2  A  B cos (180   ) R A 2  B 2 when  = 90 o  Vectors 3 But sin(180   )  sin and cos(180   )   cos Ry Ry  cos    m R R x2  R y2  R z2 B sin  tan  2  A  B cos Rz Rz  cos    n Resolution of Vector Into Components R R x2  R y2  R z2 Consider a vector R in X-Y plane as Y Where l, m, n are called Direction Cosines of the vector R and shown in fig. If we draw orthogonal vectors R x and R y along x and y axes respectively, R x2  R y2  R z2 l 2  m 2  n 2  cos 2   cos 2   cos 2   1 by law of vector addition, R  R x  R y Ry R R x2  R y2  R z2  Now as for any vector A  A nˆ so, Rx X Note : When a point P have coordinate (x, y, z) R x  ˆi R x and R y  ˆjR y Fig. 0.8 then its position vector OP  xˆi  yˆj  zkˆ  When a particle moves from point (x , y , z ) to (x , y , so R  ˆi R x  ˆjRy 1 1 1 2 2 …(i) z ) then its displacement vector 2 But from figure R x  R cos  …(ii) r  (x 2  x 1 )ˆi  (y 2  y1 )ˆj  (z 2  z1 )kˆ and R y  R sin …(iii) Scalar Product of Two Vectors Since R and  are usually known, Equation (ii) and (iii) give the (1) Definition : The scalar product (or dot product) of two vectors is defined as the product of the magnitude of two vectors with cosine of angle magnitude of the components of R along x and y-axes respectively. between them. Here it is worthy to note once a vector is resolved into its components, the components themselves can be used to specify the vector Thus if there are two vectors A and B having angle  between as them, then their scalar product written as A. B is defined as A. B (1) The magnitude of the vector R is obtained by squaring and  AB cos  adding equation (ii) and (iii), i.e. (2) Properties : (i) It is always a scalar B R R x2  R y2 which is positive if angle between the vectors is acute (i.e., < 90°) and negative if angle between  (2) The direction of the vector R is obtained by dividing equation them is obtuse (i.e. 90° v). The distance Relative Velocity through which he is carried down stream by the river is x. Which of the following statement is correct 1. Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a distance of 5 km, the speed of a car du (a) If he crosses the river in minimum time x  moving in the opposite direction if it meets these two cars at an v interval of 4 minutes, will be (a) 40 km/hr (b) 45 km/hr  18 Vectors du (a) 1 (b) 2 (b) x can not be less than v (c) 3 (d) 4 (c) For x to be minimum he has to swim in a direction making an 3. Can the resultant of 2 vectors be zero [IIIT 2000]  v  angle of  sin1   with the direction of the flow of (a) Yes, when the 2 vectors are same in magnitude and direction 2 u (b) No water (c) Yes, when the 2 vectors are same in magnitude but opposite in (d) x will be max. if he swims in a direction making an angle of sense  v  sin1 with direction of the flow of water (d) Yes, when the 2 vectors are same in magnitude making an 2 u 2 angle of with each other 11. A man sitting in a bus travelling in a direction from west to east 3 with a speed of 40 km/h observes that the rain-drops are falling 4. The sum of the magnitudes of two forces acting at point is 18 and vertically down. To the another man standing on ground the rain will appear [HP PMT 1999] the magnitude of their resultant is 12. If the resultant is at 90° with the force of smaller magnitude, what are the, magnitudes of forces [Roorkee 199 (a) To fall vertically down (a) 12, 5 (b) 14, 4 (b) To fall at an angle going from west to east (c) 5, 13 (d) 10, 8 (c) To fall at an angle going from east to west 5. A vector a is turned without a change in its length through a small (d) The information given is insufficient to decide the direction of rain. angle d . The value of | a | and a are respectively 12. A boat takes two hours to travel 8 km and back in still water. If the (a) 0, a d (b) a d , 0 velocity of water is 4 km/h, the time taken for going upstream 8 km and coming back is [EAMCET 1990] (c) 0, 0 (d) None of these (a) 2h 6. Find the resultant of three vectors OA, OB and OC shown in the (b) 2h 40 min following figure. Radius of the circle is R. C (c) 1h 20 min (a) 2R B (d) Cannot be estimated with the information given 45o (b) R(1  2 ) 45o 13. A 120 m long train is moving towards west with a speed of 10 m/s. A A bird flying towards east with a speed of 5 m/s crosses the train. The O (c) R 2 time taken by the bird to cross the train will be [Manipal 2002] (a) 16 sec (b) 12 sec (d) R( 2  1) (c) 10 sec (d) 8 sec 7. Figure shows ABCDEF as a regular hexagon. What is the value of AB  AC  AD  AE  AF E D 14. A boat crosses a river with a velocity of 8 km/h. If the resulting velocity of boat is 10 km/h then the velocity of river water is [CPMT (a) AO2001] (a) 4 km/h (b) 6 km/h F C (b) 2 AO O (c) 8 km/h (d) 10 km/h (c) 4 AO A B (d) 6 AO 8. The length of second's hand in watch is 1 cm. The change in velocity of its tip in 15 seconds is [MP PMT 1987]  (a) Zero (b) cm / sec 1. If a vector P making angles , , and  respectively with the X, Y 30 2 and Z axes respectively.   2 (c) cm / sec (d) cm / sec Then sin2   sin2   sin2   30 30 (a) 0 (b) 1 9. A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average (c) 2 (d) 3 acceleration of the particle is 2. If the resultant of n forces of different magnitudes acting at a point [CPMT 1997; IIT-JEE 1982] is zero, then the minimum value of n is [SCRA 2000]  Vectors 19 1 (d) If the assertion and reason both are false. (a) Zero (b) m / s2 N W (e) If assertion is false but reason is true. 2 1 1     (c) m / s2 N  E (d) m / s2 S W 1. Assertion : A  B is perpendicular to both A  B as well as 2 2   A  B.     10. A force F   K(yˆi  xˆj) (where K is a positive constant) acts on Reason : A  B as well as A  B lie in the plane     a particle moving in the x-y plane. Starting from the origin, the containing A and B , but A  B lies particle is taken along the positive x- axis to the point (a, 0) and   then parallel to the y-axis to the point (a, a). The total work done by perpendicular to the plane containing A and B. the forces F on the particle is 2. Assertion : Angle between ˆi  ˆj and ˆi is 45° [IIT-JEE 1998] Reason : ˆi  ˆj is equally inclined to both î and ˆj and the (a)  2 Ka 2 (b) 2 Ka 2 angle between î and ˆj is 90°   (c)  Ka 2 (d) Ka 2 3. Assertion : If  be the angle between A and B , then   11. The vectors from origin to the points A and B are AB tan     A  3ˆi  6 ˆj  2kˆ and B  2ˆi  ˆj  2kˆ respectively. The area of A.B     the triangle OAB be Reason : A  B is perpendicular to A.B      (a) 5 17 sq.unit (b) 2 17 sq.unit 4. Assertion : If | A  B |  | A  B | , then angle between A 2 5  and B is 90° 3 5     (c) 17 sq.unit (d) 17 sq.unit Reason : ABBA 5 3 5. Assertion : Vector product of two vectors is an axial vector 12. A metal sphere is hung by a string fixed to a wall. The sphere is   pushed away from the wall by a stick. The forces acting on the Reason : If v = instantaneous velocity, r = radius vector and     sphere are shown in the second diagram. Which of the following  = angular velocity, then   v  r. statements is wrong 6. Assertion : Minimum number of non-equal vectors in a plane required to give zero resultant is three. (a) P  W tan      Reason : If A  B  C  0, then they must lie in one plane (b) T  P W  0   7. Assertion : Relative velocity of A w.r.t. B is greater than the velocity of either, when they are moving in opposite (c) T 2  P2  W 2 directions. P   Reason : Relative velocity of A w.r.t. B  v A  v B (d) T  P W W   13. The speed of a boat is 5 km/h in still water. It crosses a river of 8. Assertion : Vector addition of two vectors A and B is width 1 km along the shortest possible path in 15 minutes. The commutative.     velocity of the river water is Reason : ABBA [IIT 1988; CBSE PMT 1998, 2000]     9. Assertion : A.B  B. A (a) 1 km/h (b) 3 km/h Reason : Dot product of two vectors is commutative. (c) 4 km/h (d) 5 km/h       10. Assertion :   r  F and   F  r 14. A man crosses a 320 m wide river perpendicular to the current in 4 Reason : Cross product of vectors is commutative. minutes. If in still water he can swim with a speed 5/3 times that of 11. Assertion : A negative acceleration of a body is associated with the current, then the speed of the current, in m/min is [Roorkee 1998] a slowing down of a body. (a) 30 (b) 40 Reason : Acceleration is vector quantity. (c) 50 (d) 60. 12. Assertion : A physical quantity cannot be called as a vector if its magnitude is zero. Reason : A vector has both, magnitude and direction. 13. Assertion : The sum of two vectors can be zero. Reason : The vector cancel each other, when they are equal and opposite. 14. Assertion : Two vectors are said to be like vectors if they have same direction but different magnitude. Read the assertion and reason carefully to mark the correct option out of Reason : Vector quantities do not have specific direction. the options given below: 15. Assertion : The scalar product of two vectors can be zero. (a) If both assertion and reason are true and the reason is the correct Reason : If two vectors are perpendicular to each other, their explanation of the assertion. scalar product will be zero. (b) If both assertion and reason are true but reason is not the correct 16. Assertion : Multiplying any vector by an scalar is a meaningful explanation of the assertion. operations. Reason : In uniform motion speed remains constant. (c) If assertion is true but reason is false.  20 Vectors 17. Assertion : A null vector is a vector whose magnitude is zero and direction is arbitrary. Reason : A null vector does not exist.   18. Assertion : If dot product and cross product of A and B are   zero, it implies that one of the vector A and B must be a null vector. Reason : Null vector is a vector with zero magnitude. 19. Assertion : The cross product of a vector with itself is a null vector. Reason : The cross-product of two vectors results in a vector quantity. 20. Assertion : The minimum number of non coplanar vectors whose sum can be zero, is four. Reason : The resultant of two vectors of unequal magnitude can be zero.      21. Assertion : If A.B  B.C, then A may not always be equal to  C Reason : The dot product of two vectors involves cosine of the angle between the two vectors. 22. Assertion : Vector addition is commutative.     Reason : ( A  B)  (B  A).  Vectors 21 Relative Velocity 1 b 2 b 3 c 4 c 5 d 6 a 7 c 8 c 9 d 10 ac 11 b 12 b 13 d 14 b Fundamentals of Vectors Critical Thinking Questions 1 d 2 b 3 c 4 d 5 d 1 c 2 c 3 c 4 c 5 b 6 a 7 a 8 b 9 b 10 d 6 b 7 d 8 d 9 b 10 c 11 d 12 d 13 a 14 b 15 c 11 a 12 d 13 b 14 d 16 c 17 a 18 b 19 c 20 c 21 d 22 d 23 b 24 d 25 b Assertion and Reason 26 b 27 a 28 a 29 a 30 d 1 a 2 a 3 d 4 b 5 c 31 a 32 b 33 a 34 a 6 b 7 a 8 b 9 a 10 c Addition and Subtraction of Vectors 11 b 12 e 13 a 14 c 15 a 1 a 2 b 3 d 4 b 5 b 16 b 17 c 18 b 19 b 20 c 6 a 7 b 8 a 9 d 10 b 21 a 22 c 11 d 12 c 13 a 14 c 15 c 16 c 17 c 18 c 19 c 20 b 21 a 22 d 23 d 24 a 25 c 26 b 27 b 28 a 29 b 30 a 31 c 32 c 33 c 34 d 35 a 36 c 37 d 38 a 39 c 40 d Fundamentals of Vectors 41 a 42 b 43 d 44 d 45 a 1. (d) As the multiple of ˆj in the given vector is zero therefore this 46 c 47 d 48 a 49 a 50 c vector lies in XZ plane and projection of this vector on y-axis is 51 c 52 a 53 d zero. 2. (b) If a point have coordinate (x, y, z) then its position vector Multiplication of Vectors  xˆi  yˆj  zkˆ. 1 c 2 b 3 d 4 a 5 a 3. (c) Displacement vector r  xˆi  yˆj  zkˆ 6 b 7 c 8 b 9 b 10 d  (3  2)ˆi  (4  3)ˆj  (5  5)kˆ  ˆi  ˆj 11 b 12 d 13 c 14 d 15 c 4. (d) y 16 c 17 b 18 c 19 b 20 a F cos 60o 21 a 22 c 23 a 24 b 25 c F 60o 26 d 27 d 28 b 29 b 30 b x 31 d 32 c 33 d 34 b 35 d F sin 60o The component of force in vertical direction 36 b 37 a 38 b 39 a 40 a 1 41 d 42 d 43 c 44 b 45 a = F cos   F cos 60  5   2.5 N 2 46 a 47 a 48 d 49 d 50 a 5. (d) | B |  7 2  (24 ) 2  625  25 51 b 52 b 53 d 54 a 55 c ˆ ˆ 56 d 57 a 58 b 59 c ˆ  3i  4 j Unit vector in the direction of A will be A 5 Lami's Theorem  3ˆi  4 ˆj  So required vector = 25    15ˆi  20 ˆj  5    1 c 2 a 3 b 4 c 5 b 6. (a) Let the components of A makes angles  ,  and  with x, y and z axis respectively then       22 Vectors  cos 2   cos 2   cos 2   1 23. (b) r  (a cos  t)ˆi  (a sin t)ˆj 1   3 cos 2   1  cos    dr v  a sin t ˆi  a cos  t ˆj 3 dt  A As r.v  0 therefore velocity of the particle is perpendicular  A x  A y  A z  A cos   3 to the position vector.  24. (d) Displacement, electrical and acceleration are vector quantities. 7. (a) A  2ˆi  4 ˆj  5kˆ  | A |  (2) 2  (4 )2  (5) 2  45 25. (b) Magnitude of unit vector = 1 2 4 5  cos   , cos   , cos    (0.5)2  (0.8)2  c 2  1 45 45 45 By solving we get c  0.11 8. (b) Unit vector along y axis  ˆj so the required vector 26. (b)  ˆj  [(ˆi  3ˆj  2kˆ )  (3ˆi  6 ˆj  7kˆ )]   4ˆi  2ˆj  5kˆ D C    9. (b) F3  F1  F2 300 m There should be minimum three coplaner F3 F2 vectors having different magnitude which A B should be added to give zero resultant 400 m F1 Displacement AC  AB  BC 10. (d) Diagonal of the hall = l 2  b 2  h2 AC  ( AB)2  (BC)2  (400)2  (300)2  500m  10 2  12 2  14 2 h Distance  AB  BC  400  300  700m  100  144  196 l b 27. (a) Resultant of vectors A and B  400  20m R  A  B  4ˆi  3ˆj  6kˆ  ˆi  3ˆj  8 kˆ  11. (d) Total angle = 100   2 50 R  3ˆi  6 ˆj  2kˆ So all the force will pass through one point and all forces will be balanced. i.e. their resultant will be zero. ˆ  R  3ˆi  6 ˆj  2kˆ 3ˆi  6 ˆj  2kˆ R     | R| 3 2  6 2  (2)2 7 12. (d) r  r2  r1  (2ˆi  2ˆj  0kˆ )  (4ˆi  4 ˆj  0kˆ )  28. (a)   B. A. In this formula A is a area vector.  r  6ˆi  2ˆj  0kˆ      29. (a) r  a  b  c  4ˆi  ˆj  3ˆi  2ˆj  kˆ  ˆi  ˆj  kˆ  | r |  (6)2  (2)2  0 2  36  4  40  2 10  ˆi  ˆj  kˆ ˆi  ˆj  kˆ r    1   1 2  2 rˆ    1 ˆ 1 ˆ P i j | P |      =1 | r| 1 2  1 2  (1)2  2   2 13. (a) 3  2 2     A.B 9  16  25 50  It is a unit vector. 30. (d) cos    = 1 | A || B | 9  16  25 9  16  25 50 14. (b)  ˆi  ˆj  cos   1    cos 1 (1) R 1 ˆ 1 ˆ 15. (c) R̂    i j  | R| 1 1 2 2 2 2 31. (a) r  3 t 2ˆi  4 t 2 ˆj  7kˆ   16. (c) R  3ˆi  ˆj  2kˆ at t  0 , r1  7 kˆ   Length in XY plane = R x2  Ry2  3 2  1 2  10 at t  10 sec , r2  300ˆi  400ˆj  7kˆ ,   17. (a) If the angle between all forces which are equal and lying in one r  r2  r1  300ˆi  400ˆj plane are equal then resultant force will be zero.    | r | | r2  r1 |  (300)2  (400)2  500m 18. (b) A  ˆi  ˆj  | A |  12  12  2 Ax 1 32. (b) Resultant of vectors A and B cos     cos 45    45 | A| 2 R  A  B  4ˆi  3ˆj  8ˆi  8 ˆj  12ˆi  5 ˆj 19. (c) ˆ ˆ ˆ  R  12i  5 j 12ˆi  5 ˆj 20. (c) R  | R| (12)2  (5)2 13 21. (d) All quantities are tensors.   22. (d) P  Q  PP ˆ  QQ ˆ  Vectors 23 A.B (2ˆi  3 ˆj) (ˆi  ˆj) 2  3 5 7. (b) R A 2  B 2  2 AB cos 33. (a)      | i  j| 2 2 2 By substituting, A  F, B  F and R  F we get A.B (3ˆi  4 ˆj  5 kˆ ) (3ˆi  4 ˆj  5 kˆ ) 1 34. (a) cos    cos    120 | A || B | 9  16  25 9  16  25 2 8. (a) 9  16  25    0 9. (d) If two vectors A and B are given then the resultant R max = 50 A  B  7 N and R min  4  3  1 N  cos   0 ,    90 i.e. net force on the particle is between 1 N and 7 N.  Addition and Subtraction of Vectors 10. (b) If C lies outside the plane then resultant force can not be zero. 1. (a) For 17 N both the vector should be parallel i.e. angle between 11. (d) them should be zero. For 7 N both the vectors should be antiparallel i.e. angle 12. (c) F  F12  F22  2 F1 F2 cos 90  F12  F22 between them should be 180° 13. (a) For 13 N both the vectors should be perpendicular to each 14. (c) other i.e. angle between them should be 90° 2. (b) A  B  4ˆi  3ˆj  6ˆi  8 ˆj  10ˆi  5 ˆj 15. (c) C  A2  B2  C | A  B |  (10)2  (5)2  5 5 The angle between A and B is B 2 5 1 1 90° tan       tan 1   10 2 2 A 3. (d) From figure    v1 =20 m/s 16. (c) R  A  B = 6ˆi  7 ˆj  3ˆi  4 ˆj = 9ˆi  11ˆj v 1  20 ˆj and v 2  20ˆi v2 =20 m/s  v  v 2  v 1  20(ˆi  ˆj)  | R |  9 2  11 2  81  121  202  O | v |  20 2 and direction 17. (c) R  12 2  5 2  6 2  144  25  36  205  14.31 m v – v1      tan 1 (1)  45 i.e. S–W 18. (c) A  3ˆi  2ˆj  kˆ , B  ˆi  3ˆj  5kˆ , C  2ˆi  ˆj  4 kˆ 4. (b) Let n̂1 and n̂ 2 are the two unit vectors, then the sum is  | A |  3 2  (2)2  1 2  9  4  1  14 n s  nˆ 1  nˆ 2 or ns2  n12  n22  2n1n2 cos   | B |  1 2  (3)2  5 2  1  9  25  35  1  1  2 cos  | A |  2 2  1 2  (4 )2  4  1  16  21 Since it is given that n s is also a unit vector, therefore 1 As B  A 2  C 2 therefore ABC will be right angled 1  1  1  2 cos  cos      120 2 triangle. 19. (c) Now the difference vector is nˆ d  nˆ 1  nˆ 2 or    20. (b) C  A  B. nd2  n12  n22  2n1n2 cos   1  1  2 cos(120) The value of C lies between A  B and A  B      nd2  2  2(1 / 2)  2  1  3  nd  3  | C |  | A | or | C |  | B | 5. (b) A  2 B  3C  (2ˆi  ˆj)  2(3ˆj  kˆ )  3(6ˆi  2kˆ ) 21. (a) 22. (d)  2ˆi  ˆj  6 ˆj  2kˆ  18ˆi  6kˆ = 20ˆi  5 ˆj  4 kˆ 23. (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 6. (a) P 1  m v sin ˆi  m v cos ˆj with an acceleration. 24. (a) A  B  16 (given) …(i) and P 2  m v sin ˆi  m v cos ˆj B sin tan    tan 90 So change in momentum A  B cos A  A  B cos  0  cos  …(ii) P  P 2  P 1  2 m v cos  ˆj, |  P |  2 m v cos B  24 Vectors 8 A 2  B 2  2 AB cos …(iii) 38. (a) C A2  B2 By solving eq. (i), (ii) and (iii) we get A  6 N , B  10 N    = 32  4 2  5 C 25. (c) | P |  5 , | Q |  12 and | R |  13 B    Angle between A and B is Q 12 R 2 cos   Q R 13 39. (c) A 40. (d) N  12  –v1 v1    cos 1     13  P W E B v 26. (b)  A  B  2 AB cos  2 2 …(i) 2 v2 S B sin  tan 90   A  B cos   0 A  B cos  If the magnitude of vector remains same, only direction change A by  then  cos    B v  v 2  v1 , v  v 2  (v1 ) B2 B Hence, from (i)  A2  B2  2A2  A  3 4 2   Magnitude of change in vector | v |  2v sin  A 3 2  cos        150 B 2   90  | v |  2  10  sin  = 10 2 = 14.14 m / s 27. (b) (ˆi  2ˆj  2kˆ )  (2ˆi  ˆj  kˆ )  R  i  2    Required vector R =  2ˆi  ˆj  kˆ Direction is south-west as shown in figure.       28. (a) Resultant R  P  Q  P  Q  2 P 41. (a) AC  AB  BC 20km   B C The angle between P and 2 P is zero. AC  ( AB)  (BC) 2 2 10km 29. (b) R Q  (10) 2  (20) 2 A   100  400  500  22.36 km P Q sin (b) cos   F1.F2  tan 90   P  Q cos  0 42. P  Q cos  | F1 || F2 | P P (5ˆi  10 ˆj  20kˆ ).(10ˆi  5 ˆj  15kˆ ) 50  50  300 cos      cos 1     Q  Q  25  100  400 100  25  225 525 350 30. (a) According to problem P  Q  3 and P  Q  1 1  cos      45 P 2 By solving we get P  2 and Q  1   2  P  2Q Q 43. (d) If two vectors A and B are given then Range of their resultant 31. (c) can be written as ( A  B)  R  ( A  B). 32. (c) i.e. Rmax  A  B and Rmin  A  B 33. (c) If B = 1 and A = 4 then their resultant will lies in between 3N 34. (d) F1  F2  F3  0  4ˆi  6 ˆj  F3  0 and 5N. It can never be 2N.   F3  4ˆi  6 ˆj 44. (d) A  3 N , B  2 N then R  A 2  B 2  2 AB cos    R  9  4  12 cos  …(i) 35. (a) v  2v sin   2  v  sin 90 2 Now A  6 N , B  2 N then  2  100  200 km/hr 2 R  36  4  24 cos  …(ii) 36. (c) 1 from (i) and (ii) we get cos       120 2 37. (d) Resultant velocity  20  15 2 2 45. (a) In N forces of equal magnitude works = 400  225  625  25 km/hr on a single point and their resultant is F 120° 120° F 120° F  Vectors 25 zero then angle between any two forces is given 3. (d) W  F. S  FS cos 360 360    120 1 N 3  50  10  cos 60  50  10   250 J. 2 If these three vectors are represented by three sides of triangle then they form equilateral triangle 4. (a) S  r2  r1 46. (c) Resultant of two vectors A and B can be given by W  F. S  (4ˆi  ˆj  3kˆ ). (11ˆi  11ˆj  15kˆ ) R  AB  (4  11  1  11  3  15)  100 J. | R | | A  B |  A 2  B 2  2 AB cos  5. (a) ( A  B) is perpendicular to ( A  B). Thus If   0 then | R |  A  B | A |  | B | 47. (d) Rmax  A  B  17 when   0 ( A  B). ( A  B) = 0 Rmin  A  B  7 when   180 or A 2  B. A  A. B  B 2  0 by solving we get A  12 and B  5 Because of commutative property of dot product A.B  B. A Now when   90 then R  A2  B2  A 2  B 2  0 or A  B  R  (12)2  (5)2  169  13 Thus the ratio of magnitudes A/B = 1 48. (a) If two vectors are perpendicular then their dot product must 6. (b) Let A.( B  A)  A. C be equal to zero. According to problem ( A  B).( A  B)  0  A. A  A.B  B. A  B.B  0 Here C  B  A Which is perpendicular to both vector  A2  B2  0  A2  B2 A and B  A. C  0  A  B i.e. two vectors are equal to each other in magnitude. 7. (c) We know that A  B  (B  A) because the angle between these two is always 90°. 49. (a) v y  20 and v x  10 y  But if the angle between A and B is 0 or . Then  velocity v  10ˆi  20 ˆj v vy AB  B A  0. direction of velocity with x axis vy 20 ˆi ˆj kˆ tan    2  x vx 10 vx 8. (b) AB  3 1 2 2 2 4    tan 1 (2) 50. (c) Rmax  A  B when   0  Rmax  12  8  20 N  (1  4  2  2)ˆi  (2  2  4  3)ˆj  (3  2  1  2)kˆ 51. (c) R A 2  B 2  2 AB cos   8ˆi  8 ˆj  8 kˆ If A  B  P and   120 then R  P Magnitude of A  B | A  B |  (8) 2  (8)2  (8) 2 52. (a) Sum of the vectors R  5ˆi  8 ˆj  2ˆi  7 ˆj  7ˆi  15 ˆj 8 3 magnitude of R | R |  49  225  274 53. (d) ˆi ˆj kˆ 9. (b)   r  F  3 2 3 Multiplication of Vectors 2 3 4 (c) Given vectors can be rewritten as A  2ˆi  3ˆj  8 kˆ and  (2  4 )  (3  3) ˆi  (2  3)  (3  4 )ˆj 1. B  4ˆi  4 ˆj  kˆ Dot product of these vectors should be equal to zero because  (3  3)  (2  2)kˆ  17 ˆi  6 ˆj  13 kˆ they are perpendicular. 10. (d) From the property of vector product, we notice that C must  A. B  8  12  8  0  8  4    1 / 2 be perpendicular to the plane formed by vector A and B. 2. (b) Let A  2ˆi  3ˆj  kˆ and B  4ˆi  6 ˆj  kˆ Thus C is perpendicular to both A and B and A and B are parallel to each other ( A  B) vector also, must lie in the plane formed by vector A a1 a 2 a 3 2 3 1 and B. Thus C must be perpendicular to ( A  B) also but   i.e.      2. b1 b 2 b 3 4 6   26 Vectors   the cross product ( A  B) gives a vector C which can not be F.P  0    90 perpendicular to itself. Thus the last statement is wrong.   26. (d) | A  B |  | (2ˆi  3ˆj)  (ˆi  4 ˆj)|  | 5kˆ |  5 units 11. (b) We know that, Angular momentum 27. (d)   L  r  p in terms of component becomes 28. (b) A  B  0  sin  0    0 y Two vectors will be parallel to each other. ˆi ˆj kˆ m     L x y z v 29. (b) A  B and B  A are parallel and opposite to each other. So b the angle will be . px py pz     30. (b) Vector (P  Q) lies in a plane and vector (P  Q) is x O perpendicular to this plane i.e. the angle between given vectors As motion is in x-y plane (z = 0 and Pz  0 ), so  is. 2 L  k (xp y  yp x ) Here x = vt, y = b, p x  m v and p y  0 31. (d) 2 2  3 2  2  2  3  cos  1   By solving we get   180  A  B  0  L  k vt  0  b mv   mvb kˆ   32. (c) Dot product of two perpendicular vector will be zero. 12. (d) F1.F2  (2ˆj  5kˆ )(3ˆj  4 kˆ )  AB 42  24  12 56  6  20  20  6  26 33. (d) cos    AB 36  36  9 49  16  16 9 71 13. (c) Force F lie in the x-y plane so a vector along z-axis will be perpendicular to F. 56 5  5       cos    sin  or   sin1   14. (d) A.B | A |. | B |. cos   A.B. cos 90  0 3  3  9 71   15. (c)  V1 V1 34. (b) Direction of vector A is along z-axis  A  akˆ V 'net  Vnet Direction of vector B is towards north  B  bˆj   V2 Now A  B  akˆ  bˆj  ab(ˆj)  –V2     According to problem | V1  V2 |  | V1  V2 |  The direction is A  B is along west.      |  | Vnet |  | Vnet A.B 1 1 35. (d) cos         60 So V1 and V2 will be mutually perpendicular. | A || B | 2 2 2  16. (c) W  F.r  (5ˆi  3ˆj)(2ˆi  ˆj)  10  3  7 J. 36. (d) AB  (4ˆi  5 ˆj  6kˆ )  (3ˆi  4 ˆj  5kˆ ) = ˆi  ˆj  kˆ   17. A.B (b) cos      264  0    90 CD  (4ˆi  6 ˆj)  (7ˆi  9 ˆj  3kˆ )  3ˆi  3ˆj  3kˆ | A || B | 14 21 AB and CD are parallel, because its cross-products is 0. 18. (c) (ˆi  ˆj).(ˆj  kˆ )  0  0  1  0  1     37. (a) W  F S  (4ˆi  5 ˆj)(3ˆi  6 ˆj)  12 A.B 1 1 cos         60     | A || B | 2 2 2 38. (b) | A  B |  A.B  AB sin  AB cos   tan   1  19. (b) P  F.v  20  6  15  (4 )  (5)  3    45  120  60  15  120  75  45 J/s 39. (a)   P.Q 20. (a) cos   1    0 ˆi ˆj kˆ PQ    40. (a) v    r  1  2 2  ˆi (6  8 )  ˆj(3)  4 kˆ 21. (a) W  F.s  (5ˆi  6 ˆj  4 kˆ )(6ˆi  5kˆ )  30  20  10 J   0 4 3 22. (c) A.B  0    90    23. (a) P.Q  0  a2  2a  3  0  a  3  2i  3 j  4k  24. (b) W  F.r  (2ˆi  15ˆj  6kˆ )(10ˆj)  150   | v |  (2)2  (3)2  4 2  29 unit 25. (c) Px  2 cos t , Py  2 sint  P  2 cos t ˆi  2 sint ˆj      41. (d) a. b  0 i.e. a and b will be perpendicular to each other  dP F  2 sin t ˆi  2 cos t ˆj     dt a. c  0 i.e. a and c will be perpendicular to each other  Vectors 27     b  c will be a vector perpendicular to both b and c 36  64  100 =  10 2 kg    1 So a is parallel to b  c 51. (a) Area of parallelogram  A  B 42. (d) Area  2ˆi  2ˆj  4 kˆ  4 unit.   (ˆi  2ˆj  3kˆ )  (3ˆi  2ˆj  kˆ )  43. (c) A  2ˆi  2ˆj  kˆ and B  6ˆi  3ˆj  2kˆ ˆi ˆj kˆ      C  A  B  2ˆi  2ˆj  kˆ  6ˆi  3ˆj  2kˆ   1 2 3  (8)ˆi  (8)ˆj  (8)kˆ 3 2 1 ˆi ˆj kˆ  2 2  1  ˆi  10 ˆj  18 kˆ Magnitude  64  64  64 = 8 3 6 3 2    52. (b) Radius vector r  r2  r1  (2ˆi  3ˆj  kˆ )  (2ˆi  ˆj  kˆ ) Unit vector perpendicular to both A and B   r  4 ˆj ˆi  10 ˆj  18 kˆ ˆi  10 ˆj  18 kˆ   1 2  10 2  18 2 5 17 Linear momentum p  2ˆi  3 ˆj  kˆ      44. (b) A  ˆj  3kˆ , B  ˆi  2ˆj  kˆ L  r  p  (4 ˆj)  (2ˆi  3ˆj  kˆ ) ˆi ˆj kˆ ˆi ˆj kˆ    C  AB  0 1 3  7ˆi  3 ˆj  kˆ  0 4 0  4ˆi  8 kˆ 1 2 1 2 3 1  Hence area = | C |  49  9  1  59 sq unit ˆi ˆj kˆ    53. (d) v    r  3  4 1  18ˆi  13ˆj  2kˆ ˆi ˆj kˆ 5 6 6    45. (a) L  r  p  1 2  1  ˆj  2kˆ 3 4 2 54. (a) i.e. the angular momentum is perpendicular to x-axis. 55. (c) A.B  AB cos      46. (a) A  B is a vector perpendicular to plane A  B and hence In the problem A.B   AB i.e. cos   1    180   perpendicular to A  B. i.e. A and B acts in the opposite direction.    47. (a)   r  F  (7ˆi  3ˆj  kˆ )(3