Vectors 1 y ˆj kˆ x ˆi 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 x y z not always imply that it is a vector. For it to be a vector the third condition 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 Fig. 0.2 Axial vector 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. Triangle Law of Vector Addition of Two Vectors (4) Collinear vectors : When the vectors under consideration can 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 Aˆ (read as A cap or A hat).  OBOAAB O A A Since, Aˆ  A  A AAˆ . (1) Magnitude of resultant Fig. 0.3 A vector Thus, we can say that unit vector gives us the direction. 2 Vectors (2) Direction AN In ABN, cos  AN Bcos B CN Bsin tan  BN ON ABcos sin  BN Bsin B Polygon Law of Vector Addition In OBN, we have OB2 ON2BN2 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 Fig. 0.4  R2 (ABcos)2 (Bsin)2 E C  R2 A2 B2cos22ABcosB2sin2  R2 A2 B2(cos2sin2)2ABcos E B  R2 A2 B2 2ABcos R B  R A2 B2 2ABcos O A (2) Direction of resultant vectors : If  is angle between A and Note A :  ResultanFti go. f0 t.6w o unequal vectors can not be zero. B, then  Resultant of three co-planar vectors may or may not be | AB| A2 B2 2ABcos zero  Resultant of three non co- planar vectors can not be If Rmakes an angle  with A, then in OBN, zero. BN BN tan  Subtraction of vectors ON OAAN Bsin tan Since, AB A(B) and ABcos Parallelogram Law of Vector Addition | AB| A2 B2 2ABcos If two non zero vectors are represented by the two adjacent sides of  | AB| A2 B2 2ABcos(180o ) 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, R2 ON2 CN2  | AB| A2 B2 2ABcos  R2 (OAAN)2 CN2 Rsum  AB  R2 A2 B2 2ABcos B  R | R|| AB| A2 B2 2ABcos B C  1 2 A 180 –  R  AB B B B sin B    O A N Rdiff  A(B) A B cos Fig. 0.7 Fig. 0.5 Bsin tan  Special cases : R AB when  = 0o 1 ABcos R AB when  = 180o Bsin(180) and tan  2 ABcos(180) R A2 B2 when  = 90 o Vectors 3 But sin(180)sin and cos(180)cos R R  cos y  y m  tan  Bsin R Rx2 Ry2 Rz2 2 ABcos R R Resolution of Vector Into Components  cos z  z n R R2 R2 R2 x y z 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 and R along x and y axes respectively, R2 R2 R2 x y l2 m2 n2  cos2cos2cos2 x y z 1 by law of vector addition, RRx Ry Ry R Rx2 Ry2 Rz2  Note Now as for any vector A Anˆ so, X :  When a point P have coordinate (x, y, z) R x Rx ˆiRx and Ry ˆjRy 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ˆiRx ˆjRy …(i) z) then its displacement vector 1 1 1 2 2 2 But from figure Rx Rcos …(ii) r (x x )ˆi(y y )ˆj(z z )kˆ 2 1 2 1 2 1 and Ry Rsin …(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 vectorR is obtained by squaring and ABcos adding equation (ii) and (iii), i.e. (2) Properties : (i) It is always a scalar B R R2 R2 which is positive if angle between the vectors is x y acute (i.e., < 90°) and negative if angle between  them is obtuse (i.e. 90°< < 180°). (2) The direction of the vector R is obtained by dividing equation (iii) by (ii), i.e. (ii) It is commutative, i.e. A.BB.A tan(Ry /Rx) or tan1(Ry /Rx) (iii) It is distributive, i.e. FigA. 0 .10 Rectangular Components of 3-D Vector A.(BC)A.BA.C R R R R q or RR ˆiR ˆjR kˆ (iv) As by definition A.BABcos x y z x y z Y A.B The angle between the vectors cos1   AB  (v) Scalar product of two vectors will be maximum when Ry R Rx cosmax 1, i.e. 0o, i.e., vectors are parallel X (A.B)max AB Rz (vi) Scalar product of two vectors will be minimum when Z | cos|min 0, i.e. 90o Fig. 0.9 If R makes an angle  with x axis,  with y axis and with z axis, (A.B) 0 then min i.e. if the scalar product of two nonzero vectors vanishes the vectors  cos Rx  Rx l are orthogonal. R R2 R2 R2 (vii) The scalar product of a vector by itself is termed as self dot x y z product and is given by (A)2 A.AAAcosA2 4 Vectors i.e. A A.A (viii) In case of unit vector nˆ nˆ.nˆ 11cos01 so nˆ.nˆˆi.ˆiˆj.ˆjkˆ.kˆ1 (ix) In case of orthogonal unit vectors ˆi,ˆj and kˆ, ˆi.ˆjˆj.kˆkˆ.ˆi11cos900 Fig. 0.12 (x) In terms of components The direction of AB, i.e. C is perpendicular to the plane A.B(iA jA kA ).(iB jB kB ) [A B A B A B ] containing vectors A and B and in the sense of advance of a right x y z x y z x x y y Z z (3) Example : (i) Work W : In physics for constant force work is handed screw rotated from A (first vector) to B (second vector) through the smaller angle between them. Thus, if a right handed screw whose axis is defined as, WFscos …(i) perpendicular to the plane framed by A and B is rotated from A to B But by definition of scalar product of two vectors, F.sFscos through the smaller angle between them, then the direction of advancement …(ii) of the screw gives the direction of AB i.e. C So from eqn (i) and (ii) W F.s i.e. work is the scalar product of (2) Properties force with displacement. (i) Vector product of any two vectors is always a vector (ii) Power P : perpendicular to the plane containing these two vectors, i.e., orthogonal to dW ds both the vectors A and B,though the vectors A and B may or may As W F.s or F. [As F is constant] dt dt not be orthogonal. (ii) Vector product of two vectors is not commutative, i.e., or PF.v i.e., power is the scalar product of force with ABBA [but BA]  dW ds  velocity. As Pand v  dt dt  ds  Here it is worthy to note that B (iii) Magnetic Flux : | AB|| BA|ABsin Magnetic flux through an area is  i.e. in case of vector AB and BA magnitudes are equal but given by dBdscos …(i) directions are opposite. But by definition of scalar O (iii) The vector product is distributive when the order of the vectors is strictly maintained, i.e. product B.dsBdscos ...(ii) Fig. 0.11 A(BC)ABAC So from eq (i) and (ii) we have n (iv) The vector product of two vectors will be maximum when dB.ds or  B.ds sinmax 1, i.e., 90o (iv) Potential energy of a dipole U : If an electric dipole of moment [AB] ABnˆ max p is situated in an electric field E or a magnetic dipole of moment M i.e. vector product is maximum if the vectors are orthogonal. in a field of induction B, the potential energy of the dipole is given by : (v) The vector product of two non- zero vectors will be minimum when | sin| minimum = 0, i.e., 0o or 180o U p.E and U M.B E B Vector Product of Two Vectors [AB] 0 min (1) Definition : The vector product or cross product of two vectors i.e. if the vector product of two non-zero vectors vanishes, the is defined as a vector having a magnitude equal to the product of the vectors are collinear. magnitudes of two vectors with the sine of angle between them, and (vi) The self cross product, i.e., product of a vector by itself direction perpendicular to the plane containing the two vectors in accordance with right hand screw rule. vanishes, i.e., is null vector AAAAsin0onˆ 0 CAB (vii) In case of unit vector nˆnˆ 0so that Thus, if A and B are two vectors, then their vector product ˆiˆiˆjˆjkˆkˆ 0 written as AB is a vector C defined by (viii) In case of orthogonal unit vectors, ˆi,ˆj,kˆ in accordance with CABABsinnˆ right hand screw rule : Vectors 5  abc …(ii) ˆj ˆj Pre-multiplying both sides by a kˆ a(ab)ac  0abac ˆi ˆi  abca …(iii) kˆ Fig. 0.13 Pre-multiplying both sides of (ii) by b ˆiˆjkˆ, ˆjkˆ ˆi and kˆˆi ˆj b(ab)bc  babbbc And as cross product is not commutative, ˆjˆikˆ, kˆˆjˆiand ˆikˆ ˆj  abbc  abbc …(iv) (x) In terms of components From (iii) and (iv), we get abbcca ˆi ˆj kˆ Taking magnitude, we get | ab|| bc|| ca| AB A A A x y z  absin(180)bcsin(180)casin(180) B B B x y z  absinbcsincasin ˆi(A B A B ) ˆj(A B A B )kˆ(A B A B ) y z z y z x x z x y y x Dividing through out by abc, we have (3) Example : Since vector product of two vectors is a vector, vector physical quantities (particularly representing rotational effects) like torque,  sin sin sin angular momentum, velocity and force on a moving charge in a magnetic field a b c and can be expressed as the vector product of two vectors. It is well – Relative Velocity established in physics that : (1) Introduction : When we consider the motion of a particle, we (i) Torque rF assume a fixed point relative to which the given particle is in motion. For example, if we say that water is flowing or wind is blowing or a person is (ii) Angular momentum Lrp running with a speed v, we mean that these all are relative to the earth (which we have assumed to be fixed). (iii) Velocity vr Y (iv) Force on a charged particle q moving with velocity v in a Y P magnetic field B is given by Fq(vB) r PS' (v) Torque on a dipole in a field  pE and MB r E B PS X Lami's Theorem S r S'S X In any ABC with sides a,b,c S Fig. 0.15 Now to find the velocity of a moving object relative to another sin sin sin moving object, consider a particle P whose position relative to frame S is a b c   r while relative to S is r . 180 –  PS PS   If the position of frames S relative to S at any time is rSS then    c b from figure, rPS rPSrSS Differentiating this equation with respect to time   180 –     dr dr dr 180 –  a dPtS  dPtS  dStS i.e. for any triangle the ratioF igo. f0 t.1h4e sine of the angle containing the side to the length of the side is a constant.      or v v v [as v dr /dt] PS PS SS For a triangle whose three sides are in the same order we establish the Lami's theorem in the following manner. For the triangle shown    or v v v PS PS SS abc0 [All three sides are taken in order] …(i) 6 Vectors (2) General Formula : The relative velocity of a particle P moving (5) Relative velocity of swimmer : If a man can swim relative to 1   with velocity v with respect to another particle P moving with velocity water with velocity v and water is flowing relative to ground with velocity 1 2       v2is given by, vr12= v1– v2 vR velocity of man relative to ground vM will be given by:       v v v v , i.e., v vv 1 M R M R v 2 So if the swimming is in the direction of flow of water, P 2 v vv M R P Fig. 0.16 1 And if the swimming is opposite to the flow of water, v vv (i) If both the particles are moving in the same direction then : M R   – (6) Crossing the river : Suppose, the river is flowing with velocity r12 1 2    . A man can swim in still water with velocity  . He is standing on one (ii) If the two particles are moving in the opposite direction, then : r m bank of the river and wants to cross the river, two cases arise.   r12 1 2 (i) To cross the river over shortest distance : That is to cross the (iii) If the two particles are moving in the mutually perpendicular directions, then: river straight, the man should swim making angle  with the upstream as shown. r12  1222 A vr B    (iv) If the angle between 1and 2 be , then r12 12 22 –212cos1/2. w vm v vr  (3) Relative velocity of satellite : If a satellite is moving in equatorial    plane with velocity vs and a point on the surface of earth with ve Upstream O Downstream relative to the centre of earth, the velocity of satellite relative to the surface Fig. 0.18     of earth Here OAB is the triangle of vectors, in which OAv ,AB. m r      Their resultant is given by OB. The direction of swimming makes v v v se s e angle  with upstream. From the triangle OBA, we find, So if the satellite moves form west to east (in the direction of rotation of earth on its axis) its velocity relative to earth's surface will be cosr Also sinr v v v m m se s e And if the satellite moves from east to west, i.e., opposite to the Where  is the angle made by the direction of swimming with the shortest distance (OB) across the river. motion of earth, v v (v )v v se s e s e Time taken to cross the river : If w be the width of the river, then (4) Relative velocity of rain : If rain is falling vertically with a time taken to cross the river will be given by   velocity vR and an observer is moving horizontally with speed vM the t w w    1  2 –2 velocity of rain relative to observer will be v v v m r RM R M (ii) To cross the river in shortest possible time : The man should which by law of vector addition has magnitude swim perpendicular to the bank. v  v2 v2 The time taken to cross the river will be: RM R M w direction tan1(vM /vR) with the vertical as shown in fig. t2  m A vr B  – vM   vR  vR  vR w vm vr vM vM Fig. 0.17 Upstream O Downstream Fig. 0.19 Vectors 7       Because AAA and AA is collinear with A In this case, the man will touch the opposite bank at a distance AB down stream. This distance will be given by:  Multiplication of a vector with –1 reverses its direction.   ABt  w or ABr w If AB, then A = B and Aˆ Bˆ . r 2 r     m m If AB0, then A = B but Aˆ Bˆ .  Minimum number of collinear vectors whose resultant can be zero is two.  Minimum number of coplaner vectors whose resultant is zero is three.  Minimum number of non coplaner vectors whose resultant is zero  All physical quantities having direction are not vectors. For is four. example, the electric current possesses direction but it is a scalar   quantity because it can not be added or multiplied according to the rules  Two vectors are perpendicular to each other if A.B0. of vector algebra.    A vector can have only two rectangular components in plane and  Two vectors are parallel to each other if AB0. only three rectangular components in space.  Displacement, velocity, linear momentum and force are polar  A vector can have any number, even infinite components. vectors. (minimum 2 components)  Angular velocity, angular acceleration, torque and angular  Following quantities are neither vectors nor scalars : Relative momentum are axial vectors. density, density, viscosity, frequency, pressure, stress, strain, modulus of  Division with a vector is not defined because it is not possible to elasticity, poisson’s ratio, moment of inertia, specific heat, latent heat, divide with a direction. spring constant loudness, resistance, conductance, reactance, impedance, permittivity, dielectric constant, permeability, susceptibility, refractive  Distance covered is always positive quantity. index, focal length, power of lens, Boltzman constant, Stefan’s constant, Gas constant, Gravitational constant, Rydberg constant, Planck’s constant  The components of a vectors can have magnitude than that of the etc. vector itself.  Distance covered is a scalar quantity.  The rectangular components cannot have magnitude greater than that of the vector itself.  The displacement is a vector quantity.  When we multiply a vector with 0 the product becomes a null  Scalars are added, subtracted or divided algebraically. vector.  Vectors are added and subtracted geometrically.  The resultant of two vectors of unequal magnitude can never be a  Division of vectors is not allowed as directions cannot be divided. null vector.  Unit vector gives the direction of vector.  Three vectors not lying in a plane can never add up to give a null vector.  Magnitude of unit vector is 1.  A quantity having magnitude and direction is not necessarily a  Unit vector has no unit. For example, velocity of an object is 5 ms–1 vector. For example, time and electric current. These quantities have due East. magnitude and direction but they are scalar. This is because they do not  obey the laws of vector addition. i.e. v 5ms1 due east.   A physical quantity which has different values in different vˆ  v  5ms1(East)East directions is called a tensor. For example : Moment of inertia has |v| 5ms1 different values in different directions. Hence moment of inertia is a tensor. Other examples of tensor are refractive index, stress, strain, So unit vector vˆ has no unit as East is not a physical quantity. density etc.  Unit vector has no dimensions.  The magnitude of rectangular components of a vector is always less than the magnitude of the vector  ˆi.ˆi ˆj.ˆjkˆ.kˆ 1     If AB, then A B , A B and A B .  ˆiˆi ˆjˆjkˆkˆ 0 x x y y z z            If ABC. Or if ABC0, then A, B and C lie in  ˆiˆjkˆ, ˆjkˆ ˆi, kˆˆi ˆj one plane.  ˆi.ˆjˆj.kˆ kˆ.ˆi 0  If AB C, then C is perpendicular to A as well as B.                  AA0. Also AA0 But AA AA  If | AB|| AB| , then angle between A and Bis 90°.  Resultant of two vectors will be maximum when  = 0° i.e. vectors 8 Vectors are parallel. R  P2Q22PQcos0 | PQ| max  Resultant of two vectors will be minimum when  = 180° i.e. vectors are anti-parallel. R  P2Q22PQcos180 | PQ| min Thus, minimum value of the resultant of two vectors is equal to the difference of their magnitude.  Thus, maximum value of the resultant of two vectors is equal to the sum of their magnitude. When the magnitudes of two vectors are unequal, then R PQ0 min   [| P|| Q|]   Thus, two vectors P and Q having different magnitudes can never be combined to give zero resultant. From here, we conclude that the minimum number of vectors of unequal magnitude whose resultant can be zero is three. On the other hand, the minimum number of vectors of equal magnitude whose resultant can be zero is two.    Angle between two vectors A and B is given by   A.B cos   | A|| B|    Projection of a vector A in the direction of vector B   A.B   | B|    Projection of a vector B in the direction of vector A   A.B   | A|     If vectors A,Band C are represented by three sides ab, bc and ca respectively taken in a order, then    | A| | B| | C|   ab bc ca  The vectors ˆiˆjkˆ is equally inclined to the coordinate axes at an angle of 54.74 degrees.        If ABC, then A.BC0.        If A.BC0, then A.B and C are coplanar.    If angle between A and B is 45°,     then A.B| AB|       If A A A ...... A 0 and A A A ...... A 1 2 3 n 1 2 3 n then the adjacent vector are inclined to each other at angle 2/n.      If ABC and A2 B2 C2, then the angle between A  and Bis 90°. Also A, B and C can have the following values. (i) A = 3, B = 4, C = 5 (ii) A = 5, B = 12, C = 13 (iii) A = 8, B = 15, C = 17. Vectors 9 (c) 4 (d) 5 10. A hall has the dimensions 10m12m14m.A fly starting at one corner ends up at a diametrically opposite corner. What is the magnitude of its displacement (a) 17 m (b) 26 m Fundamentals of Vectors (c) 36 m (d) 20 m 11. 100 coplanar forces each equal to 10 N act on a body. Each force 1. The vector projection of a vector 3ˆi4kˆ on y-axis is makes angle /50with the preceding force. What is the resultant [RPMT 2004] of the forces (a) 5 (b) 4 (a) 1000 N (b) 500 N (c) 3 (d) Zero (c) 250 N (d) Zero 2. Position of a particle in a rectangular-co-ordinate system is (3, 2, 5). Then its position vector will be 12. The magnitude of a given vector with end points (4, – 4, 0) and (– 2, – 2, 0) must be (a) 3ˆi 5ˆj2kˆ (b) 3ˆi 2ˆj5kˆ (a) 6 (b) 5 2 (c) 5ˆi 3ˆj2kˆ (d) None of these (c) 4 (d) 2 10 3. If a particle moves from point P (2,3,5) to point Q (3,4,5). Its displacement vector be 13. The expression  1 ˆi 1 ˆj is a   (a) ˆiˆj10kˆ (b) ˆiˆj5kˆ  2 2  (a) Unit vector (b) Null vector (c) ˆiˆj (d) 2ˆi4ˆj6kˆ (c) Vector of magnitude 2 (d) Scalar 4. A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical component be 14. Given vector A2ˆi3ˆj,the angle between A and y-axis is (a) 10 N (b) 3 N [CPMT 1993] (c) 4 N (d) 2.5 N 5. If A3ˆi 4ˆj and B7ˆi 24ˆj,the vector having the same (a) tan13/2 (b) tan12/3 magnitude as B and parallel to A is (c) sin12/3 (d) cos12/3 (a) 5ˆi 20ˆj (b) 15ˆi 10ˆj 15. The unit vector along ˆiˆj is (c) 20ˆi 15ˆj (d) 15ˆi 20ˆj (a) kˆ (b) ˆiˆj 6. VectorA makes equal angles with x, y and z axis. Value of its ˆiˆj ˆiˆj (c) (d) components (in terms of magnitude of A) will be 2 2 (a) A (b) A 16. A vector is represented by 3ˆiˆj2kˆ. Its length in XY plane is [EAMCET (Engg.) 1994] 3 2 (a) 2 (b) 14 3 (c) 3 A (d) A (c) 10 (d) 5 7. If A2ˆi4ˆj5kˆ the direction of cosines of the vector A are 17. Five equal forces of 10 N each are applied at one point and all are lying in one plane. If the angles between them are equal, the 2 4 5 1 2 3 resultant force will be [CBSE PMT 1995] (a) , and (b) , and 45 45 45 45 45 45 (a) Zero (b) 10 N 4 4 3 2 5 (c) 20 N (d) 10 2N (c) ,0and (d) , and 45 45 45 45 45 18. The angle made by the vector Aˆiˆj with x- axis is 8. The vector that must be added to the vector ˆi3ˆj2kˆ and [EAMCET (Engg.) 1999] (a) 90° (b) 45° 3ˆi6ˆj7kˆ so that the resultant vector is a unit vector along (c) 22.5° (d) 30° the y-axis is 19. Any vector in an arbitrary direction can always be replaced by two (a) 4ˆi 2ˆj5kˆ (b) 4ˆi2ˆj5kˆ (or three) (a) Parallel vectors which have the original vector as their (c) 3ˆi4ˆj5kˆ (d) Null vector resultant 9. How many minimum number of coplanar vectors having different (b) Mutually perpendicular vectors which have the original vector magnitudes can be added to give zero resultant as their resultant (a) 2 (b) 3 10 Vectors (c) Arbitrary vectors which have the original vector as their where ˆi,ˆj,kˆ are unit vectors, along the X, Y and Z-axis respectively. resultant (d) It is not possible to resolve a vector The unit vectors rˆ along the direction of sum of these vector is [Kerala CET (Engg.) 2003] 20. Angular momentum is [MNR 1986] (a) A scalar (b) A polar vector (a) rˆ  1 (ˆiˆjkˆ) (b) rˆ  1 (ˆiˆjkˆ) (c) An axial vector (d) None of these 3 2 21. Which of the following is a vector (a) Pressure (b) Surface tension (c) rˆ  1(ˆiˆjkˆ) (d) rˆ  1 (ˆiˆjkˆ) 3 2 (c) Moment of inertia (d) None of these    22. If PQthen which of the following is NOT correct 30. The angle between the two vectorsA3ˆi4ˆj5kˆ and (a) Pˆ Qˆ (b) |P||Q| B 3ˆi4ˆj5kˆ is [DPMT 2000] (c) PQˆ QPˆ (d) PQ Pˆ Qˆ (a) 60° (b) Zero  (c) 90° (d) None of these 23. The position vector of a particle is r (acost)ˆi(asint)ˆj. 31. The position vector of a particle is determined by the expression The velocity of the particle is [CBSE PMT 1995]  (a) Parallel to the position vector r 3t2ˆi4t2ˆj7kˆ (b) Perpendicular to the position vector The distance traversed in first 10 sec is [DPMT 2002] (c) Directed towards the origin (a) 500 m (b) 300 m (d) Directed away from the origin 24. Which of the following is a scalar quantity [AFMC 1998] (c) 150 m (d) 100 m (a) Displacement (b) Electric field  32. Unit vector parallel to the resultant of vectors A4ˆi3ˆjand (c) Acceleration (d) Work  B8ˆi8ˆjwill be [BHU 1995] 25. If a unit vector is represented by 0.5ˆi0.8ˆjckˆ , then the value of ‘c’ is [CBSE PMT 1999; EAMCET 1994] 24ˆi5ˆj 12ˆi5ˆj (a) (b) (a) 1 (b) 0.11 13 13 (c) 0.01 (d) 0.39 6ˆi5ˆj (c) (d) None of these 26. A boy walks uniformally along the sides of a rectangular park of size 13 400 m× 300 m, starting from one corner to the other corner diagonally opposite. Which of the following statement is incorrect 33. The component of vector A2ˆi3ˆjalong the vector ˆiˆjis [HP PMT 1999] [KCET 1997] (a) He has travelled a distance of 700 m 5 (b) His displacement is 700 m (a) (b) 10 2 2 (c) His displacement is 500 m (d) His velocity is not uniform throughout the walk (c) 5 2 (d) 5 27. The unit vector parallel to the resultant of the vectors    A4ˆi3ˆj6kˆ and Bˆi3ˆj8kˆ is [EAMCET 2000] 34. The angle between the two vectors A3ˆi4ˆj5kˆ and  (a) 1(3ˆi6ˆj2kˆ) (b) 1(3ˆi6ˆj2kˆ) B3ˆi4ˆj5kˆ will be [Pb. CET 2001] 7 7 (a) 90° (b) 0° (c) 1 (3ˆi6ˆj2kˆ) (d) 1 (3ˆi6ˆj2kˆ) (c) 60° (d) 45° 49 49 28. Surface area is [J&K CET 2002] Addition and Subtraction of Vectors (a) Scalar (b) Vector 1. There are two force vectors, one of 5 N and other of 12 N at what (c) Neither scalar nor vector (d) Both scalar and vector angle the two vectors be added to get resultant vector of 17 N, 7 N and 13 N respectively 29. With respect to a rectangular cartesian coordinate system, three (a) 0°, 180° and 90° (b) 0°, 90° and 180° vectors are expressed as (c) 0°, 90° and 90° (d) 180°, 0° and 90°    a4ˆiˆj, b 3ˆi2ˆj and c kˆ 2. If A4ˆi3ˆj and B6ˆi8ˆj then magnitude and direction of AB will be