Page 3 : Vector Algebra, , (4) Collinear or parallel vectors : Vectors having, the same or parallel supports are called collinear or, parallel vectors., (5) Co-initial vectors : Vectors having the same, initial point are called co-initial vectors., (6) Coplanar vectors : A system of vectors is said, to be coplanar, if their supports are parallel to the, same plane., Two vectors having the same initial point are, always coplanar but such three or more vectors may or, may not be coplanar., (7) Coterminous vectors : Vectors having the same, terminal point are called coterminous vectors., , Introduction, Vectors represent one of the most important, mathematical systems, which is used to handle certain, types of problems in Geometry, Mechanics and other, branches of Applied Mathematics, Physics and, Engineering., Scalar and vector quantities : Those quantities, which have only magnitude and which are not related, to any fixed direction in space are called scalar, quantities, or briefly scalars. Examples: Mass, Volume,, Density, Work, Temperature etc. Those quantities, which have both magnitude and direction, are called, vectors., Displacement,, velocity,, acceleration,, momentum, weight, force are examples of vector, quantities., , Representation of vectors, Geometrically a vector is represent by a line, segment. For example, a  AB . Here A is called the, initial point and B, the terminal point or tip., Magnitude or modulus of a is expressed as B, , | a | | AB |  AB ., , a, A, , Types of vector, (1) Zero or null vector : A vector whose magnitude, is zero is called zero or null vector and it is represented, by O ., (2) Unit vector : A vector whose modulus is unity,, is called a unit vector. The unit vector in the direction, of a vector a is denoted by â , read as “a cap”. Thus,, | aˆ |  1 ., , aˆ , , a, Vector a, , | a | Magnitude of a, , (8) Negative of a vector : The vector which has the, same magnitude as the vector a but opposite direction,, is called the negative of a and is denoted by a . Thus,, if PQ  a , then QP  a ., (9) Reciprocal of a vector : A vector having the, same direction as that of a given vector a but, magnitude equal to the reciprocal of the given vector is, known as the reciprocal of a and is denoted by a 1 ., 1, Thus, if | a |  a ,| a 1 |  ., a, (10) Localized and free vectors : A vector which is, drawn parallel to a given vector through a specified, point in space is called a localized vector. For example,, a force acting on a rigid body is a localized vector as its, effect depends on the line of action of the force. If the, value of a vector depends only on its length and, direction and is independent of its position in the, space, it is called a free vector., (11) Position vectors : The vector OA which, represents the position of the point A with respect to a, fixed point O (called origin) is called position vector of, the point A. If (x, y, z) are co-ordinates of the point A,, then OA  xi  yj  zk ., (12) Equality of vectors : Two vectors a and b are, said to be equal, if (i) | a | | b | (ii) They have the same, or parallel support and (iii) The same sense., , Properties of vectors, (1) Addition of vectors, (i) Triangle law of addition : If, , in a ABC ,, , AB  a BC  b and AC  c , then AB  BC  AC i.e., a +, b = c., C, c = a+b, , (3) Like and unlike vectors : Vectors are said to be, like when they have the same sense of direction and, unlike when they have opposite directions., , b, a, , A, , B

Page 4 : Vector Algebra, , (ii) Parallelogram law of addition : If in a, parallelogram OACB, OA  a, OB  b and OC  c, B, , C, , whose direction is the same as that of a , if m is, positive and opposite to that of a , if m is negative., Properties of Multiplication of vectors by a, scalar : The following are properties of multiplication, of vectors by scalars, for vectors a, b and scalars m, n., , c = a+b, b, , O, , A, , a, , (ii) (m)(a)  m a, , (iii) m (na)  (mn) a  n(m a), , (iv) (m  n)a  m a  na, , (v) m (a  b)  m a  m b, (4) Resultant of two forces, , Then OA  OB  OC i.e., a  b  c , where OC is a, diagonal of the parallelogram OABC., (iii) Addition in component form : If the vectors, are defined in terms of i, j and k, i.e.,, if, a  a1i  a2 j  a3 k and b  b1 i  b 2 j  b 3 k , then their sum is, , (i) m(a)  (m) a  (m a), , Let P and Q be two forces and R be the resultant, of these two forces then, R  P  Q, , , , Q, , | R |  R  P 2  Q 2  2 PQ cos , , defined as a  b  (a1  b1 )i  (a2  b 2 )j  (a3  b3 )k ., , , , R, , where | P | = P, | Q |  Q ,, , Properties of vector addition : Vector addition, has the following properties., (a) Binary operation : The sum of two vectors is, always a vector., (b) Commutativity : For any two vectors a and b,, a + b = b + a., (c) Associativity : For any three vectors a, b and, , , , Q sin, Also, tan  , P  Q cos , , , , , , P, , Deduction : When | P | | Q | ,, i.e., P = Q, tan  , ,  , , c , a  (b  c )  (a  b )  c ., , P sin , P  P cos , , , , sin , ,  tan ;, 1  cos , 2, , , 2, , (d) Identity : Zero vector is the identity for, addition. For any vector a, 0  a  a  a  0, , Hence, the angular bisector of two unit vectors a, and b is along the vector sum a  b ., , (e) Additive inverse : For every vector a its, negative vector a exists such that a  (a )  (a )  a  0, i.e., (a ) is the additive inverse of the vector a., , Position vector, , (2) Subtraction of vectors : If a and b are two, vectors, then their subtraction a  b, is defined, as a  b  a  (b ) where b is the negative of b having, magnitude equal to that of b and direction opposite to b ., If a  a1 i  a2 j  a3 k , b  b1i  b2 j  b3 k, , and P is any point, then OP is called the position, vector of P with respect to O. P, , If a point O is fixed as the origin in space (or plane), , r, , Then a  b  (a1  b1 )i  (a2  b 2 )j  (a3  b3 )k ., B, a+b, O, , If we say that P is the point r , then we mean that, the position vector of P is r with respect to some origin, O., , b, a, , A, –b, , a +(–b)=a –, b, , B, , Properties of vector subtraction, (i) a  b  b  a, (ii) (a  b )  c  a  (b  c ), (iii) Since any one side of a triangle is less than the, sum and greater than the difference of the other two, sides, so for any two vectors a and b, we have, (a) | a  b | | a | | b |, (b) | a  b | | a | | b |, (c) | a  b | | a | | b |, , O (Origi, n), , (d) | a  b | | a | | b |, , (3) Multiplication of a vector by a scalar : If a is, a vector and m is a scalar (i.e., a real number) then m a, is a vector whose magnitude is m times that of a and, , (1) AB in terms of the position vectors of points, A and B : If a and b are position vectors of points A, and B respectively. Then, OA  a , OB  b, ,  AB = (Position vector of B) – (Position vector of, A), ,  OB  OA  b  a, (2) Position vector of a dividing point : The, position vectors of the points dividing the line AB in, m b  na, the ratio m : n internally or externally are, or, m n, m b  na, ., m n, , Linear combination of vectors

Page 5 : Vector Algebra, , A vector r is said to be a linear combination of, vectors a, b, c ..... etc, if there exist scalars x, y, z etc.,, such that r  x a  y b  zc  ...., Examples : Vectors, , r1  2a  b  3c , r2  a  3b  2c, , are linear combinations of the vectors a, b, c ., (1) Collinear and Non-collinear vectors : Let a, and b be two collinear vectors and let x be the unit, vector in the direction of a . Then the unit vector in the, direction of b is x or  x according as a and b are like, or, unlike, parallel, vectors., Now,, a | a | xˆ a  (b  c )  (a  b )  c and b   | b | xˆ ., , | a | ,  | a| , | b | xˆ  a   ,  b  a  b ,, |, b, |, ,  | b| , , ,  a  , , , | a|, . Thus, if, | b|, , a, b, , where, , are collinear vectors, then, , a   b or b   a for some scalar  ., (2) Relation between two parallel vectors, (i) If a and b be two parallel vectors, then there, exists a scalar k such that a  k b i.e., there exist two, non-zero scalar quantities x and y so that x a  y b  0 ., If a and b be two non-zero, non-parallel vectors, then x a  y b  0  x  0 and y  0 ., a  0 , b  0, , or, , Obviously x a  y b  0   x  0, y  0, , or, ,  a || b, , (ii) If a  a1 i  a2 j  a3 k and b  b1 i  b2 j  b3 k then, from, , the property, a, a, a, a || b  1  2  3 ., b1 b 2 b 3, , of, , parallel, , vectors,, , we, , have, , (3) Test of collinearity of three points : Three, points with position vectors a, b, c are collinear iff, there exist scalars x, y, z not all zero such that, x a  y b  zc  0 , where x  y  z  0 . If a  a1 i  a2 j ,, , b  b1 i  b 2 j, , and, , c  c1i  c2 j , then the points with, , a1, position vector a, b, c will be collinear iff b1, c1, , a2, b2, c2, , 1, 1 0 ., 1, , (4) Test of coplanarity of three vectors : Let a, and b two given non-zero non-collinear vectors. Then, any vectors r coplanar with a and b can be uniquely, expressed as r  x a  y b for some scalars x and y., (5) Test of coplanarity of Four points : Four, points with position vectors a, b, c , d are coplanar iff, there exist scalars x, y, z, u not all zero such that, x a  y b  zc  u d  0 , where x  y  z  u  0 ., Four points with position vectors, , c  c1 i  c 2 j  c3 k ,, , a  a1 i  a2 j  a3 k , b  b1 i  b 2 j  b 3 k ,, a1, b1, d  d1 i  d 2 j  d 3 k will be coplanar, iff, c1, d1, , a2, b2, c2, d2, , a3, b3, c3, d3, , 1, 1, 0., 1, 1, , Linear independence and dependence of vectors, (1) Linearly independent vectors : A set of nonzero vectors a 1 , a 2 ,..... a n is said to be linearly, independent,, x 1 a 1  x 2 a 2  .....  x n a n  0  x 1  x 2  .....  x n  0 ., , if, , (2) Linearly dependent vectors : A set of vectors, a 1 , a 2 ,..... a n is said to be linearly dependent if there, exist scalars, , x 1 , x 2 ,......, x n, , not all zero such that, , x1a 1  x 2a 2  .....  x na n  0, Three vectors a  a1 i  a2 j  a3 k ,, , b  b1 i  b 2 j  b 3 k, , and c  c1 i  c 2 j  c 3 k will be linearly dependent vectors, a1, iff b 1, c1, , a2, b2, c2, , a3, b3  0 ., c3, , Properties, of, linearly, independent, and, dependent vectors, (i) Two non-zero, non-collinear vectors are linearly, independent., (ii) Any two collinear vectors are linearly, dependent., (iii) Any three non-coplanar vectors are linearly, independent., (iv) Any three coplanar vectors are linearly, dependent., (v) Any four vectors in 3-dimensional space are, linearly dependent., , Scalar or Dot product, (1) Scalar or Dot product of two vectors : If a and, b are two non-zero vectors and  be the angle between, them, then their scalar product (or dot product) is, denoted by a . b and is defined as the scalar | a | | b | cos  ,, where | a | and | b | are modulii of a and b respectively, and 0     . Dot product ofB two vectors is a scalar, quantity., b, , , O, , a, , A, , Angle between two vectors : If a, b be two vectors, inclined at an angle  , then a . b | a | | b | cos ,  cos  , ,  a.b, a.b,    cos 1 , | a || b|, | a | | b |, , , , , , If a  a1 i  a 2 j  a3 k and b  b1 i  b 2 j  b 3 k ; then

Page 6 : Vector Algebra, , , , , , , ., b12  b 22  b 32 , , If a number of forces are acting on a particle, then, the sum of the works done by the separate forces is, equal to the work done by the resultant force., , a1 b1  a 2 b 2  a 3 b 3, ,   cos 1 , ,  a12  a 22  a 32, , , (2) Properties of scalar product, (i) Commutativity : The scalar product of two, vector is commutative i.e., a . b  b . a ., , Vector or Cross product, (1) Vector product of two vectors : Let a, b be two, non-zero, non-parallel vectors., , (ii) Distributivity of scalar product over vector, addition The scalar product of vectors is distributive, over vector addition i.e., (a) a . (b  c )  a . b  a . c ,, (Left, , b, , distributivity), (b) (b  c ). a  b . a  c . a , (Right distributivity), (iii) Let a, a . b  0  ab ., , and, , b, , be, , the co-ordinate axes, therefore, i . j  j . i  0 ; j . k  k . j  0;, , k.i  i .k  0 ., For any vector a , a . a | a | 2 ., , As i, j, k are unit vectors along the co-ordinate axes,, therefore i . i | i | 2  1 , j . j | j | 2  1 and k. k | k | 2  1, (v) If m, n are scalars and a, b be two vectors, then, (vi), For any vectors a and b , we have, (a) a . (b)  (a . b)  (a). b, (b) (a). (b)  a . b, (vii) For any two vectors a and b , we have, (a) | a  b | 2  | a | 2  | b | 2  2a . b, (b) | a  b | 2  | a | 2  | b | 2  2a . b, , (e) | a  b | 2  | a | 2  | b | 2  a b, (f) | a  b |  | a  b |  ab, (3) Scalar product in terms of components: If, a  a1 i  a2 j  a3 k and b  b1 i  b 2 j  b 3 k , then, a . b  a1b1, , a2b2  a3b3 ., The components of b along and perpendicular to a,  a.b , , ,  a and b   a . b  a respectively., are , | a | 2 , | a | 2 , , , , , (4) Work done by a force :, , | F | | OA | cos   F . OA  F .d ,, B, , d  OA, F, , , O, , (ii) If a, b are two vectors and m, n are scalars,, (iii) Distributivity of vector product over vector, addition., Let a, b, c be any three vectors. Then, (a) a  (b  c )  a  b  a  c, , (Left distributivity), , (b) (b  c )  a  b  a  c  a, , (Right distributivity), , a b  a c ., (v) The vector product of two non-zero vectors is, zero vector iff they are parallel (Collinear) i.e.,, a  b  0  a || b, a, b are non-zero vectors., , (d) | a  b | | a | | b |  a || b, , =, , (2) Properties of vector product, (i) Vector product is not commutative i.e., if a and, b are any two vectors, then a  b  b  a , however,, a  b  (b  a ), , (iv) For any three vectors a, b, c we have a  (b  c ) , , (c) (a  b ). (a  b )  | a | 2  | b | 2, , done, , where  is the angle between a and b, η̂ is a unit, vector perpendicular to the plane of a and b such that, a , b , η̂ form a right-handed system., , then m a  nb  mn(a  b)  m (a  nb)  n(m a  b) ., , m a . nb  mn(a . b)  (mn a). b  a .(mn b), , Work, , a, , Then a  b | a | | b | sin  η̂ , and | a  b | | a || b | sin  ,, , As i, j, k are mutually perpendicular unit vectors along, , (iv), , , O, , two non-zero vectors, , A, , Work done = (Force). (Displacement), , where, , It follows from the above property that a  a  0 for, every non-zero vector a , which in turn implies that, i i  j  j  k  k  0 ., (vi), Vector product of orthonormal triad of, unit vectors i, j, k using the definition of the vector, product,, we, obtain, i  j  k, j  k  i, k  i  j ,, , j  i  k, k  j  i, i  k  j ., (3) Vector product in terms of components: If, a  a1i  a2 j  a3 k and b  b1 i  b2 j  b3 k ., i, Then, a  b  a1, b1, , j, a2, b2, , k, a3, b3, , (4) Angle between two vectors : If  is the angle, | a b|, between a and b , then sin  , ., | a || b |, (5) (i) Right handed system of vectors : Three, mutually perpendicular vectors a, b, c form a right, handed system of vector iff a  b  c , b  c  a , c  a  b

Page 7 : Vector Algebra, , Examples: The unit vectors i , j , k form a righthanded system, i  j  k, j  k  i, k  i  j, Y, , 1, | AB  AC | or, 2, , 1, 1, | BC  BA | or | CB  CA |, 2, 2, , b, , (vi) If a, b, c are position vectors of vertices of a, , j, k, , (v) The area of a triangle ABC is, , X, , i, , ABC , then its area =, , a, , c, , 1, | (a  b )  (b  c )  (c  a )|, 2, , (vii) Three points with position vectors a, b, c are, Z, , collinear if (a  b)  (b  c )  (c  a)  0 ., , (ii) Left handed system of vectors : The vectors, a, b, c mutually perpendicular to one another form a, left, handed, system, c  b  a, a  c  b, b  a  c, , of, , vector, , iff, , a, , b, , Scalar triple product, (1) Scalar triple product of three vectors : If, a, b, c are three vectors, then their scalar triple product, , c, , (6) Vector normal to the plane of two given, vectors : If a, b be two non-zero, nonparallel vectors, and, let, be, the, , where, a  b | a | | b | sin  η̂, , (8) Moment of a force : Moment of a force F about a, point O is OP  F , where P is any point on the line of, action of the force F., , angle, is, η̂, , between, a unit, , them., vector, , perpendicular to the plane of a and b such that a, b, η, , is defined as the dot product of two vectors a and b ×, c. It is generally denoted by, a . (b × c) or [a b c]., (2) Properties of scalar triple product, (i) If a, b, c are cyclically permuted, the value of, scalar triple product remains the same., (a  b). c  (b  c ). a  (c  a). b or [a b c ]  [b c a]  [c a b], , i.e.,, , form a right-handed system., , , (a  b) | a  b | η̂  ˆη , , Thus,, , ab, | a  b|, , ab, is a unit vector perpendicular to the, | ab|, , (ii) The change of cyclic order of vectors in scalar, triple product changes the sign of the scalar triple, product, but, not, the, magnitude, i.e.,, [a b c ]  [b a c ]  [c b a]  [a c b], , is also a unit, , (iii) In scalar triple product the positions of dot and, cross can be interchanged provided that the cyclic, order of the vectors remains same i.e., (a  b). c  a .(b  c ), , vector perpendicular to the plane of a and b. Vectors of, magnitude '  ' normal to the plane of a and b are given, (a  b ), by , ., | a b|, , (iv) The scalar triple product of three vectors is, zero if any two of them are equal., , plane of a and b. Note that , , ab, | a  b|, , (7) Area of parallelogram and triangle, (i) The area of a parallelogram with adjacent sides, a and b is | a  b | ., (ii) The area of a parallelogram with diagonals d 1, and d 2 is, , 1, | d1  d 2 | ., 2, , (iii) The area of a plane quadrilateral ABCD is, 1, | AC  BD | , where AC and BD are its diagonals., 2, (iv) The area of a triangle with adjacent sides a, 1, and b is | a  b |, 2, , (v) For any three vectors a, b, c and scalar  ,, , [ a b c ]  [a b c ], (vi) The scalar triple product of three vectors is, zero if any two of them are parallel or collinear., (vii) If a, b, c , d are four vectors, then [(a  b) c d] , , [a c d]  [b c d], (viii) The necessary and sufficient condition for, three non-zero non-collinear vectors a, b, c to be, coplanar is that [a b c ]  0 ., (ix) Four points with position vectors a, b, c and d, will be coplanar, if [a b c ]  [d c a]  [d a b]  [a b c ] ., (x) Volume of parallelopiped whose coterminous, edges are, a, b, c is [a b c ] or a(b × c)., (3) Scalar triple product in terms of components, (i) If a  a1i  a2 j  a3 k , b  b1i  b2 j  b3 k

Page 8 : Vector Algebra, and c  c1i  c2 j  c3 k be three vectors, , (ii) a . b  a . c   0; b . c   b . a  0; c . a  c . b  0, , a1, then, [a b c ]  a2, a3, , (iii) [a  b  c ] , , b1, b2, b3, , c1, c2, c3, , (iv) a, b, c are non-coplanar iff so are a , b , c  ., , (ii) If a  a1l  a2 m  a3 n, b  b1l  b2 m  b3 n, , c  c1l  c2 m  c3 n ,, , and, a1, [a b c ]  b 1, c1, , a2, b2, c2, , 1, [a b c ], , Vector triple product, then, , Let a, b, c be any three vectors, then the vectors, a  (b  c ) and (a  b)  c are called vector triple product, , a3, b 3 [l m n ], c3, , of a, b, c ., Thus, a  (b  c )  (a . c ) b  (a . b)c, , (iii) For any three vectors a, b and c, (a) [a  b b  c c  a]  2[a b c ], (b) [a  b b  c c  a]  0, (c) [a  b b  c c  a ]  [a b c ] 2, (4) Tetrahedron : A tetrahedron is a threedimensional figure formed by four triangle OABC is a, tetrahedron with ABC as the base. OA, OB, OC, AB, BC, and CA are known as edges of the tetrahedron., OA, BC ; OB, CA and OC, AB are known as the pairs of, opposite edges. A tetrahedron in which all edges are, equal, is called a regular tetrahedron., Any two edges of, A(a), regular tetrahedron are perpendicular to each other., D, B(b), , combination of those two vectors which are within, brackets., (ii) The vector r  a  (b  c ) is perpendicular to a, and lies in the plane of b and c ., (iii) The formula a  (b  c )  (a . c ) b  (a . b)c is true, only when the vector outside the bracket is on the left, most side. If it is not, we first shift on left by using the, properties of cross product and then apply the same, formula., Thus, (b  c )  a  {a  (b  c )} = {(a . c ) b  (a . b)c }, = (a .b )c  (a .c )b, (iv) Vector triple product is a vector quantity., (v) a  (b  c )  (a  b )  c, , a, , b, , Properties of vector triple product, (i) The vector triple product a  (b  c ) is a linear, , c, C(c), , Scalar product of four vectors, (a  b).(c  d) is a scalar product of four vectors. It is, , Volume of tetrahedron, , the dot product of the vectors a  b and c  d ., It is a scalar triple product of the vectors a, b and, , (i) The volume of a tetrahedron, , c  d as well as scalar triple product of the vectors, , 1, = (area of the base) (correspon ding altitude), 3, , a  b, c and d., , , , 1, [ AB BC AD ], 6, , (ii) If a, b, c are position vectors of vertices A, B, and C with respect to O, then volume of tetrahedron, 1, OABC = [a b c ] ., 6, (iii) If a, b, c , d are position vectors of vertices A, B,, C, D of a tetrahedron ABCD, then its volume =, 1, [b  a c  a d  a ] ., 6, (5) Reciprocal system of vectors : Let a, b, c be, three, non-coplanar, vectors,, and, let, b c, c a, ab, . a , b , c  are said to form, a , , b , , c , [abc ], [abc ], [abc ], a reciprocal system of vectors for the vectors a, b, c ., If a, b, c and a , b , c  form a reciprocal system of, vectors, then (i), , a . a   b . b  c . c   1, , (a  b ). (c  d) , , a .c, b.c, , a .d, b .d, , Vector product of four vectors, (1) (a  b) (c  d) is a vector product of four vectors., It is the cross product of the vectors a  b and, c d ., (2), , a  {b  (c  d)}, {(a  b)  c }  d, , are also different, , vector products of four vectors a, b, c and d., , Rotation of a vector about an axis, Let a  (a1 , a2 , a3 ) . If system is rotated about, (i) x-axis through an angle  , then the new, components, of, are, a, (a1, a2 cos   a3 sin ,  a2 sin   a3 cos ) .

Page 9 : Vector Algebra, , (ii) y-axis through an angle  , then the new, components, of, are, a, (a3 sin   a1 cos , a2 , a3 cos   a1 sin ) ., (iii) z-axis through an angle  , then the new, components, of, are, a, (a1 cos   a2 sin ,  a1sin   a2 cos , a3 ) ., , Application of vectors in 3-dimensional geometry, (1) Direction, , cosines, , of, , r  ai  b j  ck, , (9) Shortest distance between two parallel lines :, l1 and l2 be two lines whose equations are, , l1 : r  a1  b1 and l 2 : r  a 2  b 2 respectively., Then, shortest distance, , are, , PQ , , a b, c, ., ,, ,, | r| | r| | r|, (2) Incentre formula : The position vector of the, aa b b cc, incentre of ABC is, ., ab c, (3) Orthocentre formula : The position vector of, a tan A  b tan B  c tan C, the orthocentre of ABC is, ., tan A  tan B  tan C, (4) Vector equation of a straight line passing, through a fixed point with position vector a and, parallel to a given vector b is r  a  b ., (5) The vector equation of a line passing through, two points with position vectors a and b is, r  a  (b  a ) ., (6) If the lines, , Let, , r  a1  b1, , and, , r  a 2  b 2, , are, , coplanar, then [a1 b1 b 2 ]  [a 2 b1 b 2 ] and the equation of, the plane containing them is [r b1 b 2 ]  [a1 b1 b 2 ], , (b 1  b 2 ) . (a 2  a 1 ) [b 1 b 2 (a 2  a 1 )], , | b1  b 2 |, | b1  b 2 |, , Shortest distance between two parallel lines :, The shortest distance between the parallel lines, , r  a 1  b and r  a 2  b is given by d , , | (a 2  a 1 )  b |, ., | b|, , If the lines r  a 1  b 1 and r  a 2  b 2 intersect,, then the shortest distance between them is zero., Therefore, [b1 b 2 (a 2  a1 )]  0,  [(a 2  a1 ) b1b 2 ]  0  (a 2  a1 ).(b1  b 2 )  0 ., (10) If the lines r  a 1   b 1 and r  a 2   b 2 are, coplanar, then [a 1 b 1 b 2 ] = [a 2 b 1 b 2 ] and the equation of, the plane containing them is [r b 1 b 2 ] = [a 1 b 1 b 2 ] or, , [r b1 b 2 ]  [a 2 b1 b 2 ] ., , or, (11) Vector equation of a plane through the point, , [r b1 b 2 ]  [a 2 b1 b 2 ] ., (7) Perpendicular distance of a point from a line, , : Let L is the foot of, P(), perpendicular drawn from, , P( ) on the line r  a  b ., Since r denotes the position, vector of any point on the, line r  a  b . So, let the, B, A, , , r = a+b, position vector of L be, L  (a  b ), a  b ., ,  (a   )b , b, Then PL  a    b  (a   )  ,  | b| 2 , , , , A(a ) and perpendicular to the vector n is (r  a ).n  0 or, , r.n  a.n or r. n  d , where d  a . n . This is known as, the scalar product form of a plane., (12) Vector equation of a plane normal to unit, ˆ d ., vector n̂ and at a distance d from the origin is r.n, If n is not a unit vector, then to reduce the equation, r.n  d to normal form we divide both sides by |n| to, d, n, d, ˆ , , obtain r , or r.n, ., | n| | n|, | n|, (13) The equation of the plane passing through a, point having position vector a and parallel to b and c is, r  a  b  c or [r b c ]  [a b c ] , where  and  are, , The length PL, is the magnitude of PL , and required, length of perpendicular., , scalars., , (8) Image of a point in a straight line : Let Q( ), , is the image of P in r  a  b , then,, , (14) Vector equation of a plane passing through a, point a,b,c is r  (1  s  t) a  s b t  c, ,  2(a   ).b , b  ,  | b| 2, , , , ,   2a  , , P, (), , A, r =(a+b), , B, , L  (a  b ), , , Q( )(imag, e), , or r. (b  c  c  a  a  b )  [a b c ] .

Page 10 : Vector Algebra, (15) The equation of any plane through the, intersection of planes r.n 1  d 1 and r.n 2  d 2 is, , r.(n 1  n 2 )  d1  d 2 , where  is an arbitrary constant., , (b) a.n = d., (23) The equation of sphere with centre at C(c) and, radius ‘a’ is | r  c |  a ., The plane r.n  d, | a .n  d |, R ., | r  a |  R , if, | n|, (24), , (16) The perpendicular distance of a point having, position vector a from the plane r.n  d is given by, | a .n  d |, ., p, | n|, (17) An angle  between the planes r1 .n 1  d1 and, , r2 .n 2  d 2 is given by cos   , , touches, , the, , sphere, , (25) If the position vectors of the extremities of a, diameter of a sphere are a and b, then its equation is, (r  a ).(r  b )  0 or | r| 2  r.(a  b )  a .b  0 ., , n 1 .n 2, ., | n 1 || n 2 |, , (18) Perpendicular distance of a point P (r) from a, line passing through a and parallel to b is given by, 2, ,  (r  a ) .b  , | (r  a )  b |, PM , = (r  a ) 2  ,  , | b|, ,  | b |  , , , 1/2, , ., , (19) The equation of the planes bisecting the angles, between the planes r1 .n 1  d 1 and r2 .n 2  d 2 are, | r.n 1  d 1 |, | r.n 2  d 2 |, , | n1 |, | n2|, , or, , r.n 2  d 2, r.n 1  d 1, , | n1 |, | n2 |, , or, , r.(n 1  n 2 ) , , d1, d,  2 ., | n1 | | n 2 |, , (20) Perpendicular distance of a point P (r) from a, plane passing through a point a and parallel to b and, (r  a ).(b  c ), ., c is given by PM , | b c|, (21) Perpendicular distance of a point P(r) from a, , , , Unit vectors parallel to x-axis, y-axis and z-axis, , are denoted by i, j and k respectively., , , Two unit vectors may not be equal unless they, , have the same direction., , , A unit vector is self reciprocal., , , , The internal bisector of the angle between any, , two, , vectors, , is, , along, , the, , vector, , sum, , of, , the, , corresponding unit vectors., , , The external bisector of the angle between two, , vectors, , is, , along, , the, , vector, , difference, , of, , the, , corresponding unit vectors., , plane passing through the points a, b and c is given by, , PM , , (r  a ). (b  c  c  a  a  b ), ., | b c  c a  a  b|, , (22) Angle between line and plane : If  is the, angle between a line r  (a  b ) and the plane r.n  d ,, then sin  , , b.n, ., | b || n |, , (i) Condition of perpendicularity: If the line is, perpendicular to the plane, then it is parallel to the, normal to the plane. Therefore b and n are parallel., So, b  n  0 or b = n for some scalar ., (ii) Condition of parallelism : If the line is parallel, to the plane, then it is perpendicular to the normal to, the plane. Therefore b and n are perpendicular. So, b.n, = 0., (iii) If the line r  a  b lies in the plane r.n = d,, then, (a) b.n = 0, , , , If a, b, c, , are position vectors of vertices of a, , triangle, then position vector of its centroid is, a b c, ., 3, , , , If a, b, c, d are position vectors of vertices of a, , tetrahedron, then position vector of its centroid is, a b c d, ., 4, , , , Lagrange's identity: If a, b are any two vectors,, , then, | a  b | 2 (a . b )2 | a | 2| b | 2, , , , a . b | a | | b | ., , | a  b | 2 | a | 2| b | 2 (a . b )2 or

Page 11 : Vector Algebra, , , , a . b  0  Angle between a and b is acute., , , , a . b  0  Angle between a and b is obtuse., , , , The dot product of a zero and non-zero vector is a, , 6., , scalar zero., , , Centre, , the, , sphere, , is, , the, , centroid, , of, , , , (d) (cos 2 ) i  (sin 2 ) j, , If a  b bisects the angle between a and b, then a, and b are, , 1, , 1, ., 3, , The distance of any vertex from the opposite face, , of regular tetrahedron is, any edge., , (b), , 7, (i  2 j  2 k), 3, , 7, (d) None of these, (i  2 j  2 k), 9, If p  7 i  2 j  3 k and q  3 i  j  5 k, then the, magnitude of p  2 q is, [MP PET 1987], (c), , 8., , 2, k , k being the length of, 3, , Unlike vectors, , (d) None of these, , If a  i  2 j  2 k and b  3 i  6 j  2 k, then a vector, in the direction of a and having magnitude as |b|, is [IIT 1983], (a) 7 (i  j  k), , The angle between any two plane faces of a, , regular tetrahedron is cos, , (c) (sin 2 ) i  (cos  ) j, , (c) Equal in magnitude, , tetrahedron., , , (b) (sin  ) i  (cos  ) j, , (a) Mutually perpendicular (b), 7., , of, , (a) (cos  )i  (sin  ) j, , 9., , (a), , 29, , (b) 4, , (c), , 62  2 35, , (d), , 66, , Let a  i be a vector which makes an angle of, 120 o with a unit vector b. Then the unit vector, (a  b ) is, [MP PET 1991], , (a) , (c), , Modulus of vector, Algebra of vectors, 1., , The perimeter of a triangle with sides 3i  4 j  5 k,, 4 i  3 j  5 k and 7 i  j is, , 2., , 3., , 5., , [MP PET 1991], , (a), , 450, , (b), , 150, , (c), , 50, , (d), , 200, , 11., , If the position vectors of the vertices of a triangle, be 2i  4 j  k, 4 i  5 j  k and 3 i  6 j  3 k, then the, triangle is, (a) Right angled, , (b) Isosceles, , (c) Equilateral, isosceles, , (d) Right, , (b) 13, , (c) 25, (d) 50, If a  2i  2 j  k and | x a |  1, then x =, (a) , , 1, 3, , (b) , , 1, 4, , (c) , , 1, 5, , (d) , , 1, 6, , Which of the following is not a unit vector for all, values of , , (d), , 3, 1, i j, 2, 2, , triangle is, (a) Right angled, (b) Isosceles, (c) Equilateral, (d) None of these, The perimeter of the triangle whose vertices have, the position vectors (i  j  k), (5i  3 j  3k) and, 157, (a) 15 2004], [UPSEAT, , 12., , 3, 1, i j, 2, 2, , vectors of the vertices of a triangle be, 6i  4 j  5 k, 4 i  5 j  6 k and 5 i  6 j  4 k, then the, , (c), angled, , 1, 3, i, j, 2, 2, , (b) , , (2i  5 j  9 k ), is given by, , 15  157, , [MP PET 1993], , (b) 15  157, (d), , 15  157, , The position vectors of two points A and B are, , i  j  k and 2 i  j  k respectively. Then | AB |  [BIT Ranch, , If one side of a square be represented by the, vector 3 i  4 j  5 k, then the area of the square is, (a) 12, , 4., , 10., , 1, 3, i, j, 2, 2, , 13., , 14., , (a) 2, (b) 3, (c) 4, (d) 5, The magnitudes of mutually perpendicular forces, a, b and c are 2, 10 and 11 respectively. Then the, magnitude of its resultant is, (a) 12, (b) 15, (c) 9, (d) None, The system of vectors i, j, k is, (a) Orthogonal, (c) Collinear, , (b) Coplanar, (d) None of these, , [

Page 12 : Vector Algebra, 15., , 16., , 17., , 18., ,  1, 1, 1 , ,, ,, (a) , 3, 6 ,  2, ,  1, 1, 1 , , ,, ,, (b) , 6, 6 ,  6, ,  1, 1, 1 , ,, ,, (c)  , 6, 6, 6 , , ,  1, 1, 1 , ,, ,, (d) , 3, 3 ,  3, , The position vectors of P and Q are 5 i  4 j  ak and, i  2 j  2 k respectively. If the distance between, them is 7, then the value of a will be, (a) – 5, 1, (b) 5, 1, (c) 0, 5, (d) 1, 0, A zero vector has, (a) Any direction, (b) No direction, (c) Many directions, (d) None of these, , A unit vector a makes an angle, with z-axis. If, 4, a  i  j is a unit vector, then a is equal to, (a), , i j, k,  , 2 2, 2, , (c) , 19., , 20., , 21., , [EAMCET 1988], , The direction cosines of the resultant of the, vectors (i  j  k ), (i  j  k ), (i  j  k ) and (i  j  k ),, are, , i, j, k,  , 2 2, 2, , (b), , i, j, k,  , 2 2, 2, , (d) None of these, , A force is a, (a) Unit vector, (b) Localised vector, (c) Zero vector, (d) Free vector, If a, b, c, d be the position vectors of the points A,, B, C and D respectively referred to same origin O, such that no three of these points are collinear, and a  c  b  d, then quadrilateral ABCD is a, , 25., , with, position, i  j  k, i  j  k, i  j  k, , 26., , [EAMCET 1994], , 27., , 22., , (b) , , 162, 162, (c) – 5, (d) 11, If the resultant of two forces is of magnitude P, and equal to one of them and perpendicular to it,, then the other force is, , 23., , 24., , (b) P, , (c) P 3, (d) None of these, The direction cosines of vector a  3 i  4 j  5 k in, the direction of positive axis of x, is, 3, 4, (a) , (b), 50, 50, 3, 4, (c), (d) , 50, 50, The point having position vectors 2i  3 j  4 k,, 3 i  4 j  2 k, 4 i  2 j  3 k are the vertices of, , If OP = 8 and OP makes angles 45 o and 60 o with, (a) 8 ( 2i  j  k), , (b) 4 ( 2i  j  k), , 1, 1, [IIT 1988], (c), (d) ( 2 i  j  k), ( 2 i  j  k), 4, 8, 28. If a and b are two non-zero and non-collinear, vectors, then a + b and a – b are, (a) Linearly dependent vectors, (b) Linearly independent vectors, (c) Linearly dependent and independent vectors, (d) None of these, 29. If, the, vectors, and, 6i  2 j  3k, 2i  3 j  6 k, 3 i  6 j  2 k form a triangle, then it is, , 30., , (a) Right angled, (b) Obtuse angled, (c) Equilteral, (d) Isosceles, If the resultant of two forces of magnitudes P and, Q acting at a point at an angle of 60 o is, P/Q is, , 7 Q, then, , [Roorkee 1999], , 3, (b), 2, , (a) 1, , (c) 2, (d) 4, The direction cosines of the vector 3 i  4 j  5 k are, , 31., , [Karnataka CET 2000], , (a), (c), , [MNR 1986], , (a) P 2, , (b) 5, (d) 3, , OX-axis and OY-axis respectively, then OP , , [MNR 1989], , (a), , (a) Are collinear, (b) Form an equilateral triangle, (c) Form a scalene triangle, (d) Form a right angled triangle, If | a |  3, | b |  4 and | a  b |  5, then | a  b | , (a) 6, (c) 4, , and 5 i  2 j  4 k, then the direction cosine of AB, along y-axis is, , 5, , vectors, [IIT Screening 1994], , (a) Square, (b) Rhombus, (c) Rectangle, (d) Parallelogram, If the position vectors of A and B are i  3 j  7 k, , 4, , (a) Right angled triangle (b) Isosceles triangle, (c) Equilateral triangle (d) Collinear, Let  ,  ,  be distinct real numbers. The points, , 3 4 1, ,, ,, 5 5 5, , 3, , ,, , 2, 32., , 4, 2, , ,, , (b), , 1, , (d), , 2, , 3, 4, 1, ,, ,, 5 2 5 2, 2, 3, 5 2, , ,, , 4, 5 2, , ,, , 1, 2, , The position vectors of A and B are 2i  9 j  4 k and, 6 i  3 j  8 k respectively, then the magnitude of, , is 1991], [MPAB, PET, [MP PET 2000], , 33., , (a) 11, (b) 12, (c) 13, (d) 14, If the position vectors of P and Q are (i  3 j  7 k ), and (5 i  2 j  4 k ), then | PQ | is, (a), , 158, , (b), , [, , 160

Page 13 : Vector Algebra, , (c), 34., , 1, | a|, , 37., , 38., , 39., , [MP PET 2002], , (a)  3 (i  j  k), , Form an isosceles triangle, Form a right-angled triangle, Are collinear, Form a scalene triangle, , 18, , (b), , 72, , (c), , 33, , (d), , 288, , If the position vectors of the vertices A, B, C of a, i  6 j  6 k, triangle ABC are 7 j  10 k,, and, 4 i  9 j  6 k respectively, the triangle is, (a) Equilateral, (b) Isosceles, (c) Scalene, (d) Right angled and isosceles also, The figure formed by the four, i  j  k, 2i  3 j, 3 i  5 j  2 k and k  j is, , 2 2, 5, , 40., , (d) 30, , If, , ABCDEF, , is, , a, , If a and b be unlike vectors, then a . b =, (a) | a | | b |, (b) – | a | | b |, (c) 0, (d) None of these, If a, b, c are unit vectors such that a  b  c  0 ,, then a . b  b . c  c . a , , 3, , (d), , , 2, , If a, b, c are mutually perpendicular unit vectors,, then | a  b  c | , [Karnataka CET 2002, 05; J &, , 8., , (a) 3, (c) 1 [MP PET 2004], If | a |  | b | | c | and, between a and b is, , (a), 2, , 9., , (b) 3, (d) 0, a  b  c , then the angle, (b) , , (c) 0, (d) None of these, If a has magnitude 5 and points north-east and, vector b has magnitude 5 and points north-west,, then | a  b | , [MNR 1984], , (a) 25, hexagon, , and, , AB  AC  AD  AE  AF   AD, then   [RPET 1985], (a) 2, (c) 4, , (d) None of these, , [UPSEAT, 2004], 1, , and 6 act along, , regular, , (b) a  (b  c ), , (c) b  c or a  (b  c ), , (c) cos 1, , [Roorkee 1999], , (c) 11  2 2, , (a) b = c, , (b) 3, (d) 6, , 10., , (b) 5, , (c) 7 3, (d) 5 2, If  be the angle between the unit vectors a and b,, , then cos , [MP PET 1998; Pb., 2, CET 2002], , Scalar or Dot product of two vectors and its, applications, 1., , (a . i)i  (a . j)j  (a . k) k , (a) a, , [Karnataka CET 2004], , (b) 2 a, , that, , (a) 1, (b) 3, (c), –, 3/2, (d) 3/2, [AIEEE 2003], 6., If a, b, c are mutually perpendicular vectors of, equal magnitudes, then the angle between the, vectors a and a  b  c is, , , (a), (b), 6, 3, , points, , their resultant force is, (b) 5, , (d)  3 (i  j  k), (i  j  k), 3, If a, b, c are non-zero vectors such, a . b  a . c , then which statement is true, , K 2005], , BC, CA and AB respectively. The magnitude of, , (a) 4, , 1, (i  j  k ), 3, , [MP PET 1988; Karnataka CET 2000; UPSEAT 2003, 04], , 7., , (a) Rectangle, (b) Parallelogram, (c) Trapezium, (d) None of these, ABC is an isosceles triangle right angled at A., Forces of magnitude, , 4., , 5., , The vectors AB  3 i  4 k, and AC  5 i  2 j  4 k are, the sides of a triangle ABC. The length of the, median through A is, (a), , 3., , (b) , , 1, , (c) , , [Kurukshetra CEE, 2002], , 36., , 2., , (d) m   2, , The position vectors of the points A, B, C are, (2 i  j  k ), (3 i  2 j  k ) and (i  4 j  3 k ) respectively., These points, , (a), (b), (c), (d), , (c) 0, (d) None of these, If r. i  r. j  r. k and | r|  3, then r , , 162, , If a is non zero vector of modulus a and m is a, non-zero scalar, then ma is a unit vector if, (a) m  1, (b) m  | a |, (c) m , , 35., , (d), , 161, , (a), , 1, | a  b|, 2, , (b), , 1, | a  b|, 2, , (c), , | a  b|, | a  b|, , (d), , | a  b|, | a  b|

Page 14 : Vector Algebra, 11., , If | a |  3, | b |  4, | c |  5 and a  b  c  0, then the, , 19., , angle between a and b is [MP PET 1989; Bihar CEE, 1994], , , 6, , , (c), (d), 3, 2, If | a  b |  | a  b |, then the angle between a and b, is, (a) Acute, (b) Obtuse, , (c), (d) , 2, If a, b, c are three vectors such that a  b  c and, the angle between b and c is  / 2, then, (a) 0, , 12., , 13., , product of F1  F2  F3 and AB will be [Roorkee 1980], , (b), , 20., , 21., , (b) b 2  c 2  a 2, , (c) c 2  a 2  b 2, (d) 2 a 2  b 2  c 2, (Note : Here a  | a |, b | b |, c | c |), 14., , 15., , equal vectors, like vectors, unlike vectors, unit vectors, i  j  k makes angles ,  ,  with, , vectors i, j, k respectively, then, , 16., , 17., , (b)     , , (c)     , , (d)     , , (a) 3r 2, (b) r 2, (c) 0, (d) None of these, The value of b such that scalar product of the, vectors (i  j  k ) with the unit vector parallel to, the sum of the vectors (2i  4 j  5 k), and, (b i  2 j  3 k ) is 1, is, , (a) – 2, (b) – 1, (c) 0, (d) 1, If a unit vector lies in yz–plane and makes angles, of 30 o and 60 o with the positive y-axis and z-axis, respectively, then its components along the coordinate axes will be, (a), (c), , 3 1, , ,0, 2 2, , (b) 0 ,, , 3, 1, , 0,, 2, 2, , 1 3, (d) 0, ,, 2 2, , 3 1, ,, 2 2, , (b) 12, (d) 7, , A vector whose modulus is 51 and makes the,  4 i  3k, i  2 j  2k, ,b, same angle with a , and, 3, 5, c  j, will be, [Roorkee 1987], , (b) 5 i  j  5 k, (d)  (5i  j  5 k), , (a) 5 i  5 j  k, (c) 5 i  j  5 k, , Vector or Cross product of two vectors and its applications, , If a, b, c are any vectors, then the true statement, is, [RPET 1988], , (r . i) 2  (r . j) 2  (r . k) 2 , , [MNR 1992; Roorkee 1985, 95; Kurukshetra CEE 1998;, UPSEAT 2000], , 18., , 22., , 1., , (a)     , , equal to, [RPET 1986], (a) – 5, (b) – 4, (c) 4, (d) 5, If | a |  3, | b |  4 and the angle between a and b, (a) 25, (c) 13, , If the angle between the vectors a and b be  and, a . b  cos  , then the true statement is, (a) a and b are, (b) a and b are, (c) a and b are, (d) a and b are, If the vector, , (a) 3, (b) 6, (c) 9, (d) 12, If the moduli of a and b are equal and angle, between them is 120 o and a . b   8, then | a | is, , be 120 o , then | 4 a  3b | , , [EAMCET 2003], , (a) a 2  b 2  c 2, , If, F2  i  2 j  k,, F1  i  j  k,, F3  j  k,, , , A  4 i  3 j  2 k and B  6 i  j  3 k, then the scalar, , 2., , (a) a  (b  c )  (a  b )  c, , (b) a  b  b  a, , (c) a . (b  c )  a . b  a . c, , (d) a . (b  c )  a . b  a . c, , If a and b are unit vectors such that a  b is also a, unit vector, then the angle between a and b is, , (a) 0, (b), 3, (c), , 3., , , 2, , (d) , , The points A (a), B (b), C (c ) will be collinear if, (a) a  b  c  0, (c) a . b  b . c  c . a  0, , 4., , (a  b )  (a  b ) , , [MP PET 1987], , (b) a  b, , (a) 2( a  b), , (c) a  b, (d) None of these, If a  b  c  0 , then which relation is correct, 2, , 5., , (b) a  b  b  c  c  a  0, (d) None of these, , 2, , [RPET 1985; Roorkee 1981; AIEEE 2002], , (a) a  b  c  0, 6., , (b) a . b  b . c  c . a, , (c) a  b  b  c  c  a, (d) None of these, If  be the angle between the vectors a and b and, | a  b |  a . b, then  , [RPET 1990; MP PET 1990; UPSEAT 2003]

Page 15 : Vector Algebra, , (a) , (c), 7., , 8., , (b), , , 4, , , 2, , 16., , (d) 0, , (2a  3 b )  (5 a  7 b ) , , [MP PET 1988], , (a) a  b, (b) b  a, (c) a  b, (d) 7 a  10 b, If a and b are two vectors such that a . b = 0 and, a  b  0 , then, , 17., , 9., , (c), 10., , a.b a.b, ,, | a| | a|, , (d), , 18., , 11., , (d) a . b  b . c  c . a  0, , (c), 19., , 20., , (d), , 1, , (2i  j  k), , 6, , 36 ,  24, ,, (b) , , 5 ,  5, , 36 ,  24, ,, (c)  , , 5 ,  5, ,  24 36 , ,, (d) , ,  5 5 , , A unit vector which is perpendicular to i  2 j  2 k, and i  2 j  2 k is, [MP PET 1992], (a), , 1, , (2 i  k ), , (b), , 1, , (2i  j  k), , (d), , 5, 21., , 1, , (2i  k), , 5, , 5, , 1, , (2 i  k ), , 5, , If A (1, 2, 3), B (1, 1, 1) and C (2,  1, 3) are points on, a plane. A unit normal vector to the plane ABC is, , (d) u 2 | u | 2, , [BIT Ranchi 1988], , of, to, , vectors, vectors, , of unit length, and, a  (1, 1, 0), ,  2i  2 j  k , (a)  , , 3, , , , b  (0, 1, 1) is, [BIT Ranchi 1991; IIT 1987; Kurukshetra CEE, 1998;, DCE 2000; MP PET 2002], , 22., , (a) Three, (b) One, (c) Two, (d) Infinite, If a  (1,  1, 1) and c  (1,  1, 0), then the vector b, ,  2i  2 j  k , (b)  , , 3, , , ,  2i  2 j  k ,  2i  2 j  k , (c)  , (d)  , , , 3, 3, , , , , The unit vector perpendicular to the vectors, 6 i  2 j  3 k and 3 i  6 j  2 k, is, (a), , 2i  3 j  6 k, 7, , (b), , 2i  3 j  6 k, 7, , (a) (1, 0, 0), (b) (0, 0, 1), (c) (0, –1, 0), (d) None of these, 23., If a  b  b  c  0, where a, b and c are coplanar, vectors, then for some scalar k [Roorkee 1985; RPET 1997], (a) a  c  k b, (b) a  b  k c, , (c), , 2i  3 j  6 k, 7, , (d), , 2i  3 j  6 k, 7, , (c) b  c  k a, , (c) a 2 b 2  (a . b )2, , satisfying a  b  c and a . b  1 is, , 15., , (i  j  k ), ,  24 36 , ,, (a)  , ,  5 5 , , (c), , (c) (u  v)2  u 2 . v 2  (u . v)2, , 14., , 1, , (i  2 j  k), , 6, , If a  2i  3 j  5 k, b  m i  nj  12 k and a  b  0, then, , [MP PET 1987], , 13., , (b), , 1, , (m, n) , , Which of the following is not a property of, vectors, , The number, perpendicular, , (2i  j  k ), , 6, , (a) u  v  v  u, (b) u . v  v . u, , 12., , 1, 6, , [MP PET 1994; BIT Ranchi 1990; IIT 1982; AMU 2002], , (b) b . c  c . a  0, , (c) a  c  b, (d) None of these, A unit vector perpendicular to the plane, determined by the points (1, – 1, 2), (2, 0, – 1) and, (0, 2, 1) is, (a) , , | (a  b). c | | a | | b | | c |, if, , (c) c . a  a . b  0, , (b) (a  c ) | | b, , [IIT 1983; MNR 1984], , | a b| | a b|, ,, | a|, | b|, , (a) a . b  b . c  0, , (a) | u |, (b) | u |+| u . a |, (c) | u |+| u . b |, (d) | u |+ u . (a+b), If a  b  b  c  0 and a  c  0, then, [RPET 1999], (a) (a  c )  b, , [IIT Screening 1989; MNR 1988; UPSEAT 2000, 01], , (a) a is parallel to b, (b) a is perpendicular to b, (c) Either a or b is a null vector, (d) None of these, The components of a vector a along and, perpendicular to the non-zero vector b are, respectively, [IIT 1988], a .b | a b|, a .b | a b|, (a), (b), ,, ,, | b| | b|, | a|, | a|, , Let a and b be two non-collinear unit vectors. If, u  a  (a . b) b and v  a  b , then | v | is [IIT 1999], , [MP PET 1989], , (d) None of these, , If a  0 , b  0 , c  0 , then true statement is [MP PET 1991], 24., (a) a  (b  c )  (c  b )  a, , (b) a . (b  c )  (b  c ). a, , (c) a  (b  c )  (c  b )  a, , (d) a . (b  c )  (c  b). a, , For any two vectors a and b, (a  b ) 2 is equal to, [Roorkee 1975, 79, 81, 85], , (a) a  b, 2, , 2, , (b) a 2  b 2, (d) None of these, , The unit vector perpendicular to 3 i  2 j  k and, 12 i  5 j  5 k, is, [Roorkee 1979; RPET 1989, 91], (a), , 5i  3 j  9 k, 115, , (b), , 5i  3 j  9 k, 115

Page 16 : Vector Algebra, (c), , 5 i  3 j  9 k, , (d), , 5i  3 j  9k, , 25., , The sine of the angle between the two vectors, 3 i  2 j  k and 12 i  5 j  5 k will be, (a), , 115, 14 194, , (c), , 9., , are three non-coplanar vectors, then the value of, (a  b  c ). (p  q  r) is given by, , 51, , (b), , 14 144, , 64, , [MNR 1992; UPSEAT 2000], , (d) None of these, , 14 194, 10., Scalar triple product and their applications, , 1., , 17, 17, (d), 146, 146, b c, c a, a b, If p , , q, , r, , where a, b, c, [a b c ] 1978], [a b c ], [a b c ], [Roorkee, , (c) , , 115, , 115, , (a) 3, (b) 2, (c) 1, (d) 0, The volume of the parallelopiped whose edges are, represented by 12 i  k, 3 j  k and 2i  j  15 k is, 546. Then  , (a) 3, (c) – 3, , If a, b, c are three non-coplanar vector, then, a .b  c b .a  c, , =, [IIT 1985, 86; UPSEAT, c  a .b c .a  b, , [IIT Screening 1989; MNR 1987], , (b) 2, (d) – 2, , Vector triple product, , 2003], , 2., , (a) 0, (b) 2, (c) – 2, (d) None of these, If a, b, c be any three non-coplanar vectors, then, [RPET 1988; MP PET 1990, 02;, [a  b b  c c  a] , , 2., , Kerala (Engg.) 2002], (a) | a b c |, (c) [ a b c ], 3., , 4., , 5., , 6., , 1., , (d) 2 [ a b c ], , (c) a and b, , (d) a, b and c, , If u  i  (a  i )  j  (a  j)  k  (a  k ), then, , (a) u  0, , If the vectors 2i  3 j, i  j  k and 3i  k form three, , (b) u  i  j  k, , (c) u  2a, (d) u  a, concurrent edges of a parallelopiped, then the, 3. 2001;, If a  i  2 j  2k, b  2i  j  k and c  i  3 j  k, then, volume of the parallelopiped is [IIT 1983; RPET 1995; DCE, a  (b  c ) is equal to, [RPET 1989], IV. Kurukshetra CEE 1998; MP PET 2001], (a) 8, (b) 10, (a) 20 i  3 j  7 k, (b) 20 i  3 j  7 k, (c) 4, (d) 14, (c) 20 i  3 j  7 k, (d) None of these, If a, b, c are any three coplanar unit vectors, then, 5., If α  2i  3 j  k, β  i  2j  4 k and γ  i  j  k, then, (a) a . (b  c )  1, (b) a . (b  c )  3, (α  β).(α  γ ) is equal to, (c) (a  b) . c  0, (d) (c  a). b  1, If a and b be parallel vectors, then [a c b] =, (a) 0, (b) 1, (c) 2, (d) None of these, If the vectors 2i  j  k, i  2 j  3k and 3 i  j  5 k, , [MNR 1984; UPSEAT 2000; Orissa JEE 2005], , 6., , [Roorkee 1986; RPET 1999, 02; Kurukshetra CEE 2002], , (a) – 1, (b) – 2, (c) – 3, (d) – 4, If a, b, c are the three non-coplanar vectors and p,, q,, r, are, defined, by, the, relations, b c, c a, a b, p, ,q, , r, then (a+b) . p, [a b c ], [a b c ], [a b c ], , 7., , 8., , (b) 1, (d) 3, position, vectors are, i  j  2 k and 4 i  5 j  k, [IIT 1986; Pb. CET 2003], , (b), , 146, 17, , (b) 64, (d) – 74, , (a) | a | | b | .| c |  1, , (b) b || c, , (c) a || b, , (d) b  c, , [RPET 1995], , a  (b  c ) is equal to, , (a) (a . c ) b  (a . a) b, , (b) (a . c ) a  (b . c ) a, , (c) (a . c ) b  (a . b) c, , (d) (a . b) c  (a . c ) b, , If a  b  c , b  c  a and a, b, c be moduli of the, vectors a, b, c respectively, then, (a) a  1, b  c, (b) c  1, a  1, , [IIT 1988; BIT Mesra 1996; AMU 2002], , (a) 0, (c) 2, If the points whose, 3 i  2 j  k, 2 i  3 j  4 k,, lie on a plane, then  , 146, (a) , 17, , (a) 60, (c) 74, If a  (b  c )  0 , then, , [RPET 1995; Kurukshetra CEE 1998; MP PET 2003], , +(b+c) . q +(c+a) . r =, , 8., , (b) c and a, , 2004; UPSEAT 2000, 02; Kerala (Engg.) 2002], , 2, , be coplanar, then  , , 7., , (a) b and c, , [RPET 1989, 97; MNR 1986, 93; MP PET 1987, 98, 99,, , (b) 2 [a b c ], 2, , a  (b  c ) is coplanar with, , (c) b  2, c  2a, 9., , (d) b  1, c  a, , b  2i  j  k, a  3 i  j  2 k,, If, c  i  2 j  2 k, then (a  b )  c is equal to, , (a) 24 i  7 j  5 k, , (b) 7 i  24 j  5 k, , (c) 12 i  3 j  5 k, , (d) i  j  7 k, , and

Page 17 : Vector Algebra, , i  (j  k ) , , 10., , [RPET, , 1988;, , MP, , PET, , 1997], , (a) 1, (c) – 1, , (b) 0, (d) None of these, , 1, , a, , 2, , c, , 3, , a, , 4, , d, , 5, , b, , 6, , c, , 7, , d, , 8, , a, , 9, , b, , 10, , d, , IV. Modulus of vector, Algebra of vectors, I., , Modulus of vector, Algebra of vectors, , 1., , (a) l1  25  25  5 2 ,, l2  25  25  5 2 , l3  5 2 ., , 1, , a, , 2, , d, , 3, , d, , 4, , a, , 5, , c, , 6, , c, , 7, , b, , 8, , d, , 9, , c, , 10, , c, , 11, , a, , 12, , b, , 13, , b, , 14, , a, , 15, , d, , 16, , a, , 17, , a, , 18, , c, , 19, , b, , 20, , d, , 21, , b, , 22, , a, , 23, , c, , 24, , c, , 25, , b, , 26, , b, , 27, , b, , 28, , b, , 29, , b, , 30, , c, , 31, , b, , 32, , d, , 33, , d, , 34, , c, , 35, , c, , 36, , c, , 37, , d, , 38, , c, , 39, , b, , 40, , b, , Hence, l1  l2  l3  3 50  450 ., 2., , b  i  j  4 k | b |  1  1  16  18, c  i  2 j  2k | c |  1  4  4  9, | a | | c | and also, b 2  a 2  c 2, , Hence it is isosceles and right angled, triangle., 3., , II. Scalar or Dot product of two vectors and, its applications, 1, , a, , 2, , d, , 3, , c, , 4, , b, , 5, , c, , 6, , c, , 7, , a, , 8, , c, , 9, , d, , 10, , b, , 11, , d, , 12, , a, , 13, , a, , 14, , d, , 15, , d, , 16, , b, , 17, , d, , 18, , b, , 19, , c, , 20, , c, , 21, , b, , 22, , d, , 23, , b, , 24, , c, , 25, , c, , III. Vector or Cross product of two vectors, and its applications, 1, , d, , 2, , c, , 3, , b, , 4, , a, , 5, , c, , 6, , c, , 7, , b, , 8, , c, , 9, , b, , 10, , d, , 11, , a, , 12, , c, , 13, , b, , 14, , a, , 15, , c, , 16, , a,c, , 17, , b, , 18, , a, , 19, , c, , 20, , d, , 21, , a, , 22, , c, , 23, , c, , 24, , c, , 25, , a, , (d) a  2i  j  2k | a |  4  1  4  9, , (d) | a |  9  16  25  5 2, Area | a | 2  25  2  50 ., 1, 3, , 4., , (a) | xa | | x || a | | x | 4  4  1  1  x   ., , 5., , (c) Since sin 2 2  cos 2  is not equal to one, necessarily., , 6., , (c) Equal in magnitude, as bisector , , aˆ  bˆ, , if, , | a | | b | ., 7., ,  i  2 j  2k  7,   (i  2 j  2 k)., (b) | b | aˆ  9  36  4 , , , 8., , (d) p  2q  i  4 j  7 k, ,  14 4 , , 3, , | p  2q |  49  16  1  66 ., 9., , 1, 2, , (c) b  cos 120 i  sin 120 j or b   i , Therefore a  b  i , , 10., , 3, j., 2, , 1, 3, 1, 3, i, j i, j., 2, 2, 2, 2, , (c) Equilateral, since each side is of length, 6., , Scalar triple product and their applications, , 11., , (a) a  4 i  2 j  4 k  | a |  16  16  4  6, , 1, , a, , 2, , b, , 3, , c, , 4, , c, , 5, , a, , b  3i  2 j  12 k  | b |  144  4  9  157, , 6, , d, , 7, , d, , 8, , a, , 9, , a, , 10, , c, , c  i  4 j  8 k  | c |  64  16  1  9, , Vector triple product, , 2, , Hence perimeter is 15  157 .

Page 18 : Vector Algebra, 12., , (b) AB  i  2 j  2k | AB |  3., , 13., , (b) R  4  100  121  15 ., (a) It is a fundamental concept., (d) Resultant vector  2i  2 j  2 k., , 14., 15., , 25., , 16., , Then,, | PQ | | QR | | RP |  (   )2  (   )2  (   )2, , 2, , Hence PQR is an equilateral triangle., , or a  5, 1., , 26., , 17., , (a) Direction is not determined., , 18., , (c) Let a  li  m j  n k, where l 2  m 2  n 2  1 ., , n , , 1, , , l2  m 2 , , 2, , 1, 2, , 27., , (b) Trick, , 28., , (b) Since a and b are non-collinear, so a  b, and a  b will also be non-collinear., Hence, a + b and a – b are linearly, independent vectors., , 29., , (b) AB = Position vector of B – Position vector, , …..(i),  a  li m j, , k, 2, , a  i  j  (l  1)i  (m  1)j , , Here is the only, 4 ( 2 i  j  k ) , whose length is 8., , vector, ,  | AB |  16  25  81  122 , BC  i  3 j  4 k, , 1, Its magnitude is 1, hence (l  1)  (m  1) , 2, 2, , 2, , | BC |  1  9  16  26, , .....(ii), , and, , AC  3i  8 j  5 k, , 1, 1, From (i) and (ii), 2lm   l  m  , 2, 2, , Hence a  , ,  | AC |  98, , (b) It is a fundamental concept., , 20., , (d) Given a  c  b  d , , Since AC 2  AB 2  BC 2 , therefore ABC is, an obtuse-angled triangle., 30., , Direction cosine along y  axis, 5, , , , 5, , ., , 162, , 22., , (a) | AB | | Q |  P  P  P 2 ., , 23., , (c), , 24., , (c) Here, OA  2i  3 j  4 k, OB  3 i  4 j  2 k, , 2, , 3, 3 4 5, 2, , 2, , 2, , , , (c) R 2  P 2  Q 2  2 PQ cos , , , ( 7 Q) 2  P 2  Q 2  2 PQ cos 60 , , , , 7 Q 2  P 2  Q  PQ  P 2  PQ  6 Q 2  0, , , , P 2  3 PQ  2 PQ  6 Q 2  0, , , , P (P  3 Q )  2 Q(P  3 Q )  0, , , , (P  2Q)(P  3 Q)  0, , , , P  2Q  0 or P  3 Q  0, , From P  2Q  0 , , 2, , 3, , ., , 50, , 31., , Clearly | AB | | BC | | CA |  6, , P, 2., Q, , (b) Vector A  3i  4 j  5 k ., direction cosines of, 3, , A , , OC  4 i  2 j  3 k, , So, AB  i  j  2k, BC  i  2 j  k , CA  2i  j  k, , and, ,  AB 2  BC 2  26  122  148, , (b) AB  4 i  5 j  11 k, , 16  25  121, , BC 2  26, , 2, , 1, 1, (a  c )  (b  d), 2, 2, , Here, mid points of AC and BD coincide,, where AC and BD are diagonals. In, addition, we know that diagonals of a, parallelogram bisect each other., Hence quadrilateral is parallelogram., , AB 2  122 ,, , Therefore,, AC  98 ., , i, j, k,  , ., 2 2, 2, , 19., , , , :, , of A  (2i  3 j  6k)  (6 i  2 j  3k)  4 i  5 j  9 k, , k, 2, , 21., , (b) We have | a  b | 2  | a  b | 2  2(| a | 2  | b | 2 ),  25  | a  b | 2  2(9  16 ) | a  b |  5 ., , , with z  axis., 4, , a makes an angle, , and, ,  i  j  k respectively., , (a) 7  (5  1)  (4  2)  (a  2)  a  2  3, 2, ,  i   j  k, ,  i   j   k,, , vectors, ,  1, 1, 1 , ., Direction cosines are , ,, ,, 3, 3 ,  3, 2, , So these points are vertices of an, equilateral triangle., (b) Let P, Q and R be points having position, , , , 3 4 5, 2, , 3, , ,, , 2, , 4, , 5 2 5 2, , ,, , 2, , 1, 2, , We, , 4, , ,, , 3 4 5, 2, , ., , 2, , 2, , ,, , know, 5, 3  4 2  52, 2, , that

Page 19 : Vector Algebra, , 32., , (d), , Clearly,, AB  (6  2)i  (3  9)j  (8  4 )k  4 i  6 j  12 k, , trapezium., 39. (b) R cos   6 cos 0   2 2 cos(180 o  B)  5 cos 270 o, , | AB |  16  36  144  14 ., 33., , C, , (d) PQ  (5  1)i  (2  3)j  (4  7)k  4 i  5 j  11 k, | PQ |  16  25  121  162 ., , 34., , AB  CD So, it is a, , BC || DA, but, , 2 2, , 5, , (c) ma is a unit vector if and only if | m a | , m | a| 1, ,  m, 35., , A, , 1, ., | a|, , R cos   6  2 2 cos B, , (c) AB  (3  2)i  (2  1)j  (1  1)k  i  3 j  2 k, , …..(i), , R sin   6 sin 0   2 2 sin(180, , BC  (1  3)i  (4  2)j  (3  1)k  2i  6 j  4 k, , …..(ii), ,  25  20 2 sin B, , | BC |  4  36  16  56  2 14, ,  61  8(cos 2 B  sin 2 B)  24 2 cos B  20 2 sin B, ,  ABC is a right angled isosceles triangle, i.e., B  C  45, , | CA |  1  9  4  14, , So, | AB |  | AC | | BC | and angle between AB, and BC is 180°.  Points A, B, C can not, form an isosceles triangle., (c) P.V., , 1, ,  R 2  61  8 (1)  24 2 , , R 5., (b) By, , triangle, E, , C, , D, , B, , Therefore, AB  AC  AD  AE  AF, = 3 AD  ( AE  BD )  ( AF  CD)  3 AD, Hence   3 ,, , [Since, , Scalar or Dot product of two vectors an its applications, , | AB | | i  j  4 k |  18, | BC | | 3i  3 j|  18, | AC | | 4 i  2 j  4 k |  36, , 38., , AB  AD  BD ,, , C, , A, , | AD |  16  16  1  33 ., , , ,  25, , 2, , law,, , F, , (d) Given, position vectors of A, B and C are, 7 j  10 k ,, i  6 j  6 k, and, 4 i  9 j  6 k respectively., , 1, , AC  AD  CD, , of, , (3  5)i  (0  2) j A(4  4 )k,  4i  j  4k, AD , 2, , 37., ,  20 2 , , 2, 40., , Hence A, B, C are collinear., , D, ,  B)  5 sin 270 o, , R 2  36  8 cos 2 B  24 2 cos B  8 sin 2 B, , | AB |  1  9  4  14, , B, , o, , R sin   2 2 sin B  5, From (i) and (ii),, , CA  (2  1)i  (1  4)j  (1  3)k  i  3 j  2k, , 36., , B, , 6, , Clearly, AB  BC and ( AC )2  ( AB )2  (BC )2, Hence, triangle is right angled isosceles., (c) Let, A  (1, 1,1), B  (2, 3, 0), C  (3, 5,  2), D  (0,  1, 1), , So,, AB  (1, 2, 1), BC  (1, 2,2), CD  (3,6, 3), DA  (1, 2,  2), , a  x i  y j  zk ., (a .i )i  (a . j)j  (a .k )k  a ., , 1., , (a) Let, , 2., , (d) Let r  x i  y j  zk. Since r.i  r.j  r.k, x y z, , Then, , .....(i), , Also | r |  x 2  y 2  z 2  3  x   3 , {By (i)}, Hence the required vector r   3 (i  j  k )., Trick : As the vector  3 (i  j  k ) satisfies, both the conditions.

Page 20 : Vector Algebra, 3., , , , (c) a .b  a .c  a .b  a .c  0  a .(b  c )  0,  Either b  c  0 or a  0  b  c or a (b  c )., , 13., , (  cos   1 )., , 4., , (b) a.b   | a || b | ,, , 5., , (c) Squaring (a  b  c )  0 ,, , and c is, , 3,  2(a .b  b .c  c .a )  3  a .b  b .c  c .a   ., 2, , (c) Since, , and, , a, b, , are, , c, , Angle between a and a  b  c is, cos  , , a .(a  b  c ), | a || a  b  c |, ,   cos, ,  1 ,  and between i  j  k and k is, ,  3, ,   cos 1 , ,  1 ,  . Hence      ., ,  3, , value, , in, , (i),, ,   cos 1 , , we, , get, , 16., , (b) Let r  x i  y j  zk  r. i  x, r. j  y, r. k  z, , ., ,  (r . i)2  (r . j)2  (r . k)2  x 2  y 2  z 2  r 2 ., , (a) Three mutually perpendicular unit vectors,  a , b and c ., Therefore, and, | a | | b | | c |  1, a.b  b.c  c.a  0 ., We know that, , 17., , | c |, , 2, , (d) Parallel vector  (2  b)i  6 j  2 k, Unit vector , , 1, , 2(a . b  b . c  c . a )  1  1  1  0  3, , or | a  b  c |  3 ., (c) a  b  c | a |  | b | 2a . b | c |, 2, , 2, , , , | a  b |  (a  b )  a  b  2a . b  25  25, 2, ,  | a  b |  5 2., , 11., , , 2, ,  cos, , , 2, , , , , ,  y2  z2, , , , y, , , , , , 2, , (y j  zk). zk, 1, z, | y j  zk | | zk |, 2, , Hence the components of unit vector are, 3 1, , ., 2 2, , Trick : Since the vector lies in yz  plane,, , (a) | a  b | | a  b |, Squaring both sides, we get, , so, , it, , will, , 0i , , 1, 3, j, k., 2, 2, , be, , either, , a  b  2a . b  a  b  2a . b, 2, , 2, , 2, , (y j  zk). (y j), | y j  zk | | y j |, , 3, 3, y, ,, 2, 2, , Similarly, cos 60  , , 1, | a  b| ., 2, , 0,, , ., , y2  z2  1, , ( y 2  z 2  1 by (i)), , (d) a  b   c  | a | 2  | b | 2 2 | a || b | cos  | c | 2,  cos   0   , , 12., , y2, , (b) (a  b ).(a  b ) | a | 2  | b | 2  2a . b, or | a  b | 2  2 .2 cos 2, , condition,, , b 2  4 b  44, , Since given that cos 30  , , (d) Since a  b  a . b  0, 2, , the, , …..(i), , a . b | a || b |  cos   1    0 ., 2, , to, , (2  b)  6  2, , (b) Let unit vector be y i  z k , then, , and | a | | b | | c | | a | 2  | b | 2 2 | a || b | | c | 2, , 2, , b 2  4 b  44, ,  b 2  4 b  44  b 2  12b  36  8 b  8  b  1., 18., , 2, , (2  b)i  6 j  2k, , According, , | a  b  c | 2  (a  b  c ) . (a  b  c )  | a | 2  | b | 2, , 10., , 2, ,  a2  b2  c 2, , Similarly angle between i  j  k and j is, , 7., , 9., , a 2  b 2  c 2  0,, , , ,  (i  j  k ). i , 1  1 , , cos 1 ,     cos , , |, i, , j, , k, |, |, i, |, , ,  3, , 3, , 8., , a 2  b 2  c 2  2 | b || c | cos, , (d) Angle between i  j  k and i is equal to, ,  | a  b  c | 2  3a2 | a  b  c |  3 a, this, , or, , 15., , .....(i), ,  a2  a2  a2  0  0  0, , 1, , a2  b2  c2  2b.c, , 14., , | a  b  c | 2  a 2  b 2  c 2  2 a .b  2 b .c  2 c .a, , Putting, , , ., 2, , i.e., a 2  b 2  c 2 ., (d) Obviously a, b are unit vectors., , Now | a | | b | | c |  a, , 1, ,  90  ,, , Hence, , So,, , or, , mutually, , perpendicular, so a . b  b . c  c . a  0, , cos   0 ., , (acute)., (a) Given that a  b  c and angle between b, , we get a 2  b 2  c 2  2a .b  2 b.c  2c .a  0,  | a | 2  | b | 2  | c | 2 2(a .b  b.c  c .a )  0, , 6., , , , 4 a.b  0, , 2, , But, , the, , 0i , , 3, 1, j k, 2, 2, , vector, , or, , 3, 1, j k, 2, 2

Page 21 : Vector Algebra, , makes angle 30  with y  axis and that of, 60  with z-axis., 19., , AB  2i  4 j  k ,, , (c)  F  2j  k,,   F . AB  8  1  9., , 20., ,  0  7  14     , , | a | | b |  a(say), , (c) a .b  aa cos 120  ,,  8  , , 24., 25., , Note: Students should remember, question as a formula., (c) It is obvious., (c) r  p   q  r . q  p . q   q . q, , this, , 1, 2, , 1, 2, , Therefore, r   (i  5 j  4 k)., , a2, a4, 2, , (Negative sign does not occur in moduli)., 21., , (b) | 4 a  3b |  (4 a  3b).(4 a  3b), , Vector or Cross product of two vectors, applications, ,  16 | a | 9 | b | 24 a.b, 2, , 2, ,  1 ,  144  144  24  3  4  ,   12 .,  2 , 22., , (d) Let the required vector be   d1i  d2 j  d3 k ,, d12, , where, , , , d 22, , , , d 32, ,  51 , (given), , .....(i), , Now, each of the given vectors a, b, c is a, unit vector cos  , , d.a, d.b, d.c, , , | d || a | | d || b | | d || c |, , 1., , (d) It is obvious., , 2., , (c) | a  b |  1  | sin  |  1  sin   1   , , 3., , (b) (b  a )  (c  b )  0 or b  c  a  b  c  a  0 ., (a) (a  b )  (a  b )  a  a  b  a  a  b  b  b, , 4., , 5., , 1, 1, (d1  2d 2  2d 3 )  (4 d1  0 d 2  3 d 3 )  d 2, 3, 5, ,  d1  5d 2  2d 3  0 and 4 d1  5d 2  3d 3  0, , 6., , d1 d 2, d, ,  3   (say), 5, 1  5, , (5 i  j  5 k)., , vectors, ,  a  b  a  c  c  a, .....(i), b  (a  b  c )  0  a  b  b  c, Similarly,, .....(ii), By (i) and (ii), we get a  b  b  c  c  a., (c) | a  b |  (a.b)  ab sin  ab cos ,  tan  , , Putting d1 , d 2 and d 3 in (i), we get   1, required, , are, , 7., 8., , (b) Since a, b and c are coplanar, therefore, there exists (x , y, z not all zero) such that, x a  y b  zc  0, , and a  b  0  a || b or a  0 or b  0, Hence, either a or b is a null vector., 9., , (b) Component of a along b  a cos  , , .....(i), , to b  a sin  , 10., , | a b|, ., | b|, , (d) We have | (a  b).c | | a || b || c |,  | a || b | sin  n.c | a || b || c |, , .....(iii), ,  | a || b || c | sin  cos  | a || b || c |, , Eliminating x, y and z from (i), (ii) and, (iii),, , | sin  || cos  |  1   , , a, we get a . a, a.b, , b, a.b, b.b, , | a.b|, | b|, , Similarly component of a perpendicular, , Multiply be a scalarly, we get, ......(ii), x(a . a)  (a . b)  z(a . c )  0, and x(a . b)  y(b . b)  z(b . c )  0, , ab, , 1   ., ab, 4, , (b) 14 (a  b )  15 (b  a )  b  a ., (c) a.b  0  a  b or a  0 or b  0, , Trick : Check it with the options., 23., , ., ,  a  (a  b  c )  0  a  a  a  b  a  c  0, , Hence,, , the, , 2, , (c) Since a  b  c  0, , | d |  51 cancels out and | a |  | b | | c |  1, , Hence, , , ,  a  b  b  a  a  b  a  b  2(a  b )., , or d . a  d . b  d . c, , On solving,, , and its, , c, a .c  0 ., b .c, , , 2, , and   0, ,  ab and c || n,  ab, , a and b, , and, , c is perpendicular to both, ,  a, b, c are mutually perpendicular

Page 22 : Vector Algebra, , 11., 12., , Hence, a.b  b.c  c.a  0., (a) Vector product is not commutative., (c) The vector perpendicular to a and b is, , i, j k, a  b  1 1  3  8i  4 j  4 k, 2 2 2, , i j k, ab  1 1 0  i  jk, 0 1 1, , Hence unit vector  , , 13., , 19., , ab, 1, , (i  j  k), | a  b|, 3, , i, (c) a  b  2, m, , m , , But (i  j  k).(b1 i  b 2 j  b3 k)  1  b1  b 2  b3  1, ......(i), i, b1, , j k, 3 5, n 12, ,  (36  5n) i  (24  5m) j  (2n  3m) k  0, , Hence there are two such vectors., (b), Let b  b1 i  b 2 j  b 3 k, , and a  b  1, , j, k, 1 1, b2 b3, , 24, 36, , n, ., 5, 5, , (d) Unit vector is equal to, , 21., , (a) AB  2i  j  2 k, AC  3 i  3 j  0 k, i, j, k, AB  AC  2  1  2  (6 i  6 j  3 k ), 3 3 0, , a b  c, , Comparing the coefficients of i, j and k, respectively,, we get, …..(ii), b2  b3  1, b1  b3  1, , …..(iii), , b 2  b1  0, , …..(iv), , By solving the equations (i), (ii), (iii) and, (iv), we get b1  0, b 2  0 and b 3  1 ., ,  2i  2 j  k , ., 3, , , , Hence unit vector  , 22., , (c) Unit vector perpendicular to both the, given, vectors, is,, (6 i  2 j  3 k)  (3 i  6 j  2 k), 2i  3 j  6 k, , ., | (6 i  2 j  3 k)  (3 i  6 j  2 k)|, 7, , 23., , (c) (a  b )2  (a  b ). (a  b )  (ab sin  nˆ )(ab sin  nˆ ),  a 2 b 2 sin 2   a 2 b 2 (1  cos 2  ), , (a) Since a  b  b  c  0  a  b  b  c  0,  a  b  c  b  0  (a  c )  b  0, , 15., 16., ,  a  c is parallel to b  a  c  k b., (c) It is obvious., (a,c), Let angle between a and b be  ., ˆ, v  a  b | a || b | sin  n, ,  a 2 b 2  a 2 b 2 cos 2   a 2 b 2  (a . b )2 ., , i, , 24., , (c) a  b  3, 12, ,  | v |  sin ,, , j k, 2  1  5 i  3 j  9 k., 5 5, , Unit vector along a  b , , , (a  b ), v , ˆ , ,  | a |  1, | b |  1, n, , |, a, , b, |, |, v| , , u  a  (a.b)b  a  cos  b, , 25., , ( a.b | a || b | cos   cos  ), , i, a, , b, , 3, (a), 12, , u . u | u |  1  cos   2 cos  cos   sin 2 , 2, , a b, 1, , (2i  k)., | a b|, 5, , 20., ,  i(b2  b3 )  j(b1  b3 )  k(b2  b1 ), , 14., , ., , 6, , Since the length of this vector is 3 , the, unit vector perpendicular to a and b is, , , 2i  j  k, , 2, ,  sin  , ,  | u |  sin , , 5 i  3 j  9 k, , ., , 115, , j k, 2  1  5 i  3 j  9 k, 5 5, , 25  9  81, 14 . 194, , , , 115, 14 . 194, , ., , V. Scalar triple product and their, applications, , u . a  a .a  cos  a .b  1  cos 2   sin 2 , , u . b  a.b  cos  b.b  cos   cos   0, , u . (a  b)  (a  cos  b). (a  b),  1  cos   cos 2   cos , , 1., , 17., , 2, , Hence (a) and (c) are correct., (b) a  b  b  c  (a  c )  b  0, but a  c  0,  a  c || b., , 18., , (a) a  i  j  3k, b  2i  2 j  2 k, , (a), a .b  c b.a  c [a b c ] [b a c ] [a b c ] [a b c ], , , , , ,  0., c  a .b c .a  b [c a b ] [c a b ] [c a b ] [c a b ], ,  1  cos   sin , 2, , 2., , (b) [a  b b  c c  a]  (a  b).{(b  c )  (c  a)},  (a  b ).(b  c  b  a  c  c  c  a ),  (a  b ).(b  c  b  a  c  a ) ,, , c  c  0

Page 23 : Vector Algebra, ,  a.b  c  a.b  a  a.c  a  b.b  c b.b  a  b.c  a, , 1., , (a) b  c is a vector perpendicular to b,c ., Therefore, a  (b  c ) is a vector again in, plane of b,c ., , 2., , (c) Let a  x i  y j  zk, ,  [a b c ]  [b c a]  2[a b c ] ., 3., , OA  2 i  3 j  a, , (c) Here,, , (say), , OB  i  j  k  b (say), , and OC  3i  k  c (say), Hence, volume, , i  (a  i)  j  (a  j)  k  (a  k ), , 2 3 0, [a b c ]  a . (b  c )  1 1  1  4 ., 3 0 1, , 4., 5., 6., , (d) We have p . (a  b)  p . a  p . b, , , (b  c ) . a (b  c ) . b [b c a ] [b c b ], , , , [a b c ], [a b c ], [a b c ] [a b c ], , 10 1 ,, , 8., ,  3a  a  2a., i, , 3., , (c) a .(b  c )  0 or (a  b ).c  0 ., (a) a .(c  b )  c .(b  a )  0 , (Since a and b are, parallel), (d) If the given vectors are coplanar, then, their scalar triple product is zero., 2 1 1, 1 2  3  0    4 ., 3 , 5, , 7., ,  (i.i )a  i(a .i)  (j.j)a  j(a .j)  (k.k)a  k(a .k ), , is, , [b c a]  [a b c ]and [b c b]  0, , 1, , 4., , 5., , 10., ,  12, (c) Since 0, 2, , (b) a  (b  c )  0  a || (b  c ) or b  c  0, i.e., b || c or a  0 ., , 6., 7., , (c) It is a fundamental concept., (d) a  b  c and a  b  c,  a is perpendicular to both b and c and, c is perpendicular to both a and b.,  a, b, c are mutually perpendicular, Now, a  b  c  b  (a  b)  (b . b)a  (b . a)b, or a  b 2 a  (b . a )b  b 2 a , a  b, ˆ,  1  b 2 , c  a  b  ab sin 90  n, , Take moduli of both sides, then c  ab , but, b 1c  a ., 8., , (a) (a  b)  c  (a . c ) b  (b . c ) a,  (3  2  4 )(2i  j  k)  (2  2  2)(3 i  j  2 k), , 146,  40  5   37    94  13   25   , ., 17, b c  c a  a b, (a) p  q  r , [a b c ], , (a  b  c ) . (p  q  r) , , (d) α .[ γ  (α  β)]  α .[(γ . β)α  (γ . α)β],  (α)2 (γ . β)  (γ . α)(β . α)  14 (3)  (4 )(8 )  74 ., , 4 5 , 4, 5, , 4 5, ,  2 3  4  1 1, 2  3  2 1, 1 1 2, 3  2 1, 2 3 4, , 9., , 3, , k, 1  a  (2i  3 j  7 k ), 1, , i, j k,  1 2  2  20 i  3 j  7 k., 2 3 7, , Similarly, q.(b  c )  1 and r. (a  c )  1, Thus, required result is 1+1+1=3., (a) Let a  3 i  2 j  k, b  2i  3 j  4 k, c  i  j  2 k, and d  4 i  5 j  k., Since the points are coplanar,, So, [d b c ]  [d c a]  [d a b]  [a b c ], , 3  2 1,  2, 3 4, 1 1, 2, , j, , (a) a  (b  c )  a  2  1, , [a b c ]  [b c a ]  [c a b ], 3., [a b c ], , 0 , 3  1  546    3 ., 1  15, ,  18 i  9 j  9 k  6 i  2 j  4 k  24 i  7 j  5 k ., , 9., , (b) i  j  k  k  k  0 ., , 10., , (d) As, , we, , know,, , a  (b  c )  (a . c ) b  (a . b) c, , ......(i),  a  (b  c ) , , b, 2, , (Given), , From equation (i),, (a . c )b  (a . b )c , , 1, b, , or  a .c   b  (a . b ) c  c, 2, 2, , , Comparison on both sides of b and c, VI. Vector triple product, , a .c , , 1,  0, a . b  0, 2, , | a || c | cos  , , 1, 1,  (1)(1) cos      60 , 2, 2

Page 24 : Vector Algebra, , or a . b  0 ,    90  ., So the angle between a with b and c are, 90  and 60  respectively.

Page 25 : Vector Algebra, , of z is denoted by Re(z) and the imaginary part by, Im(z)., If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4., A complex number z is purely real if its, imaginary part is zero i.e., Im(z) = 0 and purely, imaginary if its real part is zero i.e., Re(z) = 0., , Algebraic operations with complex numbers, Let two complex numbers bc z1  a  ib and, z 2  c  id, , Addition, “Complex number is the combination of real and, imaginary numbers”, , Basic concepts of complex number, (1) Definition : A number of the form x  iy,, where, , x, y  R, , and i   1, , is called a complex, , number and 'i' is called iota., A complex number is usually denoted by z and, the set of complex number is denoted by C., i.e.,, C  {x  i y : x R, y R, i   1}, For, , example,, , 5  3i,  1  i, 0  4 i, 4  0i, , etc., , are, , complex numbers., (i), Euler was the first mathematician to, introduce the symbol i (iota) for the square root, of – 1 with property i 2  1 . He also called this, symbol as the imaginary unit., (ii) For any positive real number a, we have,  a  1 a  1 a  i a, , (iii) The property a b  ab is valid only if at, least one of a and b is non-negative. If a and b are, both negative then a b   | a | . | b | ., (2) Integral powers of iota (i) : Since i   1, hence we have i 2  1 , i 3  i and i 4  1 . To find the, value of in (n  4 ), first divide n by 4. Let q be the, quotient and r be the remainder., n  4 q  r where 0  r  3, i.e.,, in  i 4 q r  (i 4 )q . (i)r  (1)q . (i)r  ir, , i4n, , In general we have the following results,  1, i 4 n 1  i, i4 n  2  1 , i4 n  3  i , where n is any, , ( z1  z 2 ), , :, , (a  ib)  (c  id )  (a  c)  i(b  d ), , (z1  z 2 ), (a  ib)  (c  id )  (a  c)  i(b  d ), , Subtraction, , :, , Multiplication, , :, , (z1 . z 2 ), (a  ib)(c  id )  (ac  bd )  i(ad  bc ), a  ib, Division (z1 / z 2 ), :, c  id, , (where at least one of c and d is non-zero), a  ib (a  ib) (c  id), (Rationalization), , ., c  id (c  id) (c  id), a  ib (ac  bd ) i(bc  ad ),  2,  2, ., c  id, c d2, c d2, , Properties of algebraic operations on, complex numbers : Let z 1 , z 2 and z 3 are any three, complex numbers then their algebraic operations, satisfy following properties :, (i) Addition of complex numbers satisfies the, commutative and associative properties, i.e.,, and, z1  z 2  z 2  z1, (z1  z 2 )  z 3  z1  (z 2  z 3 )., , (ii) Multiplication of complex numbers, satisfies the commutative and associative, properties., i.e.,, z 1 z 2  z 2 z 1 and (z1 z 2 )z 3  z1 (z 2 z 3 )., (iii) Multiplication of complex, distributive over addition, i.e.,, z1 (z 2  z 3 )  z1 z 2  z1 z 3, , numbers, , is, , and, , (z 2  z 3 )z1  z 2 z1  z 3 z1 ., , Equality of two complex numbers, Two, , complex, , numbers, , z1  x 1  iy1, , and, , integer., , z 2  x 2  iy 2 are said to be equal if and only if their, , Real and Imaginary parts of a complex number, , real and imaginary parts are separately equal., i.e.,, z 1  z 2  x 1  iy1  x 2  iy 2  x 1  x 2 and, , If x and y are two real numbers, then a, number of the form z  x  iy is called a complex, number. Here ‘x’ is called the real part of z and ‘y’, is known as the imaginary part of z. The real part, , y1  y 2 ., , Complex numbers do not possess the property, of order i.e., (a  ib)  (or )  (c  id ) is not defined. For

Page 26 : Vector Algebra, , example, the statement (9  6 i)  (3  2i) makes no, sense., , Conjugate of a complex number, (1) Conjugate complex number : If there, exists a complex number z  a  i b, (a , b)  R, then its, Y, , Imaginary, axis, , conjugate is defined as z  a  i b ., , O, , P(z), , the distance of point P from the origin, i.e. |z| =, OP., If | z |  1 the corresponding complex number is, known as unimodular complex number. Clearly, z lies on a circle of unit radius having centre (0,, 0)., Properties of modulus, (i) z  0  z  0 if z  0 and |z|  0 if z  0 ., (ii)  z  Re (z )  z, , , , –, , (iii) z  z   z   z | zi|, , X, , Real, axis, , and  z  Im (z )  z, , 2, , (iv) zz  z, , | z | 2, , (v) z 1 z 2  z 1, , Q (z ), , z2 ., , In general z 1 z 2 z 3 ...... z n  z 1, Hence, we have Re( z ) , , zz, zz, and Im (z ) , ., 2, 2i, , (2) Properties of conjugate :, , If, , z, z 1 and, , z 2 are existing complex numbers, then we have, , (vii) | z n | | z | n , n  N, (viii), z1  z 2, , z1 z 2  z1 z 2 ,, , In, , general, , z 1 .z 2 .z 3 ..... z n  z 1 .z 2 .z 3 ..... z n, ,  z1, , z,  2, , 2, ,  (z 1 z 2  z 1 z 2 ), , z1  z 2, , 2, ,  z1, , 2, ,  z1  z 2, , 2, , 2, ,  z2 , , z1, is purely imaginary, z2, , z, Re 1,  z2, , , 0, , , , (x) z 1  z 2, ,  2,  z1, , , 2, , 2, ,  z2, , 2, , , , , (Law of parallelogram), ,  z1,   , z2 0,  z, , 2, , Argument of a complex number, Let z  a  ib be any complex number. If this, complex number is represented geometrically by, a point P, then the angle made by the line OP with, real axis is known as argument or amplitude of z, and is expressed as, , (vi) (z )n  (z n ), (vii), , z  z  2 Re(z)  2 Re(z )  purely real, , (viii), , z  z  2i Im(z)  purely imaginary, , (ix), , z z  | z | 2  purely real, , (x), ,  z2, , or, , (iii) z 1  z 2  z 1  z 2, (iv), , 2, ,  (z 1  z 2 )(z 1  z 2 )  z 1, , (ix), , (i) (z )  z, , (v), , 2, , or | z1 | 2  | z 2 | 2  2 Re( z1 z 2 ), , the following results:, (ii) z 1  z 2  z 1  z 2, , z 3 .... z n, , z1, z1, , , (z 2  0), z2, z2, , (vi), , Geometrically, the conjugate of z is the, reflection or point image of z in the real axis., , z2, , z1 z 2  z1 z 2  2 Re(z1 z 2 )  2 Re(z1 z 2 ), , (–,+), , (3) Reciprocal of a complex number : For an, existing non-zero complex number z  a  ib , the, reciprocal is given by z 1  1  z  z 2 ., z, , z.z, , Modulus of a complex number, , , -, , O, , z  a  ib, , is, , P(z), , number given by | z |  a 2  b 2 ,, where a, b real numbers., Geometrically |z| represents, O, , M, , X, , – +, , , , , X, (+,–), , (–, –), , Modulus of a complex number, Y, defined by a positive real, , (+,+), , –, , , , , X', , | z|, , Y, , 2–, Y'

Page 27 : Vector Algebra, , b, a, , arg (z )    tan 1  ,   POM Also, argument of a, complex number is not unique, since if  be a, value of the argument, so also is 2n    , where, nI ., ,   tan 1, , b, a, , , , +ve Im (z), ,  /2, , –ve Im (z), , 3 / 2 or   / 2, , respectively, (iz), , , ,   arg (z ), 2, , , –(iz), , , , arg (z )  , 2, , , Y, , –, (–,+), X', , , (+,+), O (+,–), –, , (–,–), – ( –, ), , X, , (z1 .z 2 ), , arg (z1) + arg (z2), ,  z1, , z,  2, , arg (z1) – arg (z2), , (acute angle), , and principal values of argument z will be, ,   ,     and  as the point z lies in the 1st ,, 2nd , 3rd and 4th quadrants respectively., (2) Properties of arguments, (i) arg (z1 z 2 )  arg(z1 )  arg(z 2 )  2k , (k  0 or 1 or –, 1), , Let z  a  ib be a complex number,, Then, ,  | z | a, | z | a , a  ib   , i,  , for b> 0, 2, 2 , , ,  | z | a, | z | a ,  , i,  , for b < 0., 2, 2 , , , (k  0 or 1 or  1), , To find the square root of a  ib, replace i by – i, in the above results., , (ii) arg(z1 z 2 )  arg(z1 )  arg(z 2 ), z, arg 1,  z2, , ,   arg z 1  arg z 2  2k  , (k  0, , , , or 1, , or – 1), z, arg   2 arg z  2k  , (k  0 or 1 or – 1), z, , (iv), , (v) arg(z n )  n arg z  2k  , (k  0 or 1 or – 1), z , , z , , If arg  2    , then arg  1   2k    ,,  z2 ,  z1 , , (vi), where kI, , arg z  arg z  arg, , (vii), , , , , , , Square root of a complex number, , In general arg(z1z 2 z 3 ......... zn )  arg(z1 )  arg(z 2 ), , (iii), , n. arg (z), , (z n ), , Y', , arg(z 3 )  ......  arg(z n )  2k ,, , |    |, if  is  ve and  ve, , – (z), , 1, z, , (viii), , arg(z  z )   / 2, , Various representations of a complex number, A complex number can be represented in the, following form:, (1) Geometrical representation (Cartesian, representation): The complex number z  a  ib  (a, b), is represented by a point P whose coordinates are, referred to rectangular axes XO X  and YO Y , which are called Yreal and imaginary axis, respectively. This plane is, called argand plane or, P(a,, argand diagram or complex, b), |plane, z |  a 2 or,  b 2 Gaussian, b, plane., X, , Imaginar, y axis, , (1) Principal value of, arg (z) : The value  of, the, argument,, which, satisfies the inequality, is called the,     , principal, value, of, argument,, where, , –ve Re (z), , Real, axis, , O, , , , a M, , X, , (ix) arg(z)  arg(z)  , (x) arg(z )  arg(z )  , , (If z is purely imaginary), , Y, , (xi) z 1 z 2  z 1 z 2  2 | z 1 | | z 2 | cos (1   2 ),, where 1  arg(z 1 ) and  2  arg(z 2 ), (xii) The general value of arg(z ) is 2n  arg(z) ., Table: 3.1 Value of arguments of complex numbers, Complex number, +ve Re (z), , Value of arguments, 0, , Distance of any complex number from the, origin is called the modulus of complex number, and is denoted by |z|, i.e., | z |  a 2  b 2 .

Page 28 : Vector Algebra, , Angle of any complex number with positive, direction of x–axis is called amplitude or, argument of z., i.e., amp (z)  arg (z)  tan 1  b ,  , , distance from P(z 1 ) and Q(z 2 ) then locus of R(z ) is, perpendicular bisector of PQ, i.e.,, PR = QR or| z  z 1 | | z  z 2 |,  | z  z 1 | 2 | z  z 2 | 2, , a, , After solving,, , (2) Trigonometrical (Polar) representation :, In  OPM, let OP  r , then a  r cos  and b  r sin ., Hence z can be expressed as z  r(cos   i sin  ), where r = |z| and  = principal value of, argument of z., For general values of the argument, z  rcos( 2n    )  i sin(2n    ), , Sometimes, cis ., , cos   i sin   is written in short as, , (3) Vector representation : If P is the point, (a, b) on the argand plane corresponding to the, complex number z  a  ib ., , z(z1  z 2 )  z (z1  z 2 ) | z1 | 2  | z 2 | 2, , (4) Equation of a straight line, (i) Parametric form : Equation of a straight, line joining the point having affixes z 1 and z 2 is, z  t z 1  (1  t)z 2 , when t  R, , (ii) Non-parametric form : Equation of a, straight line joining the points having affixes z 1, and z 2, , b, a, , 1, 1 0, 1, , (iii) (a) Three points z 1 , z 2 and z 3 are collinear, if,, z1, z2, z3, , (4) Eulerian representation (Exponential, form) : Since we have e i = cos   i sin and thus z, can be expressed as z  re i , where | z |  r and  , , z, z1, z2, ,  z(z 1  z 2 )  z (z 1  z 2 )  z 1 z 2  z 2 z 1  0 ., , Then OP  aˆi  bˆj ,  | OP |  a 2  b 2 | z | and, arg (z) = direction of the vector OP  tan 1  , , z, is z 1, z2, , z1, z2, z3, , 1, 1 0, 1, , arg (z)., , (b) If three points A(z 1 ), B(z 2 ), C(z 3 ) are collinear, then slope of, AB = slope of BC = slope of, , Logarithm of a complex number, , AC , , log( x  iy)  log e (re i )  log e r  log e e i  log e r  i, , y,  log e (x 2  y 2 )  i tan 1  , x, , log e (z)  log e | z |  i amp z, , (iv) General equation of a straight line: The, general equation of a straight line is of the, form az  az  b  0 , where a is complex number and, b is real number., (v) Slope of a line : The complex slope of the, , Use of complex numbers in co-ordinate geometry, (1) Distance formula : The distance between, two points P(z 1 ) and Q(z 2 ) is given by PQ | z 2  z 1 |, = |affix of Q – affix of P|, , z1  z 2 z 2  z 3 z1  z 3, , , ., z1  z 2 z 2  z 3 z1  z 3, , line az  az  b  0 is , , coeff. of z, a, and real slope, , a, coeff. of z, , of the line az  az  b  0 is , , Q(z2), , Re(a), (a  a ), .,  i, Im(a), (a  a ), , (vi) Length of perpendicular : The length of, perpendicular from a point z1 to the line, , P(z1), , (2) Section formula : If R(z) divides the line, segment joining P(z 1 ) and Q(z 2 ) in the ratio, m1 : m 2 (m1 , m 2  0) then, , az  az  b  0, , is, , given, , by, , | a z 1  az 1  b |, | a| | a |, , or, , | a z 1  az 1  b |, 2| a|, , (i) For internal division z , , m1 z 2  m 2 z1, m1  m 2, , (5) Equation of a circle : (i) The equation of a, circle whose centre is at point having affix z o and, , (ii) For external division z , , m1 z 2  m 2 z1, m1  m 2, , radius r is | z  z o |  r, , (3) Equation of the perpendicular bisector :, If P(z 1 ) and Q(z 2 ) are two fixed points and R(z ) is, moving point such that it is always at equal, P(z1), , R(z), , (ii) If the centre of the circle, is at origin and radius r, then its, equation is | z |  r ., , P(z), r

Page 29 : Vector Algebra, , (iii) | z  z 0 |  r represents interior of a circle, C(z0), of the, | z  z 0 |  r and | z  z 0 |  r represents exterior, circle | z  z o |  r ., , Clearly the complex number z 1 represents a, point Q in the argand plane, when OQ  r and, QOX     ., , (iv) General equation of a circle : The general, equation of the circle is zz  az  az  b  0 where a is, complex number and b  R ., , Clearly multiplication of z with e i rotates the, vector OP through angle  in anticlockwise, , , , | a | 2 b, , Centre and radius are – a and, , respectively., (v) Equation of circle in diametric form : If, end points of diameter represented by A(z 1 ) and, B(z 2 ), , and, , be, , P(z ), , any, , point, , on, , the, , sense. Similarly multiplication of z with e i will, rotate the vector OP in clockwise sense., (i) If z 1 , z 2 and z 3 are the affixes of the points, A,B and C such that AC  AB and CAB  C(z, . 3), Therefore, AB  z 2  z 1 , AC  z 3  z 1 ., , circle, , , then, (z  z1 ) (z  z 2 )  (z  z 2 ) (z  z1 )  0 ., which is required, diametric form., , equation, , of, , circle, , Rotational theorem i.e., angle between two, intersecting lines. This is also known as coni, method., Let z 1 , z 2 and z 3 be the affixes of three points, A, B and C respectively taken on, argand plane., Y, C(z3), B(z2), Then we have AC  z 3  z 1, , , and AB = z 2  z 1, and let arg AC  arg (z 3  z 1 )  , and AB  arg (z 2  z 1 )  , , A(z1), , , , , , AC, , =, , arg, , and, , AB, , =, , arg, ,  affix of C  affix of A , , ,  affix of B  affix of A , , (1) Complex number as a rotating arrow in, the argand plane : Let z  rcos   i sin    re i, ..…(i), r. e i, , be a complex number representing a, Y, , point P in the argand plane., Then OP | z |  r and POX  , , or z 1  re i .e i  re i  , , X', , O, , {From (i)}, Y', , Triangle inequalities, In any triangle, sum of any two sides is, greater than the third side and difference of any, two side is less than the third side. By applying, this basic concept to the set of complex numbers, we are having the following results., (1) | z1  z 2 |  | z1 |  | z 2 |, (2) | z1  z 2 |  | z1 |  | z 2 |, (3) | z1  z 2 |  | z1 |  | z 2 |, (4) | z1  z 2 |  | z1 |  | z 2 |, , Q(zei), , , , , Now consider complex number, z 1  ze i, , (ii) If A, B and C are three points in argand, plane such that AC  AB and CAB   then use the, rotation about A to find e i , but if AC  AB use, coni method., (2) If four points z1, z 2 , z3 and z4 are con-cyclic, ,  (z  z )(z  z ) , , arg (z 3  z 1 )  arg (z 2  z 1 ), , between, , z 3  z1,  e i, z 2  z1, , or arg 2 3 4 1     , 0,  (z1  z 3 )(z 4  z 2 ) , ,  z 3  z1 , , , z z , 1 ,  2, , or angle, , AC  AB e i or (z 3  z 1 )  (z 2  z 1 )e i or, , (z 4  z1 )(z 2  z 3 ), = real, (z 4  z 2 )(z1  z 3 ), , X, ,  CAB       = arg AC  arg AB, , =, , Then AC will be obtained by rotating AB, through an angle  in anticlockwise sense, and, therefore,, , then, , O, , Let CAB   ,, , A(z1), , in, , Rotation theorem, , B(z2), , P(z), X, , Standard loci in the argand plane, If z is a variable point and z 1 , z 2 are two fixed, points in the argand plane, then

Page 30 : Vector Algebra, , (i) | z  z 1 | | z  z 2 |,  Locus of z is the, perpendicular bisector of the line segment joining, z 1 and z 2 ., (ii) | z  z1 |  | z  z 2 | = constant, , | z1  z 2 | , ,  Locus of z is the line segment joining z 1, and z 2, | z  z 1 |  | z  z 2 | | z 1  z 2 |, ,  Locus of z is a straight line joining z 1 and, z 2 but z does not lie between z 1 and z 2 ., (v), , | z  z1 |  | z  z 2 |  constant, , |, , | z  z1 | 2  | z  z 2 | 2 | z1  z 2 | 2  Locus of, , z is a circle with z 1 and z 2 as the extremities of, diameter., (vii) | z  z1 |  k | z  z 2 |, ( k  1) , , Locus of z is a, , circle., (viii), ,  z  z1 ,    (fixed )  Locus of z is a, ,  z  z2 , , arg , , segment of circle., (ix), ,  z  z1 ,  =   / 2  Locus of z is a, ,  z  z2 , , arg , , circle with z 1 and z 2 as the vertices of diameter., (x), ,  z  z1, arg ,  z  z2, , ,  = 0 or   Locus of z is, , , , a straight line passing through z 1 and z 2 ., , De' Moivre's theorem, (1) If n is any rational, (cos   i sin  )n  cos n  i sin n ., , number,, , then, , (2) If z  (cos 1  i sin1 )(cos  2  i sin 2 ), (cos  3  i sin 3 ) ...... (cos n  i sinn ), , and n is a positive, , , , , , integer, then z 1 / n  r1 / n cos  2k      i sin 2k     ,, , , , , n, , Roots of a complex number, (1) nth roots of complex number (z1/n), Let z  r(cos   i sin  ) be a complex number. By, using De'moivre's theorem nth roots having n, distinct values of such a complex number are, given by, , where k  0,1, 2,....., (n  1)., Properties of the roots of z1/n, (i) All roots of z1/n are in geometrical, progression with common ratio e 2 i / n ., (ii) Sum of all roots of z1/n is always equal to, zero., (iii) Product of all roots of z 1 / n  (1)n 1 z., (iv) Modulus of all roots of z1/n are equal and, each equal to r 1 / n or | z | 1 / n ., (v) Amplitude of all the roots of z1/n are in A.P., with common difference, , 2, ., n, , (vi) All roots of z1/n lies on the circumference, of a circle whose centre is origin and radius equal, to | z | 1 / n . Also these roots divides the circle into n, equal parts and forms a polygon of n sides., (2) The nth roots of unity : The nth roots of, unity are given by the solution set of the equation, x n  1  cos 0  i sin 0  cos 2k  i sin 2k, x  [cos 2k   i sin 2k  ]1 / n, x  cos, , 2k , 2k ,  i sin, , where k  0, 1, 2, ....., (n  1) ., n, n, , (i) Let   cos, , where  1 ,  2, 3 ..... n  R ., z  r(cos   i sin  ), , , , , , (sin   i cos  )n  cos n      i sin n   , 2, 2, , , , , , Properties of nth roots of unity, , then z  cos(1  2  3  .....  n ), i sin(1   2  3  .....  n ), (3) If, , (iv), , 2k   , 2k    , , z 1 / n  r 1 / n cos,  i sin, ,, n, n, , , , z1  z 2 | , ,  Locus of z is a hyperbola., (vi), , (cos   i sin  )n  cos n  i sin n, , This theorem is not valid when n is not a, rational number or the complex number is not in, the form of cos   i sin  ., ,  Locus of z is an ellipse, (iii) | z  z 1 |  | z  z 2 | | z 1  z 2 |, , (iv), , (iii), , , , where k  0, 1, 2, 3,...( n  1) ., , , , n, , , , 2, 2,  i sin,  e i(2 / n), the nth roots of, n, n, , unity can be expressed in the form of a series, i.e., 1, ,  2 ,.....  n 1 . Clearly the series is G.P. with, common ratio  i.e., e i(2 / n ) ., (ii) The sum of all n roots of unity is zero i.e.,, 1     2  .....   n1  0., , Deductions: If n  Q, then, , (iii) Product of all n roots of unity is (1)n 1 ., , (i) (cos   i sin  )n  cos n  i sin n, , (iv) Sum of pth power of n roots of unity, , (ii) (cos   i sin  )n  cos n  i sin n

Page 31 : Vector Algebra, , 0, when p is not multiple of n, 1   p   2 p  .....   (n1) p  , n, when p is a multiple of n, th, ,  Complex number as an ordered pair : A, , (v) The n, n roots of unity if represented on a, complex plane locate their positions at the, vertices of a regular polygon of n sides inscribed, in a unit circle having centre at origin, one vertex, on positive real axis., (3) Cube roots of unity : Cube roots of unity, are the solution set of the equation x 3  1  0 , , complex number may also be defined as an, , x  (1)1 / 3, , addition., , ordered pair of real numbers and may be, denoted by the symbol (a, b). For a complex, number to be uniquely specified, we need two, real numbers in particular order., ,  0 = 0 + i.0, is the identity element for, ,  x  (cos 0  i sin 0)1 / 3, ,  1 = 1+i0 is the identity element for, , 2k ,  2k  ,  i sin,  x  cos,  , where k  0,1, 2, 3,  3 , , multiplication., , Therefore, , roots, , 2, 2, 4, 4, 1, cos,  i sin, , cos,  i sin, or 1, e 2 i / 3 , e 4  i / 3 ., 3, 3, 3, 3, ,  The additive inverse of a complex number z, are, , Alternative : x  (1)1 / 3  x 3  1  0,  (x  1)(x 2  x  1)  0, 1  i 3 1  i 3, x  1,, ,, 2, 2, , (ii)  3  1, (iii), , 1  , , 2n, , 0, if n is not a multiple of 3, , 3, if n is a multiple of 3, , (iv), The cube roots of unity, when, represented on complex plane, lie on vertices of, an equilateral triangle inscribed in a unit circle, having centre at origin, one vertex being on, positive real axis., (v) A complex number a  ib, for which, | a : b |  1 : 3 or, , 3 : 1, can always be expressed in, , terms of i,  ,  2 ., (vi), ,  For every non-zero complex number z, the, multiplicative inverse of z is, , Cube root of – 1 are  1,   ,   2 ., , (4) Fourth roots of unity : The four, fourth, roots of unity are given by the solution set of the, equation x 4  1  0,  (x 2  1)(x 2  1)  0  x   1,  i, Fourth roots of unity are vertices of a square, which lies on coordinate axes., , 1, ., z, ,  | z | | Re(z)|  Re(z) and | z | | Im(z)|  Im(z), , , If one of the complex root is , then other root, will be  2 or vice-versa., Properties of cube roots of unity, (i) 1     2  0, , n, , = a + ib is –z (i.e. – a – ib)., , z, is always a unimodular complex number, | z|, , if z  0, ,  | Re(z)|  | Im(z)|  2 | z |,  || z1 | | z 2 || | z1  z 2 | | z1 |  | z 2 |, Thus | z1 |  | z 2 | is the greatest possible value, of | z1  z 2 | and || z1 | | z 2 || is the least possible, value of | z1  z 2 |, ,  If z , , 1,  a, the greatest and least values of, z, , | z | are respectively, ,  a  a2  4, a  a2  4, and, ., 2, 2, ,  | z1  z12  z 22 |  | z 2  z12  z 22 | | z1  z 2 |  | z1  z 2 |,  If z1  z 2  z1  z 2 or arg z 1 = arg z2, , , z1  z 2  z1 |  | z 2 , , arg (z 1 )  arg (z 2 ) i.e., z1, , and z2 are parallel., , , , arg (z 1 )  arg (z 2 )  2n ,, , z1  z 2  z1 |  | z 2 , , where n is some integer., ,  | z1  z 2 |  | | z1 | | z 2 ||  arg(z1 )  arg(z 2 )  2n ,, where n is some integer.,  z 1  z 2  z 1  z 2  arg (z 1 ) – arg (z 2 )  / 2 ., ,  If | z1 | 1,| z 2 | 1 then, ,  z1 , 2, z1  z 2   z1 , , (i) z 1  z 2, (ii), , , , z1  z 2, , 2, , 2, ,  z1, , , , 2, ,  z2, , 2, , 2  (arg (z1 )  arg (z 2 ))2 ., 2, z 2   arg (z 1 )  arg (z 2 )2 ., , z2, ,  2| z 1 | | z 2 | cos( 1   2 ).

Page 32 : Vector Algebra, , , , z1  z 2, ,  If, , 2, ,  z1, , 2, ,  z2, , | z1 | | z 2 |, , 2, , is, ,  2| z 1 | | z 2 | cos( 1   2 )., , amp (z1 )  amp (z 2 )  0,, , and, , a(sec A)z 1  b(sec B)z 2  (c sec C)z 3, ., a sec A  b sec B  c sec C, , then, , z1 and z 2 are conjugate complex numbers of each, , other,  Multiplication of i: Since z  r (cos   i sin ) and,  , , , , , , , i   cos  i sin  then iz  cos      i sin   , 2, 2, 2, 2, , , , , , , , Hence, multiplication of z with i then vector for, z rotates a right angle in the positive sense. i.e.,, to multiply a vector by – 1 is to turn it through, two right angles. i.e., to multiply a vector by, (cos   i sin  ) is to turn it through the angle  in the, positive sense.,  If z 1 and z 2 are two complex numbers then, | z 1 z 2 | = r1r2 ; arg( z1 z 2 )  1   2 and,  z1, , z,  2, , ,   1   2, , , , z1, r,  1 , arg, z2, r2, , and where | z1 |  r1 ,| z 2 |  r2 , arg( z1 )  1, ,  The area of the triangle whose vertices are, 1, | z| 2 ., 2, , z, iz and z + iz is, , 3 2, | z |., 4, ,  If z1 , z 2 , z 3 be the vertices of an equilateral, and, , z o be, , the, , 1, (z1  z 2 ), 2, , centre at, , and radius, , 1, 2, , 2 k  | z1  z 2 | 2, , 1, 2, , provided k  | z1  z 2 | 2 ., ,  (iz )  iz , Re(iz)   Im(z), Im(iz)  Re(z) .,  If the complex numbers z 1 and z 2 are such, that the sum z 1  z 2 is a real number, then they, are not necessarily conjugate complex., ,  If z 1 and z 2 are two complex numbers such, that the product z 1 z 2 is a real number, then they, are not necessarily conjugate complex., z | 2  [Re( z )]2  [Im(z )]2 ,therefore, , Re(z) | z |, Im(z) | z | ., ,  For any a, b  R, ,  The area of the triangle with vertices z, wz, , triangle, , a real number) will represent a circle with, ,  Since|, , and arg( z 2 )   2 ., , and z  wz is, ,  The equation | z  z1 | 2  | z  z 2 | 2  k (where k is, , circumcentre,, , then, , z 12  z 22  z 32  3z 02 ., , (i), (ii), , a  ib  a  ib  2{ a2  b 2  a}, a  ib  a  ib  i 2{ a 2  b 2  a}, ,  The one and only one case in which, | z |  | z 2 | ......  | zn | | z1  z 2  .....  zn |, , is, , that, , the, ,  If z1 , z 2 , z 3 ..... z n be the vertices of a regular, , numbers z 1 , z 2 , ..... z n have the same amplitude., , polygon of n sides and z 0 be its centroid, then, ,  The sum and product of two complex, , z 12, , , , z 22, ,  ..... , , z n2, , , , nz 02, , ., ,  If z1 , z 2 , z 3 be the vertices of a triangle, then, the, , triangle, , is, , equilateral, , iff, , (z 1  z 2 )2  (z 2  z 3 )2  (z 3  z 1 )2  0, , or, or, , z12, , , , z 22, , , , z 32, ,  z1 z 2  z 2 z 3  z 3 z1, , (i) (a  b 2 )(a 2  b)  a2  b 2  ab, (ii) (a  b)(a  b 2 )(a 2  b )  a3  b 3, ,  If z 1 , z 2 z 3 are the vertices of an isosceles, right, , angled, , at, , z2, , then, , z 12  z 22  z 32  2 z 2 (z 1  z 3 ) ., ,  If z 1 , z 2 , z 3 are the vertices of a right-angled, isosceles, , triangle,, (z 1  z 2 ) 2  2z 1  z 3 (z 3  z 2 ) ., ,  If  and  2 are the complex cube roots of, unity, then, , 1, 1, 1, , , 0., z1  z 2 z 2  z 3 z 3  z1, , triangle,, , numbers are real simultaneously if and only if, they are conjugate to each other., , then, ,  If z 1 , z 2 , z 3 be the affixes of the vertices A, B, C, respectively of a triangle ABC, then its orthocentre, , (iii), (a  b  c 2 )(a  b 2  c)  a 2  b 2  c 2  ab  bc  ca, , (iv) (a  b  c) (a  b  c 2 )(a  b 2  c )  a3  b 3  c 3  3 abc, ,  If three points z1, z 2 , z3 connected by relation, az1  bz 2  cz 3  0 where a  b  c  0 , then the three, , points are collinear., ,  If z is a complex number, then, , ez, , is, , periodic.,  If three complex numbers are in A.P., then

Page 33 : Vector Algebra, , they lie on a straight line in the complex plane., , 51., , ASSIGNMENT ON COMPLEX NUMBERS, 1i,   1, then the least integral value of, 1i, , If , , 52., , (a) 6, (b) 10, (c) – 18, (d) – 15, The smallest positive integer n for which, (1  i)2n  (1  i)2n is, , m is, , (a)1, , (a), (c) 8, 42., , 2 (b), (d) None of these, i 592  i 590  i 588  i 586  i 584, , The value of, , i 582  i 580  i 578  i 576  i 574, , (a) 1, (c) 3, 43., , 44., , 4, , 1  i 2  i 4  i 6  .....  i 2 n is, , 53., , [EAMCET 1980], , 54., , (b) i, (d) 1, , 55., , is equal to, , (b) 1, (d) 1, , 56., 200, , 46., , If i 2  1 , then the value of, , i, , n, , (a) 50, (c) 0, , (b) – 50, (d) 100, , 57., , 13, , 47., , The value of the sum, ,  (i, , n, , (b), , (c), ,  2 23 , ,, , ,  13 13 , , If, , 6 i  3i, 4, 3i, 20, 3, , 48., , 58., , The least positive integer n which will, n, , i1 , reduce ,  to a real number, is, , 49., , (b) 3, (d) 5, 2 (d), , None of these, , (a) 16, (c) 32, 50., , (a) 32 i, (c) 24 i – 32, , (b) (1, 3), (d) (0, 0), 1a, , 1a, , (b) cot, , , , [, , , 2, , (d) i tan, , 2, , , 2, , Solving 3  2yi  9  7i , where i  1, for x and, y real, we get, [AMU 2000], (a) x  0.5 , y  3.5, (b) x  5 , y  3, 2, , 2, , , y 7, , (d) x  0, y , , The complex number, , 1  2i, 1i, , 3  7i, 2i, , lies in which, , quadrant of the complex plane, (a) First, (b) Second, (c) Third, (d) Fourth, , The value of (1  i)8  (1  i)8 is [RPET 2001; KCET 2001], (b) – 16, (d) – 32, 2, 10, (1  i) , where i  1, is equal to, , 1,  1 = x  iy , then (x, y) is[MP PET 2000], i, , [Roorkee, 1998], 1, , 59., , [Karnataka CET 1999], , (b) 1 – i, (d) i/2, , x, , (c) x , ,  i1 , , (a) 2, (c) 4, (c), , 1 ), , If a  cos   i sin , then, , (c) i cot, (b) i  1, (d) 0, , (a) i, (c) i, , (d), , If z  1  i, then the multiplicative inverse of, , n 1, , i   1 , equals, ,  2 23 ,  ,, ,  13 13 , , (a) , , (a) cot , ,  in 1 ) , where, , z1, equals[RPET 1996], z2, , If z1  (4,5) and z 2  (3,2) then, , (a) (3, 1), (c) (0, 3), , is [MP PET 1996], , n 1, , [, , (b) a  1, b  0, (d) a  1, b  2, , z2 is (where i =, (a) 2 i, (c) – i/2, , [RPET 1995], , (a) 2  i, (c) 3, , (d)4, ,  a  ib , then, ,  23 2 , ,, ,  12 13 , , be, , i 2  i 4  i 6  ...... upto (2n  1) terms =, , If i   1 , then 1  i 2  i 3  i 6  i 8, , 3, , 1i, If , , 1i, , (a) a  2, b  1, (c) a  0, b  1, , [EAMCET 1980; Kerala (Engg.) 2005], , 45., , 2, , 100, , 1 , , (b) Negative, (d) Cannot, , (a) i, (c) 1, , (b), , (c), , (b) – 2, (d) – 4, , (a) Positive, (c) Zero, determined, , [UPSEAT, , 2003], , m, , 41., , If x  3  i , then x 3  3 x 2  8 x  15 , , 1, , 60., , The real part of, , 61., , (a), 1/4, (b), (c) tan /2, (d) 1/1– cos , The statement (a  ib)  (c  id ) is true for [RPET 2002], , [AMU 2001], , (b) 64 + i, (d) None of these, , 1  cos   i sin , , is equal to, , 1

Page 34 : Vector Algebra, , 62., , 63., , (a) a 2  b 2  0, (b) b 2  c 2  0, (c) a 2  c 2  0, (d) b 2  d 2  0, The multiplication inverse of a number is the, number itself, then its initial value is, (a) i, (b) – 1, (c) 2, (d) – i, If z  x  iy, z 1 / 3  a  ib and, value of k equals, (a) 2, (c) 6, , 1., , If, , ci,  a  ib ,, ci, , x y,   k (a 2  b 2 ) then, a b, , 8., , real, then, , 9., , 10., , 1, (a) x , 5, , (c) x  iy , 3., , The conjugate of, , (d) x  iy , , 11., , 1i, 1  2i, , (2  i)2, , in the form of a +, 3i, 13,  15 ,  i , 2,  2 , , 12., , 4., , 13,  9 ,  i, , 10,  10 , , 5., , (b), , 3i, 4, 11  10 i, (c), 17, , 6., , 7., , (d), , 13,  9 ,  i , 10,  10 , , 13., , (b) 3  5i, (d) 3  5i, , 11  10 i, 17, 2  3i, (d), 4i, , z 1, is, z 1, , [, , (d) | z |  1, , If z is a complex number, then which of the, following is not true, (a) | z 2 | | z | 2, (b) | z 2 | | z | 2, , If, , (d) z 2  z 2, , z1 and z 2, , are two non-zero complex, ,  (b), , , 2, , (b), , , , , 2, , (d) 0, , arg(5  3 i) , , (a) tan 1, , 5, , (c) tan 1, , 3, 5, , , , 5 , , , , , , 3, 5, , , (b) tan 1  , 3 , , , 3, , (d) tan 1  , , 2  3i, , is [MP PET 2003], 4 i, 14. Argument, , Conjugate of 1 + i is, [RPET 2003], (a) i, (b) 1, (c) 1 – i, (d) 1 + i, The inequality | z  4 | | z  2 | represents the, region given by, [IIT, , and, , modulus, , of, , , , , , 1i, 1i, , are, , respectively, (a), (c) 0 and, 15., , 1982; RPET 1995; AIEEE 2002], , (a) Re( z )  0, (c) Re( z )  2, , If z is a complex number such that, , (a), , [EAMCET 2002], , The conjugate of complex number, (a), , (d) 2 | z1 | | z 2 |, , (c), , If z  3  5 i, then z 3  z  198 , (a) 3  5i, (c) 3  5i, , (c) | z1 | 2 | z 2 | 2, , numbers such that | z1  z 2 | | z1 |  | z 2 |, then arg, (z1 )  arg (z 2 ) is equal to, , 13  15 ,  i, , 10  2 , , (c), , (b) 2 | z1 | 2  2 | z 2 | 2, , (c) z  z, , ib, is, (a), , (a) 2 | z1 | 2 | z 2 | 2, , (c) | z |  1, , 3, (b) y , 5, 1i, 1  2i, , [MP PET 1993; RPE, , purely imaginary, then, (a) | z |  0, (b) | z |  1, , [MP PET 1996], , 2., , (a) 3/2, (b) 1, (c) 2/3, (d) 4/9, If z1 and z 2 are any two complex numbers, then | z1  z 2 | 2  | z1  z 2 | 2 is equal to, , a2  b 2 , , (a) 1, (b) 1, 2, c, (c), (d)  c 2, If the conjugate of (x  iy)(1  2i) be 1  i , then, , is a purely imaginary number, then, , [MP PET 1993], , (b) 4, (d) 1, a, b, c are, , 2 z1, 3z2, , z1  z 2, =, z1  z 2 [RPET 2003], , [DCE 2005], , where, , If, , (b) Re( z )  0, (d) None of these, , 2, , (b), , If z be the conjugate of the complex number, z , then which of the following relations is, false, [MP PET 1987], (a) | z | | z |, (b) z . z | z | 2, (c) z 1  z 2  z 1  z 2, , 16., , , and 1, 2, , (d) and 1, 2, , (d) arg z  arg z, , 5, , then z = [MP PET 1987], If | z |  4 and arg z , 6, , (a) 2 3  2i, , (b) 2 3  2i, , [

Page 35 : Vector Algebra, , (d)  3  i, , (c)  2 3  2i, 17., , If z , , 1i 3, 1i 3, , 27., , [IIT Screening 2000], , , then arg(z )  [Roorkee 1990; UPSEAT, , 19., , , , (b), , 2, , (c) , 20., , 1  3i, , The amplitude of, , 3 1, , 23., , 24., , 25., , such, , 2, , two non–zero complex, that, and, | z |  1, , , then z  is equal to, , (a), (c) i, , 2., , , 2, , 1 (b), (d) – i, , If x  iy , , –1, , a  ib, , then (x 2  y 2 )2 , c  id, ab, cd, , (a), , a2  b 2, c2  d 2, , (b), , (c), , c2  d 2, a2  b 2, ,  a2  b 2 , , (d)  2, 2 , ,  8  6i , , 2, , c d , , [Roorkee, , 1979;, , RPET, , 1992], , (a) 1  3i, (c) (1  3 i), , (b) (1  3 i), (d) (3  i), , If | z1 | | z 2 | and amp z 1  amp z 2  0 , then, (a) z1  z 2, , (b) z1  z 2, , (c) z 1  z 2  0, , (d) z1  z 2, , 3., , , 3, 2, (c), 3, , 1, z1, , (a) z 2  z 1, , (b) z 2 , , (c) arg (z 1 )  arg (z 2 ), , (d) | z 1 | | z 2 |, , 1i,  is, 1i, , Amplitude of , , (a) | z1z 2 | | z1 || z 2 |, , (b) arg (z1z 2 )  (arg z1 )(arg z 2 ), , (c) | z1  z 2 | | z1 |  | z 2 |, , (d) | z1  z 2 | | z1 |  | z 2 |, , 1 3 i, 3 i, , If three complex numbers are in A.P., then, they lie on, (a) A circle in the complex plane, (b) A straight line in the complex plane, (c) A parabola in the complex plane, (d) None of these, , 2., , In the argand diagram, if O, P and Q, represents respectively the origin, the, complex numbers z and z + iz, then the angle, OPQ is, [MP PET 2000], , is, , [Roorkee 1998], , , 4, , (c), 2, , (a), , [DCE 1999; Karnataka CET 2005], , (b) , , , 6, , (d) None of these, , The amplitude of 0 is, [RPET 2000], (a) 0, (b)  / 2, (c) , (d) None of these, , , 3, 2, (d) , 3, , (b) , , 1., [RPET 1996], , (a) –/2, (b) /2, (c), /4 (d), /6, Which of the following are correct for any, two complex numbers z1 and z 2, , The amplitude of, , If  1   3  rei , then  is equal to, (a), , | z 1  z 2 | | z 1 | | z 2 | is possible if, , , (a), 6, , (c), 3, 26., ,  are, , If z and, numbers, arg(z )  arg( ) , , is, , (b) , , (d), , 2, , [MP PET 1995], , 1., , , 3, , (d) , 6, , , 3, , (c), 6, , 22., , , 2, , 28., , (d) None of these, , (a), , 21., , , , (c) , , (a) 60 o, (b) 120 o, o, (c) 240, (d) 300 o, If arg (z)   , then arg (z) , (a) , (b) , (c)   , (d)   , The amplitude of the complex number, z  sin   i(1  cos  ) is, (a) 2 sin, , (b) , , (a) , , 2004], , 18., , If arg z  0 then arg(z)  arg(z) is equal to, , 3., , , 3, 2, (d), 3, , (b), , The points 1  3i, 5  i and 3  2i in the complex, plane are, (a), Vertices of a right, angled triangle, (b) Collinear

Page 36 : Vector Algebra, , 4., , | z1  z 2 | is, | z1 |  | z 2 |, , (a), 5., , 2| z | 2, , (c) | z | 2, , (d) sin 9  i cos 9, 5, 5, | z1 |  | z 2|, 3 i   3 i , 3., If z  , , then, , , ,  2, 2   2, 2 , , , 6, , (b), , (d) Re( z )  0, Im(z )  0, , (c) Re( z )  0, Im(z )  0, 1, | z|2, n, 2  1  cos   i sin  , ,  , 4.,  1  cos   i sin  , (a) cos n   i sin n , , (b) cos n   i sin n , , (c) sin n   i cos n , , (d) sin n   i cos n , cos 49  i sin 49, , (a), (c) cos 49  i sin 49, , 9, , [MP PET 1997], , (b) Im( z )  0, , (a) Re( z )  0, , If the area of the triangle formed by the, points z, z  iz and iz on the complex plane is, 18, then the value of | z | is, , (b), , (d) cos 21  i sin 21, , (c) 3 2, (d) 2 3, (sin   i cos  )n is equal to, 5., A point z moves on Argand diagram in such a, (a) cos n  i sin n, way that |z –3i|  2, then its locus will be [RPET 1992; MP PET 2002], (b) sin n  i cos n, (a) y  axis, (b) A straight line, (c) A circle, , 8., , (b), , 3, | z|2, 2, , (d), , (a), 7., , (b), , (c) | z1 |  | z 2 |, (d) | z1 | | z 2 |, z, , x, , iy, ,, If, then area of the triangle whose, vertices are points z, iz and z  iz is, (a), , 6., , (a) cos   i sin , (b) cos 9  i sin 9, (c) sin   i cos , , (c) Vertices of an obtuse angled triangle, (d) Vertices of an equilateral triangle, If z1 and z 2 are two complex numbers, then, , , , , ,     i sin n   , 2, , 2, , , (d) None of these, , , If arg (z  a) , , 4, , (c) cos n, , , where a  R , then the locus of, , z  C is a, (a) Hyperbola, (c) Ellipse, , [, , (d) None of these, 6., , (b) Parabola, (d) Straight line, , [MP PET 1997], , The value of, , (cos   i sin  ) (cos   i sin  ), is [RPET, (cos   i sin  ) (cos   i sin  ), , 2001], , (a) cos(       )  i sin(       ), 9., , 10., , [AIEEE 2004], , (b) cos(       )  i sin(       ), , (a) An ellipse, , (b) The imaginary axis, , (c) sin(       )  i cos(       ), , (c) A circle, , (d) The real axis, , (d) sin(       )  i cos(       ), , If | z 2  1 | | z | 2 1 , then z lies on, , If z  x  iy and  , , 1  iz, than |  |  1 shows that, z i, , 7., , in complex plane, (a)z will be at imaginary axis, (c) z will be at unity circle, 11., , 1., , If w , , z, 1, z i, 3, , (a) – 1, (b)z will be at real axis, (c) 1of these, (d), None, , and | w |  1 , then z lies on[AIEEE 2005] 8., , (a) A straight line, , (b) A parabola, , (c) An ellipse, , (d) A circle, , 3 (b), , (cos   i sin  ), is equal to, (sin   i cos  )5, 2000], , [RPET 2001], , (d) 2, ,  ,  ,   i sin  n  , then x 1 . x 2 . x 3 ....  , n, 4 , 4 , , If x n  cos , , (c), 2, , 0, , 4, , 2., , is equal to, , 1i 3, 2, , (b), , 1  i 3, 2, ,  ,  ,   i sin r  , then x1 . x 2 ......  is, r, 2 , 2 , , (d), , 8, , (b) 0, , (a), , If x r  cos , (a), (c) 1, ,  1  cos( / 8)  i sin( / 8) , , ,  1  cos( / 8)  i sin( / 8) , , [MNR 1985; UPSEAT, , 9., , 1i 3, 2, , (cos   i sin  )4, (sin   i cos  )5, , (d), , , (a) cos(4  5  )  i sin(4  5  ), (b) cos(4  5  )  i sin(4  5  ), , 1  i 3, 2, [RPET 2002]

Page 37 : Vector Algebra, , (c) sin(4  5  )  i cos( 4  5  ), 10., , (d) None of these, The value of i1/3 is, 3i, 2, , (a), , [UPSEAT 2002], , 3 i, 2, , (b), , 19., , 1i 3, 2, Re(, z, ), Given z  (1  i 3 )100 , then, equals [AMU 2002] 20., Im(z ), , (c), 11., , 18., , 1i 3, 2, , (d), , (a) 2100, , (b) 250, , 1, , (c), , (d) 3, , 3, 12., ,  1  sin   i cos , ,  1  sin   i cos , , n, , ,  =, , , [Kerala (Engg.) 2002], , 21., ,  n, ,  n, ,  n    i sin,  n , (a) cos ,  2, ,  2, , , 22., , , , , ,  2   i sin n  2 , 2, , 2, , , If n is a positive integer, then (1  i)n  (1  i)n is, equal to, , 23., , [Orissa JEE 2003], , 14., , If, ,  n , ,  4 , , (b) ( 2 )n  2 sin, ,  n , ,  4 , , (d) ( 2 )n  2 sin, , 1,  x  2 cos  ,, x, ,  n , ,  4 , ,  n , ,  4 , , then, , xn , , 1, , 15., , (c), , 1i, 2, 1, 4i, , xn, , 17., , 25., , (b) 2 sin n, (d) sin n, , (b) cos, , , 8, ,  i sin, , , 8, , 26., , (d) i, , (d)  ,  2, , If  is a cube root of unity, then (1     2 ), (1     2 ) =, (a) 1, (c) 2, , (b) 0, (d) 4, , If n is a positive integer not a multiple of 3,, then 1   n   2n =, [MP PET 2004], (a) 3, (b) 1, (c) 0, (d) None of these, Square of either of the two imaginary cube, roots of unity will be, (a) Real root of unity, (b) Other imaginary cube root of unity, (c) Sum of two imaginary roots of unity, (d) None of these, If  is a cube root of unity, then, (1  )3  (1   2 )3 , (b) , , (c) , (d) None of these, If  and  are imaginary cube roots of unity,, 1, , , , , , [IIT 1977], , (a) 3, (b) 0, (c) 1, (d) 2, If  is a complex cube root of unity, then, (1   )(1   2 ) (1   4 )(1   8 ) , (a) 0, (b) 1, (c) – 1, (d) 9, If  is a cube root of unity, then the value of, (1     2 )5  (1     2 )5 , 16 (b), , (c), , 48, , 32, , – 32, , (d), , x  a, y  b, z  c , where  is a complex, x y z, cube root of unity, then   , a b c, 2, , If, , (a) 3, (b) 1, (c) 0, (d) None of these, If  is a complex cube root of unity, then, (x  y )(x   y ) (x  2  y) , (a) x 2  y 2, , The two numbers such that each one is, square of the other, are, (a)  ,  3, (b) i, i, (c) 1, 1, , (d) None of these, , (a), , If iz 4  1  0 , then z can take the value[UPSEAT 2004], (a), , 16., , 24., , is equal to, , [UPSEAT 2001], , (a) 2 cos n , (c) cos n , , (c) 3, 3 , 3 2, , then  4   4 +, , (d) cos n, , (c) ( 2 )n  2 cos , , (b) 3, 3 i, 3 i2, , 2, ,  n, ,  n, ,  n    i cos ,  n , (c) sin,  2, ,  2, , , (a) ( 2 )n  2 cos , , (a) 3, , (a) 0, ,  n, ,  n, ,  n    i sin,  n , (b) cos , 2, 2, , , , , , 13., , (27 )1 / 3 , , (b) x 2  y 2, [MP PET 1987], , (c) x 3  y 3, (d) x 3  y 3, 27., , If, ,  is a complex cube root of unity, then, , (1   )(1   2 ) (1   4 )(1   8 )... to 2n factors = [AMU 2000], , (a) 0, (c) 1, , (b) 1, (d) None of these

Page 38 : Vector Algebra, 28., , The, , product, 3/4, , , , ,  cos  i sin , 3, 3, , , 29., , of, , (b) 1, , 3, (c), 2, , (d) , , 1, 2, , , 2, , , x 2, 1, , 1  0 is, x , , 34., , 39., , 3, , 3, , then the, , (b) 3(a3  b 3 ), , (c) 3(a2  b 2 ), , (d) None of these, , 40., ,  1, ,    3 i,  2, , 2, , , , 1  i 3, 2, , (b), , 1i 3, 2, , (c), , 1i 3, 2, , (d), , 3 i, 2, , If, ,  ( 1) is, , a, , cube root, , of, , unity, , and, , (1   )  A  B ,, , then, and B are, A, respectively, the numbers, [IIT 1995], (a) 0, 1, (b) 1, 0, (c) 1, 1, (d) 1, 1, 41., , a  b   c 2 a  b   c 2, , will be, b  c  a 2 c  a  b  2, , If  ( 1) is a cube root of unity, then, 1, 1 i2, 2, 1i, 1,  2  1 is equal to,  i  i  1, 1, , (a) 0, (c) , , 1000, , , 1, 3, i, (b) , 2, 2, , 1, 3, , i, (d) None of these, 2, 2, If  ,  ,  are the cube roots of p( p  0) , then, x   y   z, , for any x, y and z,, x   y   z, , 1, (1  i 3 ), 2, 1, (c) (1  i 3 ), 2, , (a), , 37., , (a), , [, , 7, , 44., , (c) , , 36., , 1i 3, 2, , [IIT 1995], , (b) 1, (d) i, , (a), 1 (b), –1, 42. The n th roots of unity are in, (c) 2, (d) – 2, (a) A.P., (b) G.P., The cube roots of unity when represented on, (c), H.P., (d) None of these, the Argand plane form the vertices of an, [IIT 1988; Pb. CET 2004], 2, (a) Equilateral triangle, (b)Isosceles triangle 43. If 1, ,  are the three cube roots of unity,, (c) Right angled triangle (d) None of these, then (3   2   4 )6 , [MP PET 1995], , 1, 3, i, (a) , 2, 2, , 35., , (b) , , (c) 1, (d) (b) and (c) both, One of the cube roots of unity is, , [Roorkee 1977; IIT 1970], , (a) a 3  b 3, , The value of, , 1i 3, 2, , [MNR 1990; MP PET 1999], , (a) x  1, (b) x  , 2, (c) x  , (d) x  0, If x  a  b, y  a  b  and z  a  b  , where,  and  are complex cube roots of unity,, then xyz =, (a) a 2  b 2, (b) a 3  b 3, (c) a 3 b 3, (d) a 3  b 3, 2, If x  a  b, y  a  b , z  a 2  b ,, , (d) None of these, , If  is a complex cube root of unity, then, for positive integral value of n , the product, of . 2 . 3 ........  n , will be, (a), , 2, , value of x  y  z is equal to, , 33., , (c) 1,1, ,  2, , of, , If  is a cube root of unity, then a root of, , 3, , 32., , roots, , 38., , (a) 1, , the equation, , 31., , the, , is, , x 1, , 30., , all, , If z , , (b), , 45., , 1, (1  i 3 ), 2, , (d) None of these, , 3 i, , then the value of z 69 is [RPET 2002], 2, , 46., , (a) 64, (b) 729, (c) 2, (d) 0, (1     2 )(1   2   4 )(1   4   8 )......... .. to, factors is, (a) 2 n, (b) 2 2 n, (c) 0, (d) 1, 1 , Let   2 2 2, [IIT 1989] 3, 3 3, , 2 2, 4 3, 6 4, , 2n, , where  is the cube, , root of unity, then, (a)   0, (b)   1, (c)   2, (d)   3, If n is a positive integer greater than unity, and z is a complex number satisfying the, equation z n  (z  1)n , then, , (a) i, (b) i, (a) Re( z )  0, (c) 1, (d) 1, (c) Re( z )  0, The roots of the equation x 4  1  0 , are [MP PET 1986], (a) 1, 1, i,i, (b) 1,  1, i,i, , (b) Re( z )  0, (d) None of these

Page 39 : Vector Algebra, , If  is an nth root of unity, other than unity,, then the value of 1     2  ...   n1 is, (a) 0, (b) 1, (c) 1, (d) None of these, 48. If z 1 , z 2 , z 3 ...... nn are nth, roots of unity, then, , 6, , 47., , for k  1, 2,....., n, , 49., , (a) | z k |  k | z k 1 |, , (b) | z k 1 |  k | z k |, , (c) | zk 1 | | zk |  | zk 1 |, , (d) | z k | | z k 1 |, , If 1, , , , 2, , are three cube roots of unity, then, , (a  b   c 2 )3 + (a  b  2  c )3 is equal to, if, , 57., , 58., , 59., , ab c  0, , (a) 27 abc, 50., , 51., , (b) 0, , (c) 3 abc, (d) None of these, The common roots of the equations, x 12  1  0 , x 4  x 2  1  0 are, (a) , (b)   2, 2, (c)   ,  , (d) None of these, If z 1 , z 2 z 3 , z 4 are the roots of the equation, 4, , z, , z 4  1 , then the value of, , 3, i, , is, , 60., , 52., , 53., ,  1  i 3 , , , , , 2, , , , 20, ,  1  i 3 , , , , , 2, , , , (a) 20 3 i, (c), 54., , If, , 1, 2 19,  and , , (1  i) 20, , , , (1  i 3 )15, , is equal to, , (1  i) 20, , (a) – 64, , (b) – 32, , (c) – 16, , (d), , [AMU 2000], , 1, 16, , If  is a complex root of the equation z 3  1 ,, , ... , , 0, , (d) 16 , , (c) 16 , 62., , (b), , 16, 2, , The value of (1     2 ) (1   2   )6 , where  ,  2, are cube roots of unity, (a) 128 , , (b)  128  2, , (c) 128 , , (d) 128  2, , [, , (b) 1, , If 1, ,  2 are the cube roots of unity, then, their product is, , (a), 0 (b), , (d) 1, , (c) – 1, , are imaginary cube roots of, , unity, then the value of   , , 56., , (1  i 3 )15, , (a), , 63., , 4, , 55., , (c) 2, (d) 1, If  is an imaginary cube root of unity,, (1     2 )7 equals, (a) 128 , (b), 128 , (c) 128  2, (d)  128  2, , 27, 1 3 9, [EAMCET, , ,   1989], , 20, , , , [, , then     2 8 32 128  is equal to, (a), – 1 (b), 0, (c) 9, (d) i, 61. If cube root of 1 is  , then the value of, [Kurukshetra, (3    3 2 CEE, )4 is1996], , i 1, , (a) 0, (b) 1, (c) i, (d) 1  i, If  is an imaginary cube root of unity, then, for n  N , the value of  3n1   3n3   3n5 is, (a) 1, (b) 0, (c) 1, (d) 3, , 6, ,  3 i i 3 , ,  ,  is equal to, [Karnataka,  2  CET2 1999], , ,  , , (a) 2, (b) 0, , 28, , , , 1, , , , ,is, , (a) 1, (b) 1, (c) 0, (d) None of these,  is the cube root of unity, then, If, (3  5  3 2 )2 + (3  3  5 2 )2 =, (a) 4, (b) 0, (c) – 4, (d) None of these, If  is an imaginary cube root of unity, then, , , , , , , , , 64., , (b)  1 / 2, , (c) 1 / 2, , (d) 3 / 2, , 3 i, , then z 69 is equal to, 2, , (a) 1, (c) i, 65., , [RPET 2001], , (b) – 1, (d) – i,  2 ,  2  2,   i sin,  , i  1 , then,  n ,  n , , Let  n  cos , , [MP PET 1999], , (x  y3  z3 2 ) (x  y3 2  z3 ) is equal to[AMU 2001], , (a) 0, , (b) x 2  y 2  z 2, , (c) x 2  y 2  z 2  yz  zx  xy, , the value of sin ( 10   23 )    is, 4, , (a)  3 / 2, , If z , , (d) 1, , x 2 Screening,  y 2  z 2 1994], yz  zx  xy, (d), [IIT, 66., , If z  z 1  1, then z 100  z 100 is equal to [UPSEAT 2001], (a) i, , (b) – i, , (c) 1, , (d) – 1

Page 40 : Vector Algebra, 67., , If, , 1 3i, is a root of equation x 4  x 3  x  1  0, 2, , then its real roots are, (a) 1, 1, , [EAMCET 2002], , (b) – 1, – 1, , (c) 1, – 1, , 77., , If  is a complex cube root of unity, then the, value of  99  100  101 is, (a) 1, (b) – 1, (c) 3, (d) 0, , (d) 1, 2, n, , 68., , 69., , 70., , 1  i 3 ,  is an integer, then n is [UPSEAT 2002], If , 1  i 3 , , , , (a) 1, , (b) 2, , (c) 3, , (d) 4, , Find the value of (1  2   2 )3 n  (1    2 2 )3 n , (a), 0 (b), 1, (c) , (d)  2, If  is a non real cube root of unity, then, (a  b ) (a  b  ) (a  b  2 ) is, , 71., , [Kerala (Engg.) 2002], , (a) a  b, (b) a  b, (c) a 2  b 2, (d) a 2  b 2, Which of the following is a fourth root of, 3, , 3, , 3, , 3, , 1 i 3, , 2, 2,  , cis  , 2, , (a),  , 6, , 72., , (b), ,  , cis  ,  12 , ,  , 3, , (c) cis , , (d) cis , , The value of (8)1/3 is, (a), , 1  i 3, , (b), , 1  i 3, 73., , 74., , (c) 2, (d) All of these, If  is a complex cube root of unity, then, 225  (3  8 2 )2  (3 2  8 ) 2 , (a) 72, (b) 192, (c) 200, (d) 248, If 1, ,  2 are the cube roots of unity, then, 1,  , , n, ,  2n, , n,  2n, ,  2n, , 1, , n, , 1 =, , (a) 0, (c) , , (b) 1, (d)  2,  1  3i, then (3    3 2 )4 =, 2, , 75., , If  , , 76., , (a) 16, (c) 16 , If 1,  ,  2, , (b) –16, (d) 16  2, are the roots of unity, then, , (1  2   2 )6 is equal to, , (a) 729, (c) 243, , [Pb. CET 2001], , (b) 246, (d) 81