Mathematics-24.Unit-20 Vectors

Vectors SYLLABUS Vectors and Scalars, addition of vectors, components of a vector in two dimensions and three dimensional space, scalar and vector products, scalar and vector triple product, Application of vectors to plane geometry. 1. INTRODUCTION A vector may be described as a quantity having both magnitude and direction. Geometrically a vector is represented by a directed line segment as shown in the adjacent figure. The direction is indicated by the length of the segment AB. A is called the initial point and B the terminal point of vector = . The line l of which the segment AB is a part, is called the line of support. The length is denoted by | | = | | = AB = a. A scalar is quantity, having magnitude only. In other words, a scalar is just a real number. Displacement, velocity, momentum, area are some examples of vectors while distance, speed, volume temperature are just scalars. 1.1 FREE VECTORS Vectors which are fully characterized by the magnitude and direction only are called free vectors and those are fully characterized by the magnitude, direction and also line of support are called line (or bound) vectors. Displacement, velocity are free vectors while force and moment of a force about a point are line vectors. Free Vector: A free vector is not located in any particular position. If a free vector can be represented by , it can equally be represented by , where OP and AB are equal in length and are in the same direction (i.e. = ). AB = OP = | | Also AB is in the same direction as OP. 1.2 POSITION VECTOR OF A POINT We take arbitrarily any point O in space to be called the origin of reference. The position vector (p.v.) of any point P, with respect to the origin is the vector . For any two points P and Q in space, the equality = expresses any vector in terms of the position vectors and of P and Q respectively. 1.3 ANGLE BETWEEN TWO VECTORS It is defined as the smaller angle formed when the initial points or the terminal points of two vectors are brought together. Note: • 0°    180° 1.4 MULTIPLICATION OF A VECTOR BY A SCALAR Given a vector and a scalar kR, then k (or k) denotes a vector whose magnitude is i.e., k times that of and whose direction is the same or opposite to that of according as k > 0 or k < 0 respectively. Also, 0 = , zero or null vector which has zero magnitude and arbitrary direction. 1 = , (-1) = - , etc. When we have two vectors and such that = k , kR, then and are called collinear vectors. is said to be a scalar multiple of . and are parallel if k > 0 and anti parallel if k < 0. Note also that k1 (k2 ) = (k1k2) = k2(k1 )  k1, k2  R. A vector having the magnitude as one (unity) is called a unit vector. Unit vector in the direction of is defined as = and is denoted by . 2. ADITTION OF TWO VECTORS 2.1 TRIANGLE LAW OF ADDITION Given two vectors and , their sum or resultant written as ( + ) is a vector obtained by first bringing the initial point of to the terminal point of and then joining the initial point of to the terminal point of giving a consistent direction by completing the triangle OAB 2.2 PARALLELOGRAM LAW OF ADDITION The sum can also be obtained by bringing the initial points of and together and then completing the parallelogram OACB Note that addition is commutative i.e., + = Also, + ( + ) = ( + ) + i.e. the addition of vectors obeys the associative law. If and are collinear, their sum is still obtained in the same manner although we do not have a triangle or a parallelogram in this case. 2.3 POLYGON LAW OF ADDITION For adding more than two vectors, we have a polygon law of addition which is just an extension of the triangle law. A consequence of this is that, if the terminus of the last vector coincides with the initial point of the first vector, the sum of the vectors is . To obtain (difference of two vectors), perform addition of and . Also, = ; + = ; (k1 + k2) = k1 + k2 ; k ( = k + k . 2.4 PROPERTIES OF VECTOR ADDITION  vector addition is commutative  vector addition is associative    (k1 + k2) = k1 + k2  k = Illustration 1: ABCD is a parallelogram A1 and B1 are the midpoints of side BC and CD respectively. If then find the value of . Key concept: In such type of questions related to parallelogram or triangle first we define the origin and position vector of different vertices. Solution: Let P.V. of A, B, D be respectively. Then P.V. of C = . Also P.V. of and P.V. of  , hence the value of  is 3/2. Illustration 2: In a quadrilateral PQRS , , , M is the midpoint of and X is a point on SM such that SX = SM, then prove that P, X and R are collinear. Solution: From the given information, we get . Also and  Also  , hence P, X and R are collinear. 3. FUNDAMENTAL THEOREMS OF VECTORS 3.1 FUNDAMENTAL THEOREM OF VECTORS IN TWO-DIMENSIONS If and be two non-zero non-collinear vectors, then any vector in the plane of and can be expressed uniquely as a linear combination of and i.e. there exist unique l, mR such that l +m = . This also means that if l1 + m1 = l2 + m2 then l1 = l2 and m1 = m2. 3.2 FUNDAMENTAL THEOREM OF VECTORS IN THREE-DIMENSIONS If , and be three non-zero, non-coplanar vectors in space, then any vector in space can be expressed uniquely as a linear combination of , and . i.e there exist unique l, m, n R such that l + m + n = This also means that if l1 + m1 + n1 = l2 + m2 + n2 , then l1 = l2, m1 = m2 and n1 = n2. 4. LINEAR COMBINATIONS OF VECTORS The linear combination of a finite set of vectors ,… is defined as a vector such that = + + ……+ , where k1, k2, … kn are any scalars (real numbers). LINEARLY DEPENDENT AND INDEPENDENT VECTORS A system of vectors is said to be linearly dependent if there exists a system of scalars k1, k2 …, kn (not all zero) such that k1 + k2 + … +kn = They are said to be linearly independent if every relation of the type k1 + k2 + … +kn = implies that k1 = k2 =….=kn = 0. Notes: • Two collinear vectors are always linearly dependent. • Two non-collinear non-zero vectors are always linearly independent • Three coplanar vectors are always linearly dependent. • Three non-coplanar non-zero vectors are always linearly independent. • More than 3 vectors are always linearly dependent. • Three points with position vectors are collinear if 1 with 1 + 2 + 3 = 0. • Four points with position vectors , are coplanar if 1 with 1 + 2 + 3 + 4= 0. Illustration 3: If are non–zero non coplanar vectors determine whether the vectors: and are linearly independent or dependent. Key concept: Three vector are linearly dependent if they are coplanar that means any one of them can be represented as a linear combination of other two. Solution: Let , where x and y are scalars. If the given vectors are linearly dependent then x and y will exist uniquely; otherwise not. Consider   but are non–zero, non–coplanar vectors. Hence 2x+ 3y = 4 …(i) –3x – 5y = –5 …(ii) x + 2y = 1 …(iii) Solving (i) and (ii), we get x = 5, y = –2 which also satisfy (iii)  x and y are unique.  Hence and are linearly dependent vectors. 5. ORTHOGONAL SYSTEM OF VECTORS In the orthogonal system of vectors we choose these vectors as three mutually perpendicular unit vectors denoted by , and directed along the positive directions of X, Y and Z axes respectively. Corresponding to any point P(x, y z) we can associate a vector w.r.t. a fixed orthogonal system and then this vector is the position vector (p.v.) of that point. i.e. p.v. of P = Distance of P from O = = x, y, z are called the components of the vector If a vector makes angles , ,  with the positive directions of X, Y and Z axes respectively, then cos, cos, cos are called the direction cosines (d.c.’s) of . cos = cos ; cos = So that cos2 + cos2 + cos2 = 1 Unit vector in the direction of is = . 6. SECTION FORMULA 6.1 INTERNAL DIVISION Let A and B be two points with position vectors and respectively, and C be a point dividing AB internally in the ratio m : n. Then the position vector of C is given by . Proof: Let O be the origin. The , let be the position vector of C which divides AB internally in the ratio m : n then,  n.  n(P.V. of – P.V. of ) = m(P.V. of – P.V. of )     or 6.2 EXTERNAL DIVISION Let A and B be two points with position vectors and respectively and let C be a point dividing externally in the ratio m : n. Then the position vector of is given by . Note: (i) If C is the mid–point of AB, then P.V. of C is . (ii) We have, . Hence is in the form of . where, and .Thus, position vector of any point on can always be taken as where  +  = 1. (iii) If circumcentre is origin and vertices of a triangle have position vectors , then the position vector of orthocentre will be . Illustration 4: ABC is a triangle. A line is drawn parallel to BC to meet AB and AC in D and E respectively. Prove that the median through A bisects DE. Solution: Take the vertex A of the triangle ABC as the origin. Let be the p.v. of B and C. The mid point of BC has the p.v. = The equation of the median is . Let D divide AB in the ratio 1:  p.v. of .Let E divide AC in the ratio 1:  p.v. E =   = . p.v. of the mid-point of DE = which lies on the median. Hence the median bisects DE. 7. ISECTOR OF THE ANGLE BETWEEN TWO VECTORS Consider two non–zero, non–collinear vectors and . The bisector of the angle between the two vectors and is k where k  R+. Illustration 5: If the vector bisects the angle between and , where is a unit vector then find . Key concept: Bisector of the angle between the two vectors and is k where k  R+. Solution: According to the given conditions  =  3 = 3 = (3 + 1) – (2 + 9) + (15 – 2)   9 = (3 + 1)2 + (2 + 9)2 + (15 – 2)2  3152 – 18 = 0   = 0, . If  = 0, (not acceptable) For  = , 8. SCALAR (OR DOT) PRODUCT OF TWO VECTORS The scalar product of and , written as , is defined to be the cos where  is the angle between the vectors and i.e. = abcos. Geometrical Interpretation: is the product of length of one vector and length of the projection of the other vector in the direction of former. . Projection of in direction of = .Projection of in direction of Properties: •  • • • (acos)b = (projection of ) b = (projection of ) a • (Distributive law) • •  are perpendicular to each other  • = • • If then • If are non-zero, the angle between them is given by = Illustration 6: Prove by vector method that (a1b1 + a2b2 + a3b3)2 £ (a12+ a22 + a32) (b12 + b22 + b32). Solution: Let = a1 + a2 + a3 and = . Now × = a1 b1 + a2b2 + a3b3 , also × = | | | | cosq £ | | | |. Þ ( × )2 £ | |2 | |2 Þ (a1b1 + a2b2 + a3b3)2 £ Illustration 7: If | | = 3, | | = 1, | | = 4 and , find the value of Solution: We know, Þ 0 = (Given ) Þ0 =  = – = –13 Illustration 8: In a ABC, prove by vector method that cos 2A + cos 2B + cos 2C  –3/2 Solution: As we know; …(i) and … (ii) Now using (i), we get  3R2 + 2R2 (cos 2A + cos2B + cos 2C)  0  cos 2A + cos 2B + cos 2C  –3/2 9. VECTOR (OR CROSS) PRODUCT OF TWO VECTORS The vector product of two vectors and , denoted by , is defined as the vector , where  is the angle between the vectors and and is a unit vector perpendicular to both and (i.e., perpendicular to the plane of and ).The sense of is obtained by the right hand thumb rule i.e. , and form a right-handed screw. If we curl the fingers of our right hand from to through the smaller angle (keeping the initial point of and same), the thumb points in the direction of . In this case, , and (or ), in that order are said to form a right handed system. It is evident that = absin. Properties: •  • (non-commutative) • (Distributive) • •  are collinear (if none of is a zero vector) • • If then = • Any vector perpendicular to the plane of is  ( ) where  is a real number. Unit vector perpendicular to is ± • denotes the area of the parallelogram OACB, whereas area of OAB = Area is also treated as a vector with its direction in the proper sense. Illustration 9: and are unit vectors and | | = 4. If angle between and is cos–1 and , then show that can be written as also find the value of . Key concept: If , then it can be written as =0, that means vector and ( -2 ) are collinear. Solution: Since vector and ( -2 ) are collinear, vector can be written as   16 + 4 – 4(4) (1) = 2(1)  2 = 16    4. Illustration 10: If , , are three vectors such that , , then show that = 1, Solution: Given …(i) …(ii) (1) – (2)   Now either is perpendicular to which is not possible Or = 0  . Also  .  . Equation the coefficients of , we get . Thus and . 10. SCALAR TRIPLE PRODUCT It is defined for three vectors in that order as the scalar which can also be written simply as . It denotes the volume of the parallelopiped formed by taking a, b, c as the co-terminus edges. i.e. V = magnitude of The value of scalar triple product depends on the cyclic order of the vectors and is independent of the position of the dot and cross. These may be interchange at pleasure. However and anti-cyclic permutation of the vectors changes the value of triple product in sign but not a magnitude. Properties: • If are given as etc., then • i.e. position of dot and cross can be interchanged without altering the product. Hence it is also represented by • • • • • in that order form a right handed system if Illustration 11: Show that . Key concept: In such type of questions first we reduce each term as a product of two or three vector, by substituting a group of vectors by a single vector and then solve. Solution: Let Now L.H.S. = = = = = = 0 = R.H.S. 11. VECTOR TRIPLE PRODUCT The vector product of two vectors, one of which is itself the vector product of two vectors, is a vector quantity called vector triple product. It is defined for three vectors as the vector . This vector being perpendicular to , is coplanar with i.e. Take the scalar product of this equation with a. We get 0 =  If we choose the coordinate axes in such a way that , it is easy to show that  = 1. Hence In general, ( Vector triple product is not associative ) . , if some or all of are zero vectors or are collinear. Illustration 12: Let be three mutually perpendicular vectors of the same magnitude. If the vector satisfy the equation then find Solution: Here - { or l2 where l2 { let + then , and . l2 {3 3 Hence . 12. COLLINEAR AND COPLANAR VECTORS 12.1 METHODS TO PROVE COLLINEARITY • Two vectors and are collinear if there exists kR such that . • If are collinear. • Three points A( ), B( ), C( )are collinear if there exists kR such that that is . • If then A, B, C are collinear. • A( ), B( ), C( ) are collinear if there exists scalars l, m, n, (not all zero) such that where l + m + n = 0. All the above methods are equivalent and any of them can be utilized to prove the collinearity. (of 2 vectors or 3 points) Illustration 13: Let be three non–zero vectors such that any two of them are non–collinear. If is collinear with and is collinear with , then prove that . Key concept: Two vectors and are collinear if there exists kR such that . Solution: It is given that is collinear with  for some scalar  …(i) Also is collinear with  for some scalar  …(ii) from (i) and (ii) (1 + 2) + (3 – ) = 0 1 + 2 = 0 and 3 –  = 0 { and are non–collinear vectors}   = – 1/2 and  = – 6 Substituting the values of  and  in (i) & (ii), we get 12.2 METHODS TO PROVE COPLANARITY • Three vectors are coplanar if there exists l, mR such that i.e., one can be expressed as a linear combination of the other two. • If are coplanar (necessary and sufficient condition). • Four points A , B , C and D lie in the same plane if there exist l, mR such that i.e. . • If = 0 then A, B, C, D are coplanar. • A, B, C, D are coplanar if there exists scalars k, l, m, n (not all zero), such that where k + l + m + n = o. Again all the above methods are equivalent. Choose the best amongst them depending on convenience. Illustration 14: Prove that if cos   1, cos  1 and cos   1, then the vectors can never be coplanar. Key concept: If three vectors are coplanar then . Solution: Suppose that are coplanar.  ( R2  R2 – R1 and R3  R3 – R1 ) or or cos (cos – 1)(cos – 1) – (1 – cos)(cos – 1) – (1 – cos)(cos – 1) = 0 dividing through out by (1 – cos) (1 – cos)(1 – cos); we get or –1 +   , which is not possible as Hence they can not be coplanar 13. APPLICATIONS OF VECTORS 13.1 VECTOR EQUATION OF A STRAIGHT LINE Following are the two most useful forms of the equation of a line. (i) Line passing through a given point A and parallel to a vector : where is the p.v. of any general point P on the line and  is any real number. The vector equation of a straight line passing through the origin and parallel to a vector is = n . (ii) Line passing through two given points A and B : For each particular value of , we get a particular point on the line. Each of the above equations can be written easily in Cartesian form also. For example, in case (i), writing , we get  x = a1 + b1, y = a2 + b2, z = a3 + b3. Illustration 15: Given vectors where O is the centre of circle circumscribed about ABC, then find vector . Solution: Here,  and Now [as radii of circle]  or   and …(i) Now if we take then from (i), …(ii) and …(iii)  Solving (ii) and (iii); and  13.2 SHORTEST DISTANCE BETWEEN TWO LINES Two lines in space can be parallel, intersecting or neither (called skew lines). Let be two lines. (i) They intersect if . (ii) They are parallel if are collinear. Parallel lines are of the form Perpendicular distance between them is constant and is equal to . (iii) For skew lines, shortest distance between them (along common perpendicular) is given by . 13.3 EQUATION OF A PLANE IN VECTOR FORM Following are the four useful ways of specifying a plane. (i) A plane at a perpendicular distance d from the origin and normal to a given direction has the equation or ( is a unit vector). (ii) A plane passing through the point A and normal to has the equation . (iii) Parameteric equation of the plane passing through A and parallel to the plane of vectors is given by . (iv) Parameteric equation of the plane passing through A , B C (A, B, C non-collinear) is given by  . In Cartesian form, the equation of the plane assumes the form Ax + By + Cz = D. The vector normal to this plane is and the perpendicular distance of the plane from the origin is . Angle Between a Line and a Plane: The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. Angle Between Two Planes: It is equal to the angle between their normal unit vectors . i.e. cos = 13.4 SOME MISCELLANEOUS RESULTS (i) Volume of the tetrahedron ABCD = (ii) Area of the quadrilateral with diagonals = . 13.5 RECIPROCAL SYSTEM OF VECTORS If are three non-coplanar vectors, then a system of vectors defined by is called the reciprocal system of vectors because . Further The scalar product of any vector of one system with a vector of other system which does not correspond to it is zero i.e. If is a reciprocal system to then is also reciprocal system to . Illustration 16: Show that the points and are equi-distant from the plane r  (5i + 2j – 7k) + 9 = 0 and are on the opposite sides of it. Solution: The given plane is = -9 Length of the perpendicular from to it is Length of the perpendicular from = = Thus the length of the two perpendiculars are equal in magnitude but opposite in sign. Hence they are located on opposite sides of the plane. SOLVED OBJECTIVE PROBLEMS 1. Any four non-zero vector will always be: (A) Linearly dependent (B) Linearly independent (C) Either ‘A’ or ‘B’ (D) None of these Sol. Four or more than four non zero vectors are always linearly dependent. Hence (A) is correct answer. 2. If and are reciprocal vectors, then: (A) (B) (C) (D) None of these Sol. If and are reciprocal, then and Hence (C) is correct answer. 3. If and , then: (A) (B) (C) (D) Sol. Also, Thus Hence (C) is correct answer. 4. If three unit vectors satisfy , then angle between and is: (A) (B) (C) (D) Sol. = 1 Hence (B) is correct answer. 5. Projection of on is equal to: (A) 3 (B) -3 (C) 9 (D) -9 Sol. Projection of on is Thus required projection = = Hence (B) is correct answer. 6. If and are two non-collinear unit vectors, then projection of on is equal to: (A) 2 (B) -2 (C) 1 (D) None of these Sol. Thus projection of on is zero. Hence (D) is correct answer. 7. ABCD is a parallelogram with and . Area of this parallelogram is equal to: (A) sq. units (B) sq. units (C) sq. units (D) sq. units Sol. Area vector of parallelogram = = = = Area of the parallelogram = sq. units Hence (B) is correct answer. 8. If and always make an acute angle with each other for every value of , then: (A) (B) (C) (D) Sol. = x(x + 1) + x – 1 + a = + 2x + a – 1 We must have + 2x + a – 1 > 0 4 – 4(a – 1) < 0 a > 2 Hence (B) is correct answer. 9. Let be three non zero vectors such that Then , where is equal to: (A) 1 (B) 2 (C) -1 (D) -2 Sol. Clearly and represents the sides of a triangle. It’s area vector, Thus, Hence (D) is correct answer. 12. If , where , then: (A) (B) (C) (D) Sol. Taking cross with in first equation, we get and Also, Hence (A) is correct answer. 13. Let be unit vectors such that , , , Then angle between and is : (A) (B) (C) (D) Sol. Taking dot with on both sides, we get If be the angle between and then Hence (B) is correct answer. 10. Let be pair wise mutually perpendicular vectors, such that . Then length of is equal to: (A) 2 (B) 4 (C) 3 (D) 6 Sol. = = = 1 + 4 + 4 + 0 + 0 + 0 = 9 = 3 Hence (C) is correct answer. 11. Let be three unit vectors such that Then is equal to: (A) (B) (C) (D) None of these Sol. Thus are coplaner. Hence Hence (D) is correct answer. 12. ABCD is a parallelogram and are the midpoints of side BC and CD respectively. If , then is equal to: (A) 1/2 (B) 1 (C) 3/2 (D) 2 Sol. Let P.V. of A,B,D be and respectively. Then P.V. of C = Also, P.V. of and, P.V. of Hence (C) is correct answer. 13. Two constant force and act on a particle. If particle is displaced from a point A with position vector to the point B with position vector Then work done in the process is equal to: (A) 15 units (B) 10 units (C) -15 units (D) -10 units Sol. Total force, Displacement, Work done = = 2 – 12 – 5 = - 15 units. Hence (C) is correct answer. 14. A, B, C and D are any four points in the space. If , where is the area of triangle ABC, then is equal to: (A) 2 (B) 1/2 (C) 4 (D) 1/4 Sol. Let P.V. of A, B, C and D be and , and = = = = 2( ) = 4 = 4 Hence (C) is correct answer. 15. The position vector of the points A, B, C and D are , , and . It is known that these points are coplanar, then is equal to: (A) (B) (C) (D) None of these Sol. If vector and are coplanar, then Hence (A) is correct answer. 16. Let be any three vectors. Then is always equal to: (A) (B) (C) Zero (D) None of these Sol. Hence (B) is correct answer. 17. Let be any three vectors. Then is also equal to: (A) (B) (C) Zero (D) None of these Sol. Hence (A) is correct answer. 18. is always equal to: (A) (B) (C) (D) Sol. Similarly, and Hence (C) is correct answer. 19. Value of is always equal to: (A) (B) (C) (D) None of these Sol. Hence (A) is correct answer. 20. For any four vectors the expression is always equal to: (A) (B) (C) (D) None of these Sol. = 0 Hence (D) is correct answer. 21. For any four vectors and is always equal to: (A) (B) (C) (D) Sol. Hence (B) is correct answer. 22. In the parallelogram ABCD if the internal bisectors of the angle and intersect at the point P, then is equal to: (A) (B) (C) (D) Sol. Let P.V. of B, A and C be and respectively. Now, and Hence (D) is correct answer. 23. If the vector bisects the angle between and , where is a unit vector, then: (A) (B) (C) (D) Sol. We must have For (not acceptable) For Hence (D) is correct answer. 24. Distance of from the plane is: (A) (B) (C) (D) None of these Sol. Let Q( ) be the foot of altitude drawn from P to the plane = 0, Also Required distance Hence (C) is correct answer. 25. Distance of from the line is: (A) (B) (C) (D) None of these Sol. (A) Let Q( ) be the foot of altitude drawn from P( ) to the line = and 26. Let and be unit vector that are mutually perpendicular, then for any arbitrary : (A) (B) (C) (D) None of these Sol. Let Also, and Hence (A) is correct answer. 27. The line will not meet the plane , provided: (A) (B) (C) (D) Sol. We must have and Hence (C) is correct answer. 28. The plane will contain the line , provided: (A) (B) (C) (D) Sol. We must have and Hence (C) is correct answer. 29. If the projection of point on the plane is the point , then: (A) (B) (C) (D) Sol. We have and Hence (B) is correct answer. 30. Let and be unit vectors that are perpendicular to each other, then will always be equal to: (A) 1 (B) Zero (C) -1 (D) None of these Sol. = 1 Hence (A) is correct answer. 31. If , then the vector is always equal to: (A) (B) (C) (D) Sol. Hence (D) is correct answer. 32. For any two vectors and , the expression is always equal to: (A) (B) (C) Zero (D) None of these Sol. Similarly, and, Let Hence (B) is correct answer. 33. Let and be three non zero vectors such that , angle between and is and is perpendicular to and , then where is equal to: (A) 1/2 (B) 1/4 (C) 1 (D) 2 Sol. Hence (A) is correct answer. 34. Let be three vectors such that , , Then: (A) (B) (C) (D) Sol. Hence (A) is correct answer. 35. Let and be unit vectors such that , then the value of is equal to: (A) (B) (C) (D) Sol. Now, Hence (C) is correct answer. 36. Let be three unit vectors such that If the angle between and is , then , where is equal to: (A) (B) (C) (D) None of these Sol. Hence (B) is correct answer. 37. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length units then, is always equal to: (A) (B) (C) (D) Sol. Let P.V. of P, A, B and C are and respectively and O( ) be the circumcentre of the equilateral triangle ABC. Now, Similarly, and as Hence (A) is correct answer. 38. . A vector coplanar with and , whose projection on is of magnitude is: (A) (B) (C) (D) Sol. Let the required vector be Then, and Now, (2 – 2 – 1) + (2 – 1 – 2) or 2 If , then where If then Hence (A) is correct answer. 39. If am + bm + cm , m =1, 2, 3, are pairwise perpendicular unit vectors, then is equal to (A) 0 (B) 1 or –1 (C) 3 or -3 (D) 4 or –4 Sol. = =1 Þ = ± 1 . Hence (B) is correct answer. 40. If are three non-coplanar unit vectors, then is equal to (A) (B) (C) (D) Sol. = projection of in the direction of . Hence the given vector is Hence (D) is correct answer. 41. If sec2A and are coplanar, then cot2A + cot2B + cot2C is (A) equal to 1 (B) equal to 2 (C) equal to 0 (D) not defined Sol. The vectors are co-planar Þ = 0 Þ cot2A + cot2 B + cot2 C + 1 = 0 which is not possible. Hence (C) is correct answer. 42. If a, b, c are three non - coplanar vectors and p, q, r are vectors defined by the relations r = then the value of expression (a + b).p + (b + c).q + (c + a).r is equal to (A) 0 (B) 1 (C) 2 (D) 3 Sol. Hence the given scalar expression = 1 + 1 + 1 = 3. Hence (D) is correct answer. ¬ 43. The value of |a  |2 + |a  |2 + |a  |2 is (A) a2 (B) 2a2 (C) 3a2 (D) none of these Sol.  |a|2 sin2  + |a|2 sin2  + |a|2 sin2  = 3|a2| – |a2|(cos2  + cos2  + cos2 ) = 2|a2| = 2a2 Hence (B) is correct answer. 44. If are non-coplanar vectors then is equal to (A) 3 (B) 0 (B) 1 (D) none of there Sol. = 1 – 1 = 0. Hence (B) is correct answer. 45. Consider ABC and A1B1C1 in such a way that and M, N, M1¬ , N1 be the mid points of AB, BC, A1B1 and B1C1 respectively, then (A) (B) (C) (D) Sol. Þ Þ Þ Þ Þ Þ Þ 2 Þ Þ 2 Þ . Hence (D) is correct answer. 46. Let , , where x1, x2, x3  {-3, -2, -1, 0, 1, 2}. Number of possible vectors such that are mutually perpendicular, is (A) 25 (B) 28 (C) 22 (D) none of these Sol. Þ x1 + x2 + x3 = 0 Thus we have to obtain the number of integral solution of this equation. Coefficient of x° | ( x-3 +x-2 +x-1 + x0 + x + x2)3 = x° = x9 |(1 – x6)3 ( 1– x)-3 =11C9 – 3.5C3 = 25. Hence (A) is correct answer. 47. Let a, b, c, be distinct and non-negative. If the vectors ai + aj + ck, i + k, and ci + cj + bk lie in a plane, then c is (A) A.M. of a and b (B) G.M. of a and b (C) H.M of a and b (D) equal to zero. Sol. = 0 C2  C2 – C1 –1(ab – c2) = 0  c2 = ab Hence (B) is correct answer. 48. If , and ,then is equal to (A) 320 (B) 320 (C) - 320 (D) -320 Sol. and process gives on = –320 . Hence (C) is correct answer. 49. If is the vector whose initial point divides the joining of and in the ratio k:1 and terminal point is origin. Also then the interval in which k lies (A) (–, –6]  [–1/6, ) (B) (–, –6]  [1/6, ) (C) (–, 6]  [–1/6, ) (D) (, 6]  [–1/6, ) Sol: The point that divides and in the ratio of k : 1 is given by Also   On squaring both sides, we get Or 12k2 + 74k + 12  0 (6k + 1) (k + 6)  0 Hence k  (–, –6]  [–1/6, ). Hence (A) is correct answer. 50. If 'a' is real constant and A, B, C are variable angles and, then the least value of is : (A) 10 (B) 11 (C) 12 (D) 13 Sol: The given relation can be re–written as  (as, a.b = |a| |b| cos)   … (i) also, (as, ) … (ii) from (i) and (ii), Hence least value of . Hence (C) is correct answer. 51. The vector and are collinear for (A) unique value of x , 0 < x < /6 (B) unique value of x , /6 < x < /3 (C) no value of x (D) infinity many value og x, 0 < x < /2 Sol: Since and are collinear, for some , we can write .    cosx = x Here we will get only one unique value of x which belongs to Hence (B) is correct answer. 52. The vectors have their initial points at (1, 1), the value of  so that the vectors terminate on one straight line is (A) 0 (B) 3 (C) 6 (D) 9 Sol: Since initial point of is , their terminal points will be , and . Now given all the vectors terminate on one straight line. Hence  1 = 1 and  = 9 Hence (D) is correct answer. 53. Given that is a perpendicular to and p is a non-zero scalar, then a vector satisfying is given by (A) (B) (C) (D) none of these Sol: We have . Taking dot by vector , we get Þ Þ Þ Þ . Hence (A) is correct answer. 54. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length ' ' units then is always equal to: (A) (B) (C) (D) Sol: Let P.V. of P, A, B and C are and respectively and O( ) be the circumcentre of the equilateral triangle ABC.  Now Similarly, and  = Hence (A) is correct answer. 55. Let and are two non collinear vector such that . The angle of a triangle whose two sides are represented by the vector and are (A) (B) (C) (D) none of these Sol: Let , clearly and are mutually perpendicular as is coplanar with and and is at right angle to the plane of and . And  = = Also, = =  Thus angles are Hence (B) is correct answer. 56. E and F are the interior points on the sides BC and CD of a parallelogram ABCD. Let and . If the line EF meets the diagonal AC in G then where  is equal to (A) (B) (C) (D) Sol: (D) Let P.V. of A.B. and D be . Then  and Equation of EF : Equation of AC : For point G we must have,   Hence (D) is correct answer. 57. If and the vector , and are non-coplanar, then the vectors and are: (A) coplanar (B) none coplanar (C) collinear (D) non collinear Sol: Given [since X, Y, Z are non-coplanar] Hence and are coplanar. Hence (A) is correct answer. 58. If b and c are any two perpendicular unit vectors and a is any vector, then (A) b (B) a (C) c (D) b + c Sol: Consider three non-coplanar vectors b, c and . Any vector a can be written as ……(i). Taking dot product with in (i) we get .Taking dot product with b in (i) Taking dot product with c in (i), we get, a.c = y Thus Hence (B) is correct answer. 59. If the lines and intersect (t and s are scalars) then. (A) (B) (C) (D) none of these Sol: For the point of intersection of the lines  Hence (B) is correct answer. 60. If and then (A) (B) (C) (D) none of these Sol: Also , and  are mutually perpendicular vectors.  and  Hence (C) is correct answer. 61. The position vector of a point P is where x, y, z Î N and . If = 10, then the number of possible positions of P is (A) 30 (B) 72 (C) 66 (D) 9C2 Sol: Given = 10 Þ x + y + z = 10, x, y, z ³1 The number of possible positions of P = coefficient of x10 in (x + x2 + x3 + … )3 = coefficient of x7 in (1 – x)-3 = 3 + 7 – 1C7 = 9C7 = 9C2 = 36 Hence (D) is correct answer. 62. If vectors ax and x make an acute angle with each other, for all x  R, then a belongs to the interval (A) (B) ( 0, 1) (C) (D) Sol: Since vectors make an acute angle with each other so their dot product must be positive i.e. ax2 – 10 ax + 6 > 0  x  R - ax2 + 10ax – 6 < 0  x  R  –a < 0 and 100a2 < 24a Hence (C) is correct answer.