Question 1

Passage: Let a b = c and b c = a . Question: Consider the following statements : I. ( a b ) c +( b c ) a =( c a ) b II. \ ( a b ) ( b c )\ b =1 Which of the statements given above is/are correct ?

Accepted Answer

Given a b = c and b c = a . Taking the dot product of the first equation with c : ( a b ) c = c c [ a b c ] = | c |^2 Taking the dot product of the second equation with a : ( b c ) a = a a [ b c a ] = | a |^2 Using the properties of the scalar triple product, [ a b c ] = [ b c a ] = [ c a b ]. Thus, | a |^2 = | c |^2 = [ a b c ]. Evaluating Statement I: The left-hand side is ( a b ) c + ( b c ) a = [ a b c ] + [ b c a ] = 2[ a b c ]. The right-hand side is ( c a ) b = [ c a b ] = [ a b c ]. For

Question 2

Passage: Let a , b , c , a + b , b + c , a + b + c be unit vectors. Question: What is the angle between a and b ?

Accepted Answer

Given | a | = 1, | b | = 1, and | a + b | = 1. Squaring the magnitude of a + b , we get: | a + b |^2 = | a |^2 + | b |^2 + 2| a || b | Substituting the given values: 1^2 = 1^2 + 1^2 + 2(1)(1) 1 = 2 + 2 2 = -1 = - 1 2 = 2 3 Answer: 2 3

Question 3

Passage: Let a , b , c be unit vectors. Further, a is perpendicular to b ; c makes an angle 3 with both a and b ; and c =p a +q b +r( a b ). Question: What is the value of r^2 ?

Accepted Answer

Given a , b , c are unit vectors, so | a | = | b | = | c | = 1. Since a is perpendicular to b , a b = 0. The vector c makes an angle of 3 with both a and b , giving c a = 1 1 ( 3 ) = 1 2 and c b = 1 1 ( 3 ) = 1 2 . We are given c = p a + q b + r( a b ). Taking the dot product of c with a : c a = p( a a ) + q( b a ) + r(( a b ) a ) 1 2 = p(1) + q(0) + r(0) p = 1 2 Taking the dot product of c with b : c b = p( a b ) + q( b b ) + r(( a b ) b ) 1 2 = p(0) + q(1) + r(0) q = 1 2 Taking the dot product

Question 4

Passage: Let a , b , c , a + b , b + c , a + b + c be unit vectors. Question: What is the angle between a and c ?

Accepted Answer

Given | a | = | b | = | c | = 1 | a + b | = 1 | a |^2 + | b |^2 + 2 a b = 1 1 + 1 + 2 a b = 1 a b = - 1 2 | b + c | = 1 | b |^2 + | c |^2 + 2 b c = 1 1 + 1 + 2 b c = 1 b c = - 1 2 | a + b + c | = 1 | a |^2 + | b |^2 + | c |^2 + 2( a b + b c + c a ) = 1 1 + 1 + 1 + 2 (- 1 2 - 1 2 + c a ) = 1 3 + 2(-1 + c a ) = 1 1 + 2 c a = 1 c a = 0 Since the dot product of a and c is 0, the angle between them is 2 . Answer: 2

Question 5

Passage: Let a b = c and b c = a . Question: Consider the following statements : I. a , b , c are orthogonal in pairs. II. a , b , c are unit vectors. Which of the statements given above is/are correc

Accepted Answer

Given a b = c and b c = a . From the properties of the cross product, c is perpendicular to both a and b . Similarly, a is perpendicular to both b and c . This implies that a , b , and c are mutually orthogonal. Thus, Statement I is correct. Since the vectors are mutually orthogonal, the angle between any two of them is 90^ . Taking the magnitude of the first equation: | a b | = | c | | a || b | = | c | Taking the magnitude of the second equation: | b c | = | a | | b || c | = | a | Substituting

Question 6

Passage: Let a , b , c be unit vectors. Further, a is perpendicular to b ; c makes an angle 3 with both a and b ; and c =p a +q b +r( a b ). Question: What is the value of (p+q) ?

Accepted Answer

Given | a | = | b | = | c | = 1 and a b = 0. Since c makes an angle 3 with both a and b : c a = | c || a | ( 3 ) = 1 2 c b = | c || b | ( 3 ) = 1 2 We are given the relation: c = p a + q b + r( a b ) Taking the dot product of both sides with a : c a = p( a a ) + q( b a ) + r(( a b ) a ) Since a a = 1, b a = 0, and ( a b ) a = 0, we get: p = 1 2 Taking the dot product of both sides with b : c b = p( a b ) + q( b b ) + r(( a b ) b ) Since b b = 1, a b = 0, and ( a b ) b = 0, we get: q = 1 2 Theref

Question 7

What is the area of the parallelogram whose sides are represented by the vectors i + 2 j + 3 k and 2 i + j + 2 k ?

Accepted Answer

Let a = i + 2 j + 3 k and b = 2 i + j + 2 k . The area of the parallelogram with adjacent sides a and b is given by | a b |. a b = vmatrix i & j & k \ 1 & 2 & 3 \ 2 & 1 & 2 vmatrix = i (4 - 3) - j (2 - 6) + k (1 - 4) = i + 4 j - 3 k The magnitude of the cross product is: | a b | = 1^2 + 4^2 + (-3)^2 = 1 + 16 + 9 = 26 The area of the parallelogram is 26 square units. Answer: 26 square units

Question 8

Let p = a - b , q = a + b . If | a | = | b | = 2 and a b = 2, then what is the value of | p q |?

Accepted Answer

p q = ( a - b ) ( a + b ) = a a + a b - b a - b b Since a a = 0 , b b = 0 , and b a = - a b : p q = 2( a b ) | p q | = 2| a b | Using Lagrange's identity: | a b |^2 = | a |^2 | b |^2 - ( a b )^2 Substituting the given values | a | = 2, | b | = 2, and a b = 2: | a b |^2 = (2)^2 (2)^2 - (2)^2 | a b |^2 = 16 - 4 = 12 | a b | = 12 = 2 3 Therefore, | p q | = 2(2 3 ) = 4 3 Answer: 4 3

Question 9

A force F = 2 i - j + 5 k is applied at the point A(1, 2, 5). If its moment about the point B(-1, -2, 3) is 16 i - 6 j + 2 k , then what is the value of ?

Accepted Answer

The position vector of the point of application A with respect to point B is r = r _A - r _B. r = (1 - (-1)) i + (2 - (-2)) j + (5 - 3) k = 2 i + 4 j + 2 k The moment of the force about point B is given by M = r F . M = vmatrix i & j & k \ 2 & 4 & 2 \ 2 & - & 5 vmatrix M = i (20 + 2 ) - j (10 - 4) + k (-2 - 8) M = (20 + 2 ) i - 6 j - (2 + 8) k It is given that M = 16 i - 6 j + 2 k . Comparing the coefficients of i and k : 20 + 2 = 16 2 = -4 = -2 -2 - 8 = 2 4 = -8 = -2 Both equations yield = -2

Question 10

ABC is a triangle right-angled at B. If A(k, 1, -1), B(2k, 0, 2) and C(2+2k, k, 1) are the vertices of the triangle, then what is the value of k?

Accepted Answer

The direction ratios of the line segment AB are (k - 2k, 1 - 0, -1 - 2) = (-k, 1, -3). The direction ratios of the line segment BC are (2+2k - 2k, k - 0, 1 - 2) = (2, k, -1). Since the triangle is right-angled at B, the line segments AB and BC are perpendicular to each other. Therefore, the dot product of their direction ratios is zero. (-k)(2) + (1)(k) + (-3)(-1) = 0 -2k + k + 3 = 0 k = 3 Answer: 3

Question 11

Consider the following statements in respect of a vector d = ( a b ) c : I. d is coplanar with a and b . II. d is perpendicular to c . Which of the statements given above is/are correct?

Accepted Answer

Given d = ( a b ) c Using the vector triple product expansion: d = ( a c ) b - ( b c ) a This shows that d is expressed as a linear combination of a and b . Thus, d lies in the plane containing a and b , making it coplanar with a and b . Statement I is correct. By the definition of the cross product, the vector resulting from the cross product of two vectors is perpendicular to both of them. Since d is the cross product of ( a b ) and c , it must be perpendicular to both ( a b ) and c . Therefor

Question 12

The position vectors of three points A, B and C are a , b and c respectively such that 3 a - 4 b + c = 0 . What is AB : BC equal to?

Accepted Answer

Given 3 a - 4 b + c = 0 Rearranging the terms: 3 a - 3 b = b - c -3( b - a ) = -( c - b ) Using the definition of position vectors, AB = b - a and BC = c - b : -3 AB = - BC BC = 3 AB Taking the magnitude on both sides: | BC | = 3| AB | BC = 3AB AB BC = 1 3 Thus, AB : BC = 1 : 3. Answer: 1:3

Question 13

The position vectors of three points A, B and C are a , b and c respectively, where c = ( ^2 ) a + ( ^2 ) b . What is ( a b ) + ( b c ) + ( c a ) equal to?

Accepted Answer

Given c = ( ^2 ) a + ( ^2 ) b Substituting the value of c in the second term of the expression ( a b ) + ( b c ) + ( c a ): b c = b (( ^2 ) a + ( ^2 ) b ) b c = ( ^2 )( b a ) + ( ^2 )( b b ) Since b b = 0 and b a = -( a b ): b c = -( ^2 )( a b ) Substituting the value of c in the third term: c a = (( ^2 ) a + ( ^2 ) b ) a c a = ( ^2 )( a a ) + ( ^2 )( b a ) Since a a = 0 : c a = -( ^2 )( a b ) Adding the three terms together: ( a b ) + ( b c ) + ( c a ) = ( a b ) - ( ^2 )( a b ) - ( ^2 )( a b )

Question 14

Let a , b , ( a b ) be unit vectors. What is ( a b ) equal to?

Accepted Answer

Given | a | = 1, | b | = 1, and | a b | = 1. Using Lagrange's identity, | a b |^2 + ( a b )^2 = | a |^2 | b |^2. Substituting the given values, 1^2 + ( a b )^2 = 1^2 1^2. 1 + ( a b )^2 = 1. ( a b )^2 = 0 a b = 0. Answer: 0

Vector Algebra — Previous Year Questions & Solutions