MULTIPLE PRODUCTS OBJECTIVES. If a i j,b j k,c i k, = + = + = + then a. ( b c) ) 8 ) 6 3) 4 5). If a = 3i j+ k and b 3i j k = = +, then a. ( a b) = ) 0 ) 3) 3 4) not defined { } 3. The scalar a. ( b c) ( a b c) + + is equal to ) 0 ) a b c + b ca 3) a b c 4) a b c 4. If a is perpendicular to b and c, a =, b = 3, c = 4 and the angle between b and c is π, then abc 3 = ) ) 3 3) 3 4) 5. The vector a lies in the plane of vectors b and c. Which of the following is correct? ) a. ( b c) = 0 ) ( ) a. b c = 3) a. ( b c) = 4) ( ) a. b c = 3 6. For three vectors u, υ, w which of the following expressions is not equal to any of the remaining three? ) u. ( υ w) ) ( υ w ).u 3) υ.( u.w) 4) ( υ) u.w

7. Volume of the parallelepiped whose coterminous edges are i 3j+ 4k,i + j k, 3i j+ k ) 5 cu. Units ) 6 cu. Units 3) 7 cu. Units 4) 8 cu. Units 8. If a b c =, then the volume of the parallelepiped whose coterminous edges are a + b,b + c and c + a is ) 9 cu. Units ) 8 cu. Units 3) 8 cu. Units 4) 6 cu. Units 9. If a b c = 4, then the volume of the parallelepiped with a + b,b + c and c + a as coterminous edges is ) 6 ) 7 3) 8 4) 5 0. If a b c =, then a + b b + c c + a = ) 4 ) 36 3) 48 4) 6. If a, b,c are linearly independent, then ) 9 ) 8 3) 7 4) 6 a b b c c a + + + = a b c ( a + b) ( b + c ).( 5c + a). If a, b, c are linearly independent and ) 0 ) 4 3) 8 4) = k, then k is a.b c 3. If a, b, c are three non-coplanar vectors, then ( a + b + c) ( a + b) ( a + c) to ) 0 ) a b c is equal 3) a b c 4) a b c 4. If a, b, c be three non-coplanar vectors and a b b c c a ) ) 0 3) - 4) = k a b c, then k =

5. Let a = i k, b = xi + j+ ( x) k and ( ) on ) Only x ) Only y 3) Neither x nor y 4) Both x and y c = yi + x j+ + x + y k. Then a b c depends 6. The value of p such that the vectors i + 3j k, i j+ 4k and 3i + j + pk are coplanar, is ) 4 ) 3) 8 4) 0 7. The vectors λ i + j+ k, i + λ j k and i j+ λ k are coplanar if λ = ),± 5 ),± 6 3),± 3 4) 3,± 8. If ( ) ( ) ( ) ) 6 a λb. b c c + 3a = 0 then λ = ) 3) 6 5 4) 5 9. If a, b,c are unit vectors perpendicular to each other, then ) ) 3 3) 4) 4 a bc = 0. If a = i + 3 j, b = i + j+ k and c = λ i + 4 j+ k are coterminous edges of a.. Let parallelepiped of volume cu. Units, then a value of λ is ) ) 3) 3 4) 4 ( ) ( ) ( ) ( ) a. b c b. a c + c a.b c. a b ) 0 ) 3) - 4) a c p =, a b c is c a a b q =, r = a bc a b c Then p( a + b) + q. ( b + c) + r. ( c + a) is equal to ) -3 ) 3 3) 0 4) - a b c being any three non-coplanar vectors

3. If a + 4b c d = λ acd + µ b cd, then λ + µ = ) 6 ) -6 3) 0 4) 8 4. The position vectors of the points A,B,C,D are 3i j k, i + 3 j 4 k, i + j+ k and 4i + 5 j+ λ k respectively. If A,B,C,D are coplanar, then λ = ) 46 ) 46 7 7 3) 46 5 4) 46 5 5. If the points (,0,3), (-,3,4), (,,) and (a,,5) are coplanar, then a = ) ) - 3) 4) - 6. Let a be a unit vector and b a non zero vector not parallel to a. Then the angle between the vectors u = 3 ( a b) and ( ) ) 4 π ) 3 π 3) π 4) 6 π υ = a b a is 7. The volume of the tetrahedron having vertices i + j k, i + 3j+ k, i + j+ 3k and k is ) 7 cu. Units 3 3) 7 cu. Units 8 ) 7 cu. units 6 4) 5 cu. units 8 8. The volume of the tetrahedron with vertices (,,),(4,3,3),(4,4,4) and (5,5,6) is ) cu. Units ) cu. units 4 3) cu. units 3 4) cu. units 5 9. Volume of the tetrahedron with vertices at (0,0,0), (,0,0), (0,,0) and (0,0,) is ) cu.units 6 ) cu.units 4 3) cu.units 3 4) cu.units 5

30. The volume of the tetrahedron whose vertices are (,-6,0), (-,-3,7), (5,-, λ ) and (7,-4,7) is cu. Units, then the value of λ is ) or 6 ) 3 or 4 3) or 7 4) 5 or 6 3. If i, j, k are orthonormal unit vectors, then i( a i) + j ( a j) + k ( a k) is ) a ) a 3) 3a 4) 4a 3. If a = i j 3k, b = i + j k, and c i 3j k p+q+r = ) -4 ) 4 3) 4) - 33. If a = i + j+ k, b i j,c i = + = and ( ) ) ) 0 3) - 4) 34. If a = i 3j+ 4k, b = i + j k and c i j k ) a ) b 3) c 4) a b { } 35. a a ( a b) is equal to ) ( a.a )( b a) ) ( a.a )( a b) 3) a.a 4) a b 36. ( ) ( ) a b a c.d is equal to ) a b c 3) a b c ( c.d) ) a b c ( c.d ) 4) b a c ( b.d) = + and a ( b c) a b c = λ a + µ b, then λ + µ = = +, then a ( b c) 37. Which of the following statement is not true? ) a ( b + c) = a.b + a.c ) ( a b) c = a ( b c) 3) a. ( b c) = ( a b ).c 4) a ( b + c) = ( a b) + ( a c) { } { ( )} { ( )} 38. ( ) a. b i i + a. b j j + a. b k k = ) ( a b) ) 3( a b) 3) a b 4) ( a b) = pi + q j+ rk, then is perpendicular to

39. The position vectors of three non- collinear points A,B,C are a, b,c respectively. The distance of the origin from the plane through A,B,C is ) ( a b ) + ( b c ) + ( c a ) 3) ( a b ) + ( b c ) + ( c a ) a bc a bc ) ( a b ) + ( b c ) + ( c a ) 3 a bc 4) ( a b ) + ( b c ) + ( c a ) 40. The shortest distance between the lines r ( 3i 8 j 3k) s( 3i j k) ( ) ( ) r = 3i 7j + 6k + t 3i + j+ 4k is ) 30 ) 30 3) 3 30 4) 4 30 = + + + + and 4. If the four points a, b,c,d are coplanar, then b cd + c a d + a b d = ) 0 ) a b c ) a b c 4) 3 a b c 4. Let a = ai + a j+ a3 k,, b = bi + b j+ b3 k, and c = ci + c j+ c3 k be three non-zero and non coplanar vectors such that c is a unit vector perpendicular to both a a b c π and b. If the angle between a and b is, then a b c = 6 ) 0 ) ± ( a + a + a3 )( b + b + b3 ) 3) 4) - a b c 3 3 3 43. If a b are non-zero and non-collinear vectors, then a b i i a b j j a b k + + k is ) a + b ) a b 3) a b 4) b a

44. If three unit vectors a, b, c are such that ( ) makes with b and c respectively the angles ) 40, 80 ) 45, 45 3) 30, 60 4) 90, 60 b b c = b, then the vector a 45 Let a, b,c be three non-coplanar vectors each having a unit magnitude. If ( ) ( ) a b c = b + c, then the angle between a and b ) 60 ) 30 3) 35 4) 0 46. Let a, b,c be three vectors having magnitude, and respectively. If ( ) a a c + b = O, then the acute angle between a and c is ) 30 ) 60 3) 45 4) 75 47. Let p,q,r be the three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the equation ( ), then x is given by ) ( p q r ) 3) ( p q r ) 3 + ) + + 4) 48. If ( ) ( ) ( ) ( + + ) p q r ( + ) p q r 3 { } {( ) } {( ) } p x r p + q x r q + r x p r = O a α β + b β γ + c γ α = O and at least one of a a,b and c is non-zero, then vectors α, β γ ) Parallel ) Mutually Perpendicular 3) Coplanar 4) None of these 49. If ( b c) ( c a) 3c =, then b c c a a b = ) ) 7 3) 9 4)

50. Vector c is perpendicular to a a = i + 3j k and b = i j+ 3k. Also c. ( i j k) 6 + =. Then the value of a b c is ) 36 ) -36 3) -63 4) 63 5. If b and c are any two non-collinear unit vectors and a is any vector, then ( ) ( ) ( ) ( ) a.b b + a. b c a.c c + b c = b c ) a ) 3a 3) a 4) 4a 5. The position vectors of vertices of ABC are a, b,c. Given that a.a = b.b = c.c = 9 and a b c = o. Then the position vector of the orthocentre of ABC is ) a b + c ) a + b c 3) a + b + c 4) b + c a 53. a ( 3b + c ), b ( c c ), c ( a 3b) ) a b,b c,c a 3) 8 a bc ) 8 a bc 4) 6 a b, b c, c a

Scalar and Vector Triple Products. (). (4) 3. () 4. () MULTIPLE PRODUCTS 0 a (b c)[a b c = 0 0 = ( 0) (0 ) = + 4 = 6. HINTS AND SOLUTIONS a b is a scalar. Hence a (a b) i.e. cross product between a vector and a scalar is not defined. G.E. = a {(b c) (a + b + c)} = a {(b a) + (b b) + (b c) + (c a) + (c b) + (c c)} = a {(b a) + (b c) + (c a) + (c b)} = a (b a) + a (b c) + a (c a) + a (c b) = [a b a] + [a b c] + [a c a] + [a c b] = 0 + [a b c] + 0 [a b c] = 0. a b, a c a //(b c) Now b c = b c sin0 3 = 3(4) = 6 3. Then [a b c] = a (b c)

5. () 6. (3) 7. (3) 8. (3) = a b c cos(a, b c) = (6 3) cos 0 = 3. Given a, b, c are coplanar. But b c is a vector perpendicular to the plane containing b and c. Hence b c is also perpendicular to a a (b c) = 0. v (u w) is not equal to the remaining three. Volume = 3 4 3 = ( + ) + 3( 6) + 4( 6) = 7 Volume = 0 0 [a b c] 0 = [(4 0) 0 + ( 0)]() = 8. 9. (3)We have 0. () [a + b b + c c + a] = [a b c] = (4) = 8. [a + b b + c c + a] = [a b c] = () = 4

. () [a + b b + c c + a] 0 = 0 [a b c] 0 = 9[a b c] 9[a b c] G.E. = = 9. [a b c]. (4) 0 0 [a b c] 3. (4) 4. () 0 5 [a b c] = k = k G.E. = (a + b + c) [(a + b) (a + c)] (a + b + c) [(a b) + (a c) + (b a) + (b c)] = a(a c) + a(b a) + a(b c) + b(a + c) + b(b + a) + b(b c) + c(a c) + c(b a) + c(b c) ( a a = 0) = 0 + 0 + [a b c] [b a c] + 0 + 0 + 0 + [c b a] + 0 = [a b c] [b c a] [c a b] = [b c a] [b c a] [a b c] = [a b c] 0 0 [a b c] = k[a b c] 0 0 = k[a b c] { } k = 0 [a b c] 0

5. (3) 6. () 7. (3) 0 0 0 [a b c] = x x = x y x + x y y x + x Applying: C 3 C 3 + C = ( + x).x = + x x = (Expanding along R ) This depends neither on x nor on y. Given vectors are coplanar 3 4 = 0 3 p ( p 8) (3p + 4) + 3( ) = 0 7p + 4 = 0 p =. Given vectors are coplanar λ λ = 0 3 λ λ( λ ) ( λ + ) + ( λ ) = 0 λ 6λ 4 = 0 By inspection one value of λ is. ( λ + )( λ λ ) = 0. Now ± 4 + 8 λ λ = 0 λ =

8. () 9. () 0. (4) λ =,± 3 ± 3 = = ± 3 [a λb b c c + 3a] = 0 λ 0 0 [a b c] = 0 3 0 ( 0) + λ (0 + 6) = 0 Since [a b c] 0 λ = / 6. Given a = b = c = and a b, b c, c a. Now b c = b c sin 90 n = n n is b and c But a is to b and c a n a (b c) = a n = a n cos 0 or a n cos π = or. [a b c] = ± [a b c] =. Volume = 3 0 = (given) λ 4 ( 4) 3( λ ) + 0 = 3λ 0 = 3λ 0 = 3λ = λ = 4.

. (). () 3. () 4. () [a b c] [b a c] G.E. = + [c a b] [c a b] [a b c] [a b c] = = 0 [c a b] [c a b] (b c) (a + b) G.E. = + [a b c] (c a) (b + c) (a b) (c + a) + [a b c] [a b c] a (a c) + b (b c) + b (c a) + c (c a) + c(a b) + a(a b) = [a b c] [a b c] + 0 + [b c a] + 0 + [c a b] + 0 = [a b c] 3[a b c] = = 3. [a b c] (a + 4b) (c d) = (a c d) + 4(b c d) = [a c d] + 4[b c d] + λ [a b c] + µ [b c d] λ =, µ = 4 λ + µ = 6. Given A(3,, ), B(, 3, 4), C = (,, ), D = (4, 5, λ) AB = (, 5, 3), AC = ( 4,3, 3), AD = (, 7, λ + ) Given A, B, C, D are coplanar

5. () 6. (3) 7. () 5 3 [AB AC AD] = 0 4 3 3 = 0 7 λ + 7λ = 46 λ = 46 /7. Let A = (, 0, 3), B = (, 3, 4), C = (,, ), D = (a,, 5) Then AB = (, 3,), AC = (0,, ) AD = (a,, ) Given points are coplanar [AB AC AD] = 0 8a = 8 a =. If θ is the angle between the given vectors, then 3(a b) (a b)a = = 0 3 a b (a b) a θ = 90 = π /. Let A = (,, ), B = (, 3, ), C = (,, 3), D = (0, 0, ) AB = (,3, + ) = (,,3) AC = (, 0, 4), AD = (,,3) Volume = [AB AC AD] 6 cos θ = 3(a b) {(a b) a} 3 a b (a b) a

8. (3) 9. () 30. (3) 3 7 = 0 4 = (4) = cu.units 6 6 3 3 Let A = (,, ), B = (4, 3, 3), C = (4, 4, 4) and D = (5, 5, 6), be the vertices. Then volume of the tetrahedron = [AB AC AD] 6 = [(,,),(,,),(3,3, 4)] 6 = = () = cu.units. 6 6 3 3 3 4 Volume = [OA OB OC] where O, A, B, C are the given points. 6 0 0 V = 0 0 = () = cu.units. 6 6 6 0 0 Let A = (, 6, 0), B = (, 3, 7), C = (5,, λ), D = (7, 4, 7) Then AB = (,3, 3), AC = (4,5, λ 0) and AD = (6,, 3) Volume = 4 6 3 5 = 6 3 λ 0 3 λ 88 = ± 66 λ = 7 or

3. () G.E. = ( i i )a ( i a) + ( j j)a ( j a) j + (k k)a (k a)k = 3a ( i a) i + ( j a) j + (k a)k = 3a a = a 3. () (a c)b (a b)c = p i + q i + rk ( 6 + 6)( i + j k) ( + 3)( i + 3 j k) = p i + qj + rk i 8 j + 5k = p i + qj + rk p + q + r = 8 + 5 = 4 33. () 34. () (c a)b (c b)a = λ a + µ b λ = (c b) = and µ = c a = λ + µ = + = 0 a (b c) = (a c)b (a b)c = ( 3 + 4)( i + j k) ( + 3 4)( i j + k) = i + 4 j 4k Now a (b c) a sin ce[a (b c)] a = 4 + 6 = 0 35. () a {a (a b)} = a{(a b)a (a a)b} (a a)(a b) (a a)(a b) = 0 + (a a)(b a) = (a a)(b a)

36. (3) 37. () 38. (3) 39. (3) (a b) (a c) d = {(a b)c}a {(a b)a}c d = {[a b c]a [a b a]c}d = [a b c](a d). (a b) c a (b c). Hence () is not true. G.E. = {a (b i )}i + {a (b j)} j + {a (b k)}k Interchanging and, we get { } { } { } = (a b) i i + (a b) j j + (a b) k k Let a b = x i + yj + zk (a b) i = x,(a b) j = y,(a b) k = z Hence G.E. = [a b i ]i + [a b j] j + [a b k]k = x i + yj + zk = a b. The equation of the plane passing through the points A(a), B(b),C(c) is r {(a b) + (b c) + (c a)} = [a b c] Distance from the origin to the plane [a b c] =. (a b) + (b c) + (c a)

40. (3) 4. () 4. () Here a = 3i + 8 j + 3k, b = 3i j + k, c = 3i 7 j + 6k, d = 3i + j + 4k i j k b d = 3 3 4 = i ( 4 ) j( + 3) + k(6 3) = 6 i 5 j + 3k Points A(a),B(b),C(c),D(d) are coplanar. AB, AC, AD are coplanar [AB, AC, AD] = 0 [b a c a d a] = 0 (b a) (c a) (d a) = 0 {(b c) (b a) (a c)} (d a) = 0 {(b c) (a b) + (c a)} (d a) = 0 (b c) d (b c) a + (a b) d (a b) a + (c a) d (c a) a = 0 [b c d] + [a b d] + [c a d] = [a b c] since [a b c] = 0 and [c a a] = 0 and (b c) a = a (b c). c (a b) and c =. Also (a, b) = 30

a a a 3 b b b = [a b c] = c (a b) 3 c c c 3 = c. a b cos0 or cos80 = ± a b sin 30 = ± a + a + a3 b + b + b3 43. () 44. (4) = ± a + a + a b + b + b 3 3 Let a b = x i + yj + zk (a b) i = x,(a b) j = y,(a b) k = z [a b i ]i + [a b j] j + [a b k]k Given b a (b c) = = x i + yj + zk = a b b (a c) (a b)c = () Taking dot product with b, (a c)(b b) (a b)(c b) = (b b) a c (a b)(c b) = () In () take dot product with c : (a c)(b c) (a b)(c c) = c b

c b (a c)(b c) (a b) = c b + (a b)(b c) (b c) a b = Using () b c b c (a c)(b c) a b = 45. (3) (b c) = b c = or a b = 0 a c = cos(a, c) = (a c) = 60 and a b = 0 (a, b) = 90 Given a (b c) = (b + c) (a c)b (a b)c = (b + c) Taking cross product with b : (a c)(b b) (a b)(b c) = (b b + b c) (a b)(b c) = (b c) (b c) a b + = 0 a b = ( c b 0) cos θ = θ = 35

46. () 47. () Given a =, b = and c = Also given a (a c) + b = 0 (a c)a (a a)c + b = 0 (a c)a c = b () a (a c) = b a (a c) = b = π But (a, a c) = π / a a c sin = a c sin θ = sin θ = θ = 30 (a c)a c = b [From ()] {(a c)a c} {(a c)a c} = b b (a c) (a c) + c c (a c)(a c) = (a c) + 4 (a c) = (a c) = 3 a c = 3 a c cos(a, c) = 3 3 π cos(a, c) = = cos. 6 (a, c) = 30 Given p = q = r = λ (say) and p q = 0, p r = 0, q r = 0

48. (3) 49. (3) 50. (4) p {(x q) p} + q {(x r ) q} + r {(x p) r} = 0 (p p)(x q) {p (x q)} p + (q q)(x r ) {q (x r )} + (r r )(x p) {r (x p)} r = 0 λ (x q + x r + x p) (p x)p + (p q)p (q x)q + (q r )q ( r x) r + ( r p) r = 0 λ {3x (p + q + r )} [(p x)p + (q x)q + (r x)r] = 0 Clearly this is satisfied by x = (p + q + r ) Given a( α β ) + b( β γ ) + c( γ α ) = O Taking dot product with α, β, γ respectively a[ α β γ ] = 0, b[ α β γ ] = 0,c[ α β γ ] = 0 Also given that at least one of a, b, c is non-zero. Hence α, β, γ are coplanar. Given (b c) (c a) = 3c { } { } (b c) a c (b c) c a = 3c [b c a]c [b c c]a = 3c [b c a]c = 3c { } [b c c] = 0 [b c a] = 3 [b c, c a, a b] = [a b c] = [b c a] = 9 Given c is to a and b c is parallel to a b

5. (3) i j k a b = 3 3 = i (9 ) j(6 + ) + k( 4 3) = 7 i 7 j 7k = 7( i j k) c is parallel to i j k. Let c = λ( i j k) Also given c ( i j + k) = 6 λ( i j k) ( i j + k) = 6 λ ( + ) = 6 λ = 6 λ = 3 Hence c = 3( i j k) Now [a b c] = (a b) c = 7( i j k) 3( i j k) = ( + + ) = 63 We know that b, c and b c are mutually vectors. Any vector a can be expressed in terms of b, c, b c a = xb + yc + z(b c) () Taking dot product on () with b c, we get a (b c) = x{b (b c)} + y{c (b c)} + z(b c) = x(0) + y(0) + z(b c) = z(b c) a (b c) z = b c Given b =, c = Again taking dot product on () with b and c

a b = x(b b) + y(c b) + z(b c) c = x() + y(0) + z[b c c] = x + y(0) + z(0) = x Also a c = x(b c) + y(c c) + z(b c) c = x(0) + y() + z[b c c] = 0 + y + 0 = y a (b c) a = (a b)b + (a c)b +. b c 5. (3)O, A, B, C are coplanar since [a b c] = 0 53. (3) OA = OB = OC a = b = c = 3 ( a = b = c = 9 given) Hence origin O is the circumcentre. P.V. op G i.e., centroid = a + b + c 3 We know that orthocenter H divides GO in the ratio 3 : externally G.E. = [3(a b) + (a c), b c (b a), (c a) 6(c b)] Let a b = p, b c = q, c a = r G.E. = [3p r, q + p,r + 6q] 3 0 = 0 [p q r ] 0 6 = [3( 0) 0 ( 0)][p q r ] = 8[p q r ] = 8[a b, b c, c a] = 8[a b c] 3(a b c) (O) P.V.of H = 3 = a + b + c 3