Representative problems and their solutions We discuss problems, which highlight certain aspects of the vector product. For this reason, questions are categorized in terms of the characterizing features of the subject matter : Condition of parallel vectors Unit vector of cross product Nature of vector product Evaluation of vector product Area of parallelogram

Condition of parallel vectors Problem : Determine whether vectors 2i – j + 2k and 3i – 3j + 6k are parallel to each other? Solution : If the two vectors are parallel, then ratios of corresponding components of vectors in three coordinate directions are equal. Here, a x b x = 2 3 a y b y = 1 3 a z b z = 1 3 The ratios are, therefore, not equal. Hence, given vectors are not parallel to each other.

Unit vector of cross product Problem : Find unit vector in the direction perpendicular to vectors i + j – 2k and 2i –j + 3k. Solution : We know that cross product of two vectors is perpendicular to each of vectors. Thus, unit vector in the direction of cross product is perpendicular to the given vectors. Now, unit vector of cross product is given by :

Vector product Unit vector in the direction of vector product. n = a ✗ b | a ✗ b | Here, a ✗ b = | i j k 1 1 - 2 2 - 1 3 | ⇒ a ✗ b = { 1 x 3 - ( - 2 x - 1 ) } i + { ( 2 x - 2 ) - 1 x 3 } j + { ( 1 x - 1 ) - 1 x 2 } k ⇒ a ✗ b = i - 7 j - 3 k ⇒ | a ✗ b | = √ { 1 2 + ( - 7 ) 2 + ( - 3 ) 2 } ⇒ n = 1 √ ( 59 ) x ( i - 7 j - 3 k )

Nature of vector product Problem : Verify vector equality B = C, if AxB = AxC. Solution : Let θ 1 and θ 2 be the angles for first and second pairs of cross products. Then, A ✗ B = A ✗ C ⇒ AB sin θ 1 n 1 = AC sin θ 2 n 2 ⇒ B sin θ 1 n 1 = C sin θ 2 n 2 It is clear that B=C is true only when sin θ 1 n 1 = sin θ 2 n 2 . It is always possible that the angles involved or the directions of cross products are different. Thus, we can conclude that B need not be equal to C.

Evaluation of vector product Problem : If a.b = |axb| for unit vectors a and b, then find the angle between unit vectors. Solution : According to question, a . b = | a ✗ b | ⇒ ab cos θ = ab sin θ ⇒ tan θ = 1 x 1 x tan 45 ° ⇒ θ = 45 ° Problem : Prove that : | a . b | 2 - | a ✗ b | 2 = a 2 x b 2 x cos 2θ Solution : Expanding LHS, we have : | a . b | 2 - | a ✗ b | 2 = ( ab cos θ ) 2 - ( ab sin θ ) 2 = a 2 b 2 ( cos 2 θ - sin 2 θ ) ⇒ | a . b | 2 - | a ✗ b | 2 = = a 2 b 2 cos 2 θ

Area of parallelogram Problem : The diagonals of a parallelogram are represented by vectors 3i+j+k and i-j-k. Find the area of parallelogram. Solution : The area of parallelogram whose sides are formed by vectors a and b, is given by : Area = | a ✗ b | However, we are given in question vectors representing diagonals – not the sides. But, we know that the diagonals are sum and difference of vectors representing sides of a parallelogram. It means that :

Diagonals of a parallelogram The vectors along diagonals are sum and difference of two vectors representing the sides. a + b = 3 i + j + k and a - b = i - j - k Now the vector product of vectors representing diagonals is : ( a + b ) ✗ ( a - b ) = a ✗ a - a ✗ b + b ✗ a + b ✗ ( - b ) = - a ✗ b + b ✗ a Using anti-commutative property of vector product, ( a + b ) ✗ ( a - b ) = - 2 a ✗ b Thus, a ✗ b = - 1 2 x ( a + b ) ✗ ( a - b ) a ✗ b = - 1 2 | i j k 3 1 1 1 - 1 - 1 | a ✗ b = - 1 2 x { ( 1 x - 1 - 1 x - 1 ) i + ( 1 x 1 - 3 x - 1 ) j + ( 3 x - 1 - 1 x 1 ) k } a ✗ b = - 1 2 x ( 4 j - 4 k ) a ✗ b = - 2 j + 2 k The volume of the parallelogram is : | a ✗ b | = √ { ( - 2 ) 2 + 2 2 } = 2 √ 2 units