Let ABCD be a parallelogram with adjacent sides a = AB and b = AD . Let d₁ = AC and d₂ = BD be its diagonals. If | a | = 3 , | b | = 2 and the angle between a and b is 3 , then the value of | d₁ ( d₂ - 3 b )|^2 + |2 d₁ - d₂ |^2 is equal to :

Step-by-step solution

By the parallelogram law of vector addition, the diagonals are given by d₁ = a + b and d₂ = b - a . First, simplify the cross product term: d₂ - 3 b = ( b - a ) - 3 b = - a - 2 b d₁ ( d₂ - 3 b ) = ( a + b ) (- a - 2 b ) = - a a - 2( a b ) - ( b a ) - 2( b b ) = 0 - 2( a b ) + ( a b ) - 0 = -( a b ) The squared magnitude of this cross product is: |-( a b )|^2 = | a |^2 | b |^2 ^2 ( 3 ) = (3)^2 (2)^2 ( 3 2 )^2 = 9 4 3 4 = 27 Next, simplify the second term: 2 d₁ - d₂ = 2( a + b ) - ( b - a ) = 3 a + b The squared magn