Let a and b be two vectors such that | a | = 2 , | b | = 3 , and | a b | = 3 3 . If the angle between a and b is acute, and c = a ( a b ) + b ( a b ) , then the value of | c |^2 is equal to :

Step-by-step solution

Using Lagrange's identity: | a b |^2 + ( a b )^2 = | a |^2| b |^2 (3 3 )^2 + ( a b )^2 = (2)^2(3)^2 27 + ( a b )^2 = 36 ( a b )^2 = 9 Since the angle between a and b is acute, a b > 0 . Thus, a b = 3 . Now, expand the vector triple products in the expression for c : a ( a b ) = ( a b ) a - ( a a ) b = 3 a - 4 b b ( a b ) = ( b b ) a - ( b a ) b = 9 a - 3 b Adding these gives c : c = (3 a - 4 b ) + (9 a - 3 b ) = 12 a - 7 b Now, find the squared magnitude of c : | c |^2 = (12 a - 7 b ) (12 a - 7 b ) | c |^2 = 144| a