Let a and b be two vectors such that | a | = 2 and 3 | a b | = a b . The scalar projection of b on a is 3 . If c = a ( a b ) + 3 b , then the value of | c |^2 is

Step-by-step solution

The scalar projection of b on a is given by a b | a | = 3 . a b = 3| a | = 3(2) = 6 Given 3 | a b | = a b = 6 | a b | = 2 3 Using Lagrange's identity, | a |^2| b |^2 = ( a b )^2 + | a b |^2 : (2)^2| b |^2 = (6)^2 + (2 3 )^2 4| b |^2 = 36 + 12 = 48 | b |^2 = 12 Now, expand the vector triple product in c : c = ( a b ) a - ( a a ) b + 3 b c = 6 a - | a |^2 b + 3 b = 6 a - 4 b + 3 b = 6 a - b Now, compute | c |^2 : | c |^2 = |6 a - b |^2 = 36| a |^2 - 12( a b ) + | b |^2 | c |^2 = 36(4) - 12(6) + 12 = 144 - 72 + 12 = 8