Let a and b be two vectors such that. | a |=1,| b |=4, a b =2 . If c =(2 a b )-3 b , then the angle between b and c is

Step-by-step solution

Given vectors a and b with | a |=1 , | b |=4 , and a b =2 , and c = (2 a b ) - 3 b , the angle between b and c is determined. The dot product b c evaluates to 2[ b ( a b )] - 3( b b ) . The scalar triple product [ b , a , b ] is zero, simplifying to b c = -3| b |^2 = -48 . To compute | c | , expand | c |^2 = |2 a b |^2 + |-3 b |^2 - 2((2 a b ) (3 b )) . The scalar triple product term is zero, yielding | c |^2 = 4| a b |^2 + 9| b |^2 . Using | a b |^2 = | a |^2| b |^2 - ( a b )^2 = 12 , we find | c |^2 = 4(12) + 9(1