Let a and b be two vectors such that | a | = 1 , | b | = 2 , and the angle between them is 6 . If c = ( a b ) - 3 b , where > 0 , and the angle between b and c is 3 4 , then the value of is equal to :

Step-by-step solution

Given | a | = 1 , | b | = 2 , and the angle between them is 6 . The magnitude of their cross product is: | a b | = | a || b | ( 6 ) = (1)(2) ( 1 2 ) = 1 We are given c = ( a b ) - 3 b . Taking the dot product of c with b : b c = b ( a b ) - 3| b |^2 Since b ( a b ) = 0 , we get: b c = -3(2)^2 = -12 Now, find the squared magnitude of c : | c |^2 = ^2| a b |^2 + 9| b |^2 - 6 ( a b ) b | c |^2 = ^2(1)^2 + 9(4) - 0 = ^2 + 36 The angle between b and c is 3 4 . Using the dot product formula: ( 3 4 ) = b c | b || c | - 1