Let a , b , c be three vectors such that a b = a c . If | a | = 1 , | b | = 2 3 , | c | = 2 , the angle between a and b is 6 , and the angle between a and c is acute, then the scalar projection of b on c is equal to

Step-by-step solution

a b = a c a ( b - c ) = 0 . This implies b - c = a for some scalar . Taking the dot product with a : a b - a c = | a |^2 = We can find a b : a b = | a || b | ( 6 ) = 1 2 3 3 2 = 3 So, 3 - a c = . Taking the dot product of b - c = a with b : | b |^2 - b c = ( a b ) 12 - b c = 3 b c = 12 - 3 Now, squaring the relation b - c = a : | b |^2 + | c |^2 - 2( b c ) = ^2 | a |^2 12 + 4 - 2(12 - 3 ) = ^2 16 - 24 + 6 = ^2 ^2 - 6 + 8 = 0 ( - 2)( - 4) = 0 = 2 or = 4 . We are given that the angle between a and c is acute, meaning