If a , ~b , c are three vectors such that | a |= 31 , 4| ~b |=| c |=2 and 2( a b )=3 ( c a . )and if the angle between b and c is 2 3 then | a c a b |^2=

Step-by-step solution

Given vector magnitudes: | a | = 31 | b | = 1 2 | c | = 2 The angle between b and c is = 2 3 . The vector relation 2( a b ) = 3( c a ) becomes 2( a b ) + 3( a c ) = 0 after using c a = -( a c ) . Factoring gives a (2 b + 3 c ) = 0 , implying 2 b + 3 c = k a for some scalar k . The magnitude squared yields: |2 b + 3 c |^2 = 4| b |^2 + 9| c |^2 + 12( b c ) = 4( 1 4 ) + 9(4) + 12(- 1 2 ) = 1 + 36 - 6 = 31 Since |k a |^2 = k^2| a |^2 = k^2 31 , we find k^2 = 1 , so k = 1 . For both cases, | a c |^2 = 4| b c |^2 = 4(| b