Let a, b, c be three vector such that a 0 and a b =2 a c ,|a|=|c|=1,|b|=4 and | b c |= 15 . If b -2 c = a then equals to

Step-by-step solution

Given a b = 2 a c , we can rewrite this as a ( b - 2 c ) = 0 . It is given that b - 2 c = a . Substituting this into the cross product equation, we get a ( a ) = ( a a ) = 0 , which is consistent for any . Now, consider the magnitude of b - 2 c = a . Squaring both sides, we get | b - 2 c |^2 = | a |^2 . | b |^2 + 4| c |^2 - 4( b c ) = ^2 | a |^2 . Given | a | = 1, | b | = 4, | c | = 1 , we have 16 + 4(1) - 4( b c ) = ^2(1) , so 20 - 4( b c ) = ^2 . We are also given | b c | = 15 . Using the identity | b c |^2 = | b