Let p , q , r be three vectors such that p q = 2( p r ) . If | p | = 1 , | q | = 4 , | r | = 2 , and | p r | = 1 , then the angle between q and r (given that q is not parallel to r ) is

Step-by-step solution

p q = 2( p r ) p ( q - 2 r ) = 0 . This implies that q - 2 r is collinear with p , so q - 2 r = p for some scalar . Squaring both sides gives: | q |^2 + 4| r |^2 - 4( q r ) = ^2 | p |^2 16 + 4(4) - 4( q r ) = ^2 32 - 4( q r ) = ^2 Taking the dot product of q - 2 r = p with r yields: q r - 2| r |^2 = ( p r ) q r - 8 = ( 1) Squaring this equation gives: ( q r - 8)^2 = ^2 Equating the two expressions for ^2 : ( q r - 8)^2 = 32 - 4( q r ) Let t = q r . Then: t^2 - 16t + 64 = 32 - 4t t^2 - 12t + 32 = 0 (t - 4)(t - 8) =