The magnitude of scalar and vector products of 2 vectors are 48√3 and 144 respectively . what is the angle between the 2 vectors?

Let vectors be A and B.Given,A*B=48√3 A.B=144 A.B=|A||B|cos©[where © is angle]Therefore,cos©=A.B/|A||B| cos©=144/48√3 cos©=3/√3As value of cos© cannot be more than 1 therefore the question is wrong

Given :- |A X B | = 144A.B = 48√3Sol :- | A X B | = |A| |B| sin x = 144. ...............(1)Also A.B = |A| |B| cos x =48√3 ...............(2)Dividing eqn 1 by 2tanx = 144 / 48 √ 3 = 3 / √3 = √3→ x = tan·¹ ( √3) x = 60°