Q 11-03-018NEETJEE MainMedium
Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. If $|\vec{A} + \vec{B}| = \sqrt{3}\,|\vec{A} - \vec{B}|$, the angle between them is
Answer: (C) $60°$
With $|\vec{A}| = |\vec{B}| = A$:
$$|\vec{A} + \vec{B}|^2 = 2A^2(1 + \cos\theta), \qquad |\vec{A} - \vec{B}|^2 = 2A^2(1 - \cos\theta)$$
$$1 + \cos\theta = 3(1 - \cos\theta) \;\Rightarrow\; \cos\theta = \frac{1}{2} \;\Rightarrow\; \theta = 60°$$
Solution by Sreeraj P, M.Sc Physics