Answer: (1)

Solution

Let magnitude of two vectors $\vec{A}$ and $\vec{B}$ = $a$ $|\vec{A} + \vec{B}| = \sqrt{a^2 + a^2 + 2a^2\cos\theta}$ and $|\vec{A} - \vec{B}| = \sqrt{a^2 + a^2 - 2a^2\cos(180° - \theta)}$ $= \sqrt{a^2 + a^2 - 2a^2\cos\theta}$ and according to question, $|\vec{A} + \vec{B}| = n|\vec{A} - \vec{B}|$ or, $\frac{a^2 + a^2 + 2a^2\cos\theta}{a^2 + a^2 - 2a^2\cos\theta} = n^2$ $\Rightarrow \frac{\cancel{a^2}(1 + 1 + 2\cos\theta)}{\cancel{a^2}(1 + 1 - 2\cos\theta)} = n^2$ $\frac{(1 + \cos\theta)}{(1 - \cos\theta)} = n^2$ using componendo and dividendo theorem, we get $\theta = \cos^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$