Q 11-03-136JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of $(\vec{X} - \vec{Y})$ is $n$ times the magnitude of $(\vec{X} + \vec{Y})$. The angle between $\vec{X}$ and $\vec{Y}$ is:
Answer: (B) $\cos^{-1}\left(\frac{n^2-1}{-n^2-1}\right)$
Let $|\vec{X}| = |\vec{Y}| = a$ and the angle be $\theta$.
$|\vec{X} - \vec{Y}|^2 = 2a^2(1 - \cos\theta)$ and $|\vec{X} + \vec{Y}|^2 = 2a^2(1 + \cos\theta)$.
$$1 - \cos\theta = n^2(1 + \cos\theta) \Rightarrow \cos\theta = \frac{1 - n^2}{1 + n^2} = \frac{n^2 - 1}{-n^2 - 1}$$
Solution by Sreeraj P, M.Sc Physics