Q 11-03-132JEE MainJEE Main 2021 (20 Jul, Shift 2)Medium
Two vectors $\vec P$ and $\vec Q$ have equal magnitudes. If the magnitude of $\vec P + \vec Q$ is $n$ times the magnitude of $\vec P - \vec Q$, then angle between $\vec P$ and $\vec Q$ is
Answer: (D) $\cos^{-1}\left(\frac{n^2-1}{n^2+1}\right)$
With $|\vec P| = |\vec Q| = P$: $|\vec P + \vec Q|^2 = 2P^2(1 + \cos\theta)$ and $|\vec P - \vec Q|^2 = 2P^2(1 - \cos\theta)$.
$$\frac{1 + \cos\theta}{1 - \cos\theta} = n^2 \Rightarrow \cos\theta = \frac{n^2 - 1}{n^2 + 1}$$
Solution by Sreeraj P, M.Sc Physics