Q 11-03-154JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $\left(\vec A + \vec B\right)$ is $n$ times the magnitude of $\left(\vec A - \vec B\right)$. The angle between $\vec A$ and $\vec B$ is:
Answer: (A) $\cos^{-1}\left[\dfrac{n^2-1}{n^2+1}\right]$
With $|\vec A| = |\vec B| = a$ and angle $\theta$:
$$|\vec A+\vec B|^2 = 2a^2(1+\cos\theta),\qquad |\vec A-\vec B|^2 = 2a^2(1-\cos\theta)$$
$$\frac{1+\cos\theta}{1-\cos\theta} = n^2 \;\Rightarrow\; \cos\theta = \frac{n^2-1}{n^2+1}$$
Solution by Sreeraj P, M.Sc Physics