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Two vectors $\overrightarrow{{X}}$ and $\overrightarrow{{Y}}$ have equal magnitude. The magnitude of $(\overrightarrow{{X}}-\overrightarrow{{Y}})$ is ${n}$ times the magnitude of $(\overrightarrow{{X}}+\overrightarrow{{Y}})$. The angle between $\overrightarrow{{X}}$ and $\overrightarrow{{Y}}$ is -
$\cos ^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)$
$\cos ^{-1}\left(\frac{{n}^{2}-1}{-{n}^{2}-1}\right)$
$\cos ^{-1}\left(\frac{-n^{2}-1}{n^{2}-1}\right)$
$\cos ^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)$
Solution
Given $X=Y$
$\sqrt{ X ^{2}+ Y ^{2}-2 \times Y \cos \theta}$
$= n \sqrt{ X ^{2}+ Y ^{2}+2 \times Y \cos \theta}$
Square both sides
$2 X ^{2}(1-\cos \theta)= n ^{2} \cdot 2 X ^{2}(1+\cos \theta)$
$1-\cos \theta= n ^{2}+ n ^{2} \cos \theta$
$\cos \theta=\frac{1- n ^{2}}{1+ n ^{2}}$
$\theta=\cos ^{-1}\left[\frac{ n ^{2}-1}{- n ^{2}-1}\right]$
