Question

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:

• A. ${\cos }^{-1}\left(\frac{n^2+1}{n^2-1}\right)$
• B. ${\cos }^{-1}\left(\frac{n^2+1}{-n^2-1}\right)$
• C. ${\cos }^{-1}\left(\frac{n^2-1}{-n^2-1}\right)$
• D. ${\cos }^{-1}\left(\frac{-n^2-1}{n^2-1}\right)$

Answer 1

The correct option is C ${\cos }^{-1}\left(\frac{n^2-1}{-n^2-1}\right)$

Given that, $X=Y$

Also, $|\overrightarrow{X}-\overrightarrow{Y}|=n|\overrightarrow{X}+\overrightarrow{Y}|$

$\Rightarrow \sqrt{X^2+Y^2-2XY\cos \theta }=n\sqrt{X^2+Y^2+2\times Y\cos \theta }$

Squaring both sides,

$2X^2(1-\cos \theta )=n^{2}2X^2(1+\cos \theta )$

$\Rightarrow 1-\cos \theta =n^2+n^2\cos \theta$

$\Rightarrow \cos \theta =\frac{1-n^2}{1+n^2}$

$\Rightarrow \theta ={\cos }^{-1}\left[\frac{n^2-1}{-n^2-1}\right]$

Hence, option $\left(B\right)$ is correct.