# Two vectors

Question:

Two vectors $\overrightarrow{\mathrm{X}}$ and $\overrightarrow{\mathrm{Y}}$ have equal magnitude. The magnitude of $(\vec{X}-\vec{Y})$ is $n$ times the magnitude of $(\overrightarrow{\mathrm{X}}+\overrightarrow{\mathrm{Y}})$. The angle between $\overrightarrow{\mathrm{X}}$ and $\overrightarrow{\mathrm{Y}}$ is :

1. $\cos ^{-1}\left(\frac{-n^{2}-1}{n^{2}-1}\right)$

2. $\cos ^{-1}\left(\frac{n^{2}-1}{-n^{2}-1}\right)$

3. $\cos ^{-1}\left(\frac{n^{2}+1}{-n^{2}-1}\right)$

4. $\cos ^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)$

Correct Option: , 2

Solution:

Given $\mathrm{X}=\mathrm{Y}$

$\sqrt{\mathrm{X}^{2}+\mathrm{Y}^{2}-2 \times \mathrm{Y} \cos \theta}$

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

Square both sides

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

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

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

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