Physics - Vectors Question with Solution | TestHub

Given: $\left|\overrightarrow{A}+\overrightarrow{B}\right|=n\left|\overrightarrow{A}-\overrightarrow{B}\right|$, so, the magnitudes of this will be given as,

$\sqrt{A^2+B^2+2AB\mathrm{c}\mathrm{o}\mathrm{s}\theta }=n\left(\sqrt{A^2+B^2-2AB\mathrm{c}\mathrm{o}\mathrm{s}\theta }\right)$;where $\theta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

Also, $\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|$,

$\sqrt{2A^2+2A^2\mathrm{c}\mathrm{o}\mathrm{s}\theta }=n\sqrt{2A^2-2A^2\mathrm{c}\mathrm{o}\mathrm{s}\theta }$

Squaring both sides:

$2A^2\left(1+\mathrm{c}\mathrm{o}\mathrm{s}\theta \right)=n^{2}2A^2\left(1-\mathrm{c}\mathrm{o}\mathrm{s}\theta \right)$

$\mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{n^2-1}{n^2+1}$

i.e., $\theta ={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}\left[\frac{n^2-1}{n^2+1}\right]$