Two vectors → A and → B have equal magnitudes. The magnitude of ( →...

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

$\sqrt{{A}^{2} + {B}^{2} + 2ABcos\theta} = n\left(\sqrt{{A}^{2} + {B}^{2} - 2ABcos\theta}\right)$;where $\theta$ is the angle between $\overset{\to}{A}$ and $\overset{\to}{B}$.

Also, \left|\overset{\to}{A}\right| = \left|\overset{\to}{B}\right|$,</p><p>$\sqrt{2{A}^{2} + 2{A}^{2}cos\theta} = n\sqrt{2{A}^{2} - 2{A}^{2}cos\theta}$</p><p>Squaring both sides:<br /><br />$2{A}^{2}\left(1 + cos\theta\right) = {n}^{2}2{A}^{2}\left(1 - cos\theta\right)$</p><p>$cos\theta = \dfrac{{n}^{2} - 1}{{n}^{2} + 1}$</p><p>i.e.,$\theta = \cos^{- 1}\left[\dfrac{{n}^{2} - 1}{{n}^{2} + 1}\right]