Two vectors overrightarrow{{X}} and overright | vedclass

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Question

Given $X=Y$

$\sqrt{ X ^{2}+ Y ^{2}-2 	imes Y \cos 	heta}$

$= n \sqrt{ X ^{2}+ Y ^{2}+2 	imes Y \cos 	heta}$

Square both sides

$2 X ^{2

Vedclass · Accepted Answer

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

Vedclass · Answer

Given $X=Y$

$\sqrt{ X ^{2}+ Y ^{2}-2 	imes Y \cos 	heta}$

$= n \sqrt{ X ^{2}+ Y ^{2}+2 	imes Y \cos 	heta}$

Square both sides

$2 X ^{2}(1-\cos 	heta)= n ^{2} \cdot 2 X ^{2}(1+\cos 	heta)$

$1-\cos 	heta= n ^{2}+ n ^{2} \cos 	heta$

$\cos 	heta=\frac{1- n ^{2}}{1+ n ^{2}}$

$	heta=\cos ^{-1}\left[\frac{ n ^{2}-1}{- n ^{2}-1}ight]$

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 -

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