If vec A and vec B are two non-zero vect | vedclass

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Question

$\sqrt{A^{2}+B^{2}+2 A B \cos 	heta}=\frac{1}{2} 	imes \sqrt{A^{2}+B^{2}-A B \cos 	heta}$

$4\left(A^{2}+B^{2}+2 A B \cos 	hetaight)=A^{2}+B^{

Vedclass · Accepted Answer

$cos^{-1}(-3/4)$

Vedclass · Answer

$\sqrt{A^{2}+B^{2}+2 A B \cos 	heta}=\frac{1}{2} 	imes \sqrt{A^{2}+B^{2}-A B \cos 	heta}$

$4\left(A^{2}+B^{2}+2 A B \cos 	hetaight)=A^{2}+B^{2}-2 A B \cos 	heta$

$3 A^{2}+3 B^{2}+10 A B \cos 	heta=0$

$12 \mathrm{B}^{2}+3 \mathrm{B}^{2}+20 \mathrm{B}^{2} \cos 	heta=0$

$20 \mathrm{B}^{2} \cos 	heta=-15 \mathrm{B}^{2}$

$\cos 	heta=-\frac{3}{4}$

$	heta=\cos ^{-1}\left(-\frac{3}{4}ight)$

If $\vec A$ and $\vec B$ are two non-zero vectors such that $\left| {\vec A + \vec B} \right| = \frac{{\left| {\vec A - \vec B} \right|}}{2}$ and $\left| {\vec A} \right| = 2\left| {\vec B} \right|$ then the angle between $\vec A$ and $\vec B$ is

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