The vectors vec{A} and vec{B} are such that
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Question

The formula for $|\vec{A}+\vec{B}|^{2}$ is,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}$

$=A+B+2 A B \cos 	heta$

Vedclass · Accepted Answer

$90$

Vedclass · Answer

The formula for $|\vec{A}+\vec{B}|^{2}$ is,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}$

$=A+B+2 A B \cos 	heta$ And The formula for $|\vec{A}-\vec{B}|^{2}$ is,

$|\vec{A}-\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}-2 \vec{A} \cdot \vec{B}$

$=A+B-2 A B \cos 	heta$

It is given that,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}$

$A+B+2 A B \cos 	heta=A+B-2 A B \cos 	heta$

$4 A B \cos 	heta=0$

$\cos 	heta=0$

$	heta=90^{\circ}$

The vectors $\vec{A}$ and $\vec{B}$ are such that

$|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$

The angle between the two vectors is

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