3.1 vectors — Physics STD 11 Science — Question

$|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}$

$=A+B+2 A B \cos \theta$ And The formula for $|\vec{A}-\vec{B}|^{2}$ is,

$|\vec{A}-\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}-2 \vec{A} \cdot \vec{B}$

$=A+B-2 A B \cos \theta$

It is given that,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}$

$A+B+2 A B \cos \theta=A+B-2 A B \cos \theta$

$4 A B \cos \theta=0$

$\cos \theta=0$

$\theta=90^{\circ}$