The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle b

Question

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\

Professor PC Thomas Classes · Answer

(b) : Let $	heta$ be angle between $\vec{A}$ and $\vec{B}$\ $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$, then $|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}$ 
$ A^{2}+B^{2}+2 A B \cos 	heta=A^{2}+B^{2}-2 A B \cos 	heta $
 or $4 A B \cos 	heta=0$ or $\cos 	heta=0$ or $	heta=90^{\circ}$\