Question

The sum and difference of two vectors are equal in magnitude:
$|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$. Prove that $\overrightarrow{A}\perp \overrightarrow{B}.$

Answer 1

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

squaring both sides-

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

$⟹A^2+B^2+2AB\cos \theta =A^2+B^2-2AB\cos \theta$

$⟹4AB\cos \theta =0⟹\cos \theta =0$

Hence, $\theta ={90}^o$

Hence, $\overrightarrow{A}\perp \overrightarrow{B}$