Question

• Show that the non-zero vectors A and B are perpendicular to each otger if a vector+b vector=a vector minus b vector​

Answer 1

Given :

➠ $\sf\:|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$

To Prove :

⟶ Both vectors (A and B) are perpendicular to each other.

Formula :

$\dag\bf\:|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2AB\cos\theta}$

$\dag\bf\:|\vec{A}-\vec{B}|=\sqrt{A^2+B^2-2AB\cos\theta}$

Explanation :

$:\implies\sf\:|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$

$:\implies\sf\:\sqrt{A^2+B^2+2AB\cos\theta}=\sqrt{A^2+B^2-2AB\cos\theta}$

$:\implies\sf\:2AB\cos\theta=-2AB\cos\theta$

$:\implies\sf\:4AB\cos\theta=0$

$:\implies\sf\:\cos\theta=0$

$:\implies\sf\:\theta=\cos^{-1}(0)$

$:\implies\underline{\boxed{\bf{\theta=90\degree}}}$

Hence Proved !!