Answer:
We
know that and \[|\vec{A}+\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos
\theta }\]
\[|\vec{A}-\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos
\theta }\]
As per questions \[\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos
\theta }=\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }\]
Squaring both the sides, we get \[\text{4 AB cos}\theta
=0\] or \[\text{cos}\theta =0\]or\[\theta =\text{9}0{}^\circ \]. It means, the
vectors \[\vec{A}\] and \[\vec{B}\]are perpendicular to each other.
You need to login to perform this action.
You will be redirected in
3 sec