Physics - Vectors Question with Solution | TestHub

Given, $|A+B|=|A-B|$ $\therefore (A+B{)}^2=(A-B{)}^2$ $A^2+B^2+2AB\cos \theta =A^2+B^2-2AB\cos \theta$ $2AB\cos \theta +2$ AB $\cos \theta =0$ $4AB\cos \theta =0$ Here, $A\ne 0,B\ne 0$ $\therefore \cos \theta =0$ $\theta =90^{\circ }$