Physics - Vectors Question with Solution | TestHub

Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively. Then magnitude of their sum is given by: $|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2AB\cos \theta }\to (1)(\theta =$ angle between the vectors $)$ Magnitude of their difference is given by: $|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{AB}\cos \theta }\to (2)$ As $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$ $\Rightarrow {\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{AB}\cos \theta ={\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{AB}\cos \theta$ $\Rightarrow 4\mathrm{AB}\cos \theta =0$ or $\cos \theta =0$ $\Rightarrow \theta =90^{\circ }$.