Mathematics - Vector Question with Solution | TestHub

$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\mid$ $\Rightarrow \sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{AB}\cos \theta }=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{AB}\cos \theta }$ Squaring both the sides, we get ${\overrightarrow{\mathrm{A}}}^2+{\overrightarrow{\mathrm{B}}}^2+2\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}}\cos \theta ={\overrightarrow{\mathrm{A}}}^2+{\overrightarrow{\mathrm{B}}}^2-2\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}}\cos \theta$ or $4\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}}\cos \theta =0$ or $\cos \theta =0$ (since the scalar or dot product is zero). Therefore angle between $\overrightarrow{\mathrm{A}}$ to $\overrightarrow{\mathrm{B}}$ is $90^{\circ }$