If |$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$| = |$\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$|, then the angle between $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$ will be

Let $\mathrm{\theta }$ be the angle between the vectors $\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}$.

Then 

|$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$| = $\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta }}$

|$\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$| = $\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{ABcos\theta }}$

|$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$| = |$\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$| 

$\therefore$ $\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta }}$ = $\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{ABcos\theta }}$

Squaring on both sides, we get

${\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta } = {\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{ABcos\theta }$

4AB$\mathrm{cos\theta }$ = 0

$\therefore$$\mathrm{cos\theta }$ = 0 or $\mathrm{\theta } = \frac{\mathrm{\pi }}{2}$