The magnitude of the sum of the two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is equal to the magnitude of
the difference of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). The angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is

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