If \(\mathbf{a}\) and \(\mathbf{b}\) are vectors such that \(|\mathbf{a}+\mathbf{b}|\) \(=\mid \mathbf{a}-\mathbf{b}\), then the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is

We have, \(\quad|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|\)
On squaring both sides, we get
\(\quad|\mathbf{a}+\mathbf{b}|^{2}=|\mathbf{a}-\mathbf{b}|^{2}\) \(\Rightarrow|\mathbf{a}|^{2}+\left.\left|\mathbf{b}^{2}+2 \mathbf{a} \cdot \mathbf{b}=\right| \mathbf{a}\right|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}\) \(\Rightarrow \quad 4 \mathbf{a} \cdot \mathbf{b}=0\) \(\Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=0\) \(\Rightarrow \mathbf{a}\) and \(\mathbf{b}\) are perpendicular to each other. So, angle between them is \(90^{\circ} .\) Alternative \(\because \quad|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|\) \(\therefore \mathbf{a}\) and \(\mathbf{b}\) are perpendicular to each other So, angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(90^{\circ} .\)
So, angle between them is \(90^{\circ}\).
Alternative
\(\therefore\) a and \(\mathbf{b}\) are perpendicular to each other So, angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(90^{\circ}\).