Explanation

Given, $A = \hat{i} + \hat{j}$, $B = \hat{i} - \hat{j}$ As we know that $\vec{A} \cdot \vec{B} = |A||B|\cos \theta$ $(\hat{i} + \hat{j}) \cdot (\hat{i} - \hat{j}) = (\sqrt{1^{2} + 1^{2}})(\sqrt{1^{2} + 1^{2}})\cos \theta$ $(i + j)(i - j) = \sqrt{2} \times \sqrt{2} \cos \theta$ where $\theta$ is the angle between A and B $\cos \theta = (1 - 0 + 0 - 1)/(\sqrt{2}\sqrt{2}) = 0 \therefore \theta = 90^{\circ}$