$A = \sqrt{2} \cos{\alpha} \hat{i} - \sqrt{2} \sin{\alpha} \hat{j}$,  $B = - \sqrt{2} \cos{\alpha} \hat{j} + \sqrt{2} \sin{\alpha} \hat{k}$   হলে A ও B এর অন্তর্ভুক্ত কোণ কোনটি?

$\begin{aligned} \mathbf{A . B} & =(\sqrt{2} \cos \alpha \hat{\mathrm{i}}-\sqrt{2} \sin \alpha \hat{\mathrm{j}}) \cdot(-\sqrt{2} \cos \alpha \hat{\mathrm{j}}+\sqrt{2} \sin \alpha \hat{k}) \\ & =0+2 \sin \cos \alpha+0 \\ & =2 \sin \alpha \cos \alpha \\ |\mathbf{A}| & =\sqrt{2 \cos ^{2} \alpha+2 \sin ^{2} \alpha}=\sqrt{2} \\ |\mathbf{B}| & =\sqrt{2 \cos ^{2} \alpha+2 \sin ^{2} \alpha}=\sqrt{2}\end{aligned}$

$\begin{aligned} \therefore \text { A ও B এর অন্তর্ভুক্ত কোণ } & =\cos ^{-1}\left(\frac{2 \sin \alpha \cos \alpha}{\sqrt{2} \cdot \sqrt{2}}\right) \\ & =\cos ^{-1}(\sin \alpha \cdot \cos \alpha)\end{aligned}$