ভেক্টর দুটির মধ্যবর্তী কোণ $\theta$ হলে, আমরা জানি, $$ \begin{aligned} \cos \theta &= \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \end{aligned} $$ এখানে, $$ \begin{aligned} \vec{A} \cdot \vec{B} &= (\sqrt{2} \cos \alpha \hat{i} - \sqrt{2} \sin \alpha \hat{j}) \cdot (-\sqrt{2} \cos \alpha \hat{j} + \sqrt{2} \sin \alpha \hat{k}) \\ &= 0 + 2 \sin \alpha \cos \alpha + 0 = 2 \sin \alpha \cos \alpha \end{aligned} $$ ভেক্টর দুটির মান, $$ \begin{aligned} |\vec{A}| &= \sqrt{(\sqrt{2} \cos \alpha)^2 + (-\sqrt{2} \sin \alpha)^2} = \sqrt{2} \\ |\vec{B}| &= \sqrt{(-\sqrt{2} \cos \alpha)^2 + (\sqrt{2} \sin \alpha)^2} = \sqrt{2} \end{aligned} $$ অতএব, $$ \begin{aligned} \cos \theta &= \frac{2 \sin \alpha \cos \alpha}{\sqrt{2} \cdot \sqrt{2}} = \sin \alpha \cos \alpha \\ \therefore \theta &= \cos^{-1}(\sin \alpha \cos \alpha) \end{aligned} $$