Two adjacent sides of a parallelogram \(\mathrm{ABCD}\) are given by \(\overline{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}\) and \(\overline{\mathrm{AD}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\). The side \(\mathrm{AD}\) is potated by an acute angle \(\alpha\) in the plane of parallelogram so that \(\mathrm{AD}\) becomes \(\mathrm{AD}^{\prime}\). If \(\mathrm{AD}^{\prime}\) makes a right angle with side \(\mathrm{AB}\), then the cosine of the angle \(\alpha\) is given by

Let \(\theta\) be the angle between \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AD}}\)
\(\therefore \cos \theta =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{AD}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{AD}}|} \)
\( =\frac{(2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\sqrt{4+100+121} \sqrt{1+4+4}} \)
\(=\frac{-2+20+22}{\sqrt{225} \sqrt{9}} \)
\( =\frac{40}{45} \)
\( =\frac{8}{9} \)
\( \therefore \sin \theta =\sqrt{1-\left(\frac{8}{9}\right)^2}=\frac{\sqrt{17}}{9} \)
\(\alpha\) is the angle of rotation of \(A D\).
\(\therefore\) The angle between side \(\mathrm{AB}\) and \(\mathrm{AD}\)
\( =\alpha+\theta\)
\(=90^{\circ}\)
\(\therefore \cos (\alpha+\theta)=\cos \left(90^{\circ}\right) \)
\( \therefore \cos \alpha \cos \theta-\sin \alpha \sin \theta=0 \)
\( \therefore 8 \cos \alpha=\sqrt{17} \sin \alpha \)
\( \therefore 64 \cos ^2 \alpha=17\left(1-\cos ^2 \alpha\right) \)
\( \therefore 81 \cos ^2 \alpha=17 \)
\( \therefore \cos \alpha=\frac{\sqrt{17}}{9}\)