Two adjacent sides of a parallelogram $\mathrm{ABCD}$ are given by $\overline{\mathrm{AB}}=2…
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
$\frac{8}{9}$
$\frac{\sqrt{17}}{9}$
$\frac{1}{9}$
$\frac{4 \sqrt{5}}{9}$
Solution
Let $\theta$ be the angle between $\overline{\mathrm{AB}}$ and $\overline{\mathrm{AD}}$
$\begin{aligned}
\therefore \quad \cos \theta & =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{AD}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{AD}}|} \\
& =\frac{(2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\sqrt{4+100+121} \sqrt{1+4+4}} \\
& =\frac{-2+20+22}{\sqrt{225} \sqrt{9}} \\
& =\frac{40}{45} \\
& =\frac{8}{9} \\
\therefore \quad \sin \theta & =\sqrt{1-\left(\frac{8}{9}\right)^2}=\frac{\sqrt{17}}{9}
\end{aligned}$
$\alpha$ is the angle of rotation of $A D$.
$\therefore \quad$ The angle between side $\mathrm{AB}$ and $\mathrm{AD}$
$\begin{aligned}
& =\alpha+\theta \\
& =90^{\circ}
\end{aligned}$
$\begin{aligned}
& \therefore \quad \cos (\alpha+\theta)=\cos \left(90^{\circ}\right) \\
& \therefore \quad \cos \alpha \cos \theta-\sin \alpha \sin \theta=0 \\
& \therefore \quad 8 \cos \alpha=\sqrt{17} \sin \alpha \\
& \therefore \quad 64 \cos ^2 \alpha=17\left(1-\cos ^2 \alpha\right) \\
& \therefore \quad 81 \cos ^2 \alpha=17 \\
& \therefore \quad \cos \alpha=\frac{\sqrt{17}}{9}
\end{aligned}$