Two adjacent sides of a parallelogram $A B C D$ are given by $\overrightarrow{A B}=2 \hat{i}+10 \hat

Question

Two adjacent sides of a parallelogram $A B C D$ are given by $\overrightarrow{A B}=2 \hat{i}+10 \hat{j}+11 \hat{k}$ and $\overrightarrow{A D}=-\hat{i}+2 \hat{j}+2 \hat{k}$. The side $A D$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $A D$ becomes $A D^{\prime}$. If $A D^{\prime}$ makes a right angle with the side $A B$, then the cosine of the angle $\alpha$ is given by\

Professor PC Thomas Classes · Answer

(b) : We have,\ $\cos \phi=\frac{-2+20+22}{\sqrt{2^{2}+10^{2}+11^{2}} \sqrt{1^{2}+2^{2}+2^{2}}}$\ $=\frac{40}{15 	imes 3}=\frac{8}{9}$\ $\Rightarrow \quad \cos \alpha=\sin \phi=\frac{\sqrt{17}}{9}$\