Question

Two adjacent sides of a parallelogram ABCD are given by  $\overrightarrow{AB}=2\hat{i}+10\hat{j}+11\hat{k} and \overrightarrow{AD}=-\hat{i}+2\hat{j}+2\hat{k}$.The side AD is rotated by an acute angel $\alpha$ in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by;

• (A) $\frac{8}{9}$
• (B) $\frac{\sqrt{17}}{9}$
• (C) $\frac{1}{9}$
• (D) $\frac{4\sqrt{5}}{9}$

Answer 1

The Correct Option is B

Explanation:
Given: Two adjacent sides of a parallelogram ABCD are$\overrightarrow{AB}=2\hat{i}+10\hat{j}+11\hat{k} and \overrightarrow{AD}=-\hat{i}+2\hat{j}+2\hat{k}$       .....(i)We have to find cosine of the acute angel α through which side AD of parallelogram is rotated to AD' such that AD'  becomes perpendicular to AB. 

Let θ be the angle between $\overrightarrow{AB},\overrightarrow{AD}$ and AD' is perpendicular to AB.Then, α + θ = π$\frac{\mathrm{\pi }}{2}$[Clear from the above diagram]   ⇒ α = $\frac{\mathrm{\pi }}{2}-\theta$Using the definition of dot product , we have$\cos \theta =\left|\frac{\overrightarrow{AB}\cdot \overrightarrow{AD}}{|\overrightarrow{AB}||\overrightarrow{AD}|}\right|$$=\mid \frac{(2\hat{i}+10\hat{j}+11\hat{k})\cdot (-\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{(2{)}^2+(10{)}^2+(11{)}^2}\sqrt{(-1{)}^2+(2{)}^2+(2{)}^2}}$$[\text{ using (i) }]$$=|\frac{-2+20+22}{\sqrt{225}\sqrt{9}\mid }=|\frac{40}{{15}^{\ast }3}\mid$$=\left|\frac{8}{9}\right|=\frac{8}{9}\dots$ (iii)Using trigonometric identities, we have$\sin \theta =\sqrt{1-{\cos }^2\theta }$$=\sqrt{1-\frac{64}{81}}=\sqrt{\frac{17}{81}}[\text{ using }(\mathrm{iii})]$$=\frac{\sqrt{17}}{9}$Taking cosine on both sides of (ii), we get$\cos a=\cos (\frac{\pi }{2}-\theta )=\sin \theta =\frac{\sqrt{17}}{9}$Hence, the correct option is (B).