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 angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $A{D}^{\prime }$. If $A{D}^{\prime }$ 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

Answer: (B)

$\frac{\sqrt{17}}{9}$

$\overrightarrow{AB}\cdot \overrightarrow{AD}>0$
So, ${}^{\prime }{\theta }^{\prime }$ is acute
$cos\theta =\frac{\overrightarrow{AB}\cdot \overrightarrow{AD}}{\mid \overrightarrow{AB}\mid \mid \overrightarrow{AD}\mid }=\frac{8}{9}$
$cos(\alpha +\theta )=0\Rightarrow cos\alpha \cdot cos\theta -sin\alpha \cdot sin\theta =0$
$8cos\alpha =\sqrt{17}sin\alpha$
$64co{s}^2\alpha =17(1-co{s}^2\alpha )$
$co{s}^2\alpha =\frac{17}{81}$
$cos\alpha =\frac{\sqrt{17}}{9}$