Question

Two adjacent sides of a parallelogram ABCD are given by $\stackrel{⟶}{AB}=2\hat{i}+10\hat{j}+11\hat{k}$ and $\stackrel{⟶}{AD}=-\hat{i}+2\hat{j}+2\hat{k}.$
The side AD is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes 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 $\frac{\sqrt{17}}{9}$

$\stackrel{⟶}{AB}=2\hat{i}+10\hat{j}+11\hat{k}\stackrel{⟶}{AD}=-\hat{i}+2\hat{j}+2\hat{k}$
Angle ${}^{\prime }{\theta }^{\prime }$ between $\stackrel{⟶}{AB}$ and $\stackrel{⟶}{AD}$ is
$cos\left(\theta \right)=\left|\frac{\stackrel{⟶}{AB}.\stackrel{⟶}{AD}}{|\stackrel{⟶}{AB}||\stackrel{⟶}{AD}|}\right|=\left|\frac{-2+20+22}{\left(15\right)\left(3\right)}\right|=\frac{8}{9}\Rightarrow sin\left(\theta \right)=\frac{\sqrt{17}}{9}Since,\alpha +\theta ={90}^{\circ }\therefore cos\left(\alpha \right)=cos({90}^{\circ }-\theta )=sin\left(\theta \right)=\frac{\sqrt{17}}{9}$