Question

Let $2$ planes are being contained by the vectors $\alpha \hat{i}+3\hat{j}-\hat{k},\hat{i}+\left(\alpha -1\right)\hat{j}+2\hat{k}$ and $3\hat{i}+5\hat{j}+2\hat{k}.$ If the angle between these $2$ planes is $\theta$ , then the value of $co{s}^2\theta$ is equal to

• A. $\frac{15}{17}$
• B. $\frac{289}{717}$
• C. $\frac{289}{2151}$
• D. $\frac{17}{2151}$

Answer 1

Answer: (C)

All given vectors are coplaner
$\left|\begin{array}{ccc}\alpha & 3& -1\\ 1& \alpha -1& 2\\ 3& 5& 2\end{array}\right|=0$
$\Rightarrow \alpha \left(2\alpha -2-10\right)-3\left(2-6\right)-1\left(5-3\alpha +3\right)=0$
$\Rightarrow 2{\alpha }^2-12\alpha +12-8+3\alpha =0$
$\Rightarrow 2{\alpha }^2-9\alpha +4=0$
$\Rightarrow \alpha =4,\frac{1}{2}$
A normal vector to the plane is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& \alpha -1& 2\\ 3& 5& 2\end{array}\right|$
$=\hat{i}\left(2\alpha -12\right)-\hat{j}\left(-4\right)+\hat{k}\left(8-3\alpha \right)$
If $\alpha =4$ , then a normal vector to one of the plane is $\overrightarrow{n_1}=-4\hat{i}+4\hat{j}-4\hat{k}=-4\left(\hat{i}-\hat{j}+\hat{k}\right)$
If $\alpha =\frac{1}{2}$ , then a normal vector to the other plane is $\overrightarrow{n_2}=-11\hat{i}+4\hat{j}+\frac{13}{2}\hat{k}$
$cos\theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}=\frac{-11-4+\frac{13}{2}}{\sqrt{3}\sqrt{121+16+\frac{169}{4}}}$
$\Rightarrow cos\theta =\frac{-17}{\sqrt{3}\sqrt{717}}$
$\Rightarrow co{s}^2\theta =\frac{289}{2151}$