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 $cos^{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
$\begin{vmatrix} \alpha & 3 & -1 \\ 1 & \alpha -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}=0$
$\Rightarrow \alpha \left(2 \alpha - 2 - 10\right)-3\left(2 - 6\right)-1\left(5 - 3 \alpha + 3\right)=0$
$\Rightarrow 2\alpha ^{2}-12\alpha +12-8+3\alpha =0$
$\Rightarrow 2\alpha ^{2}-9\alpha +4=0$
$\Rightarrow \alpha =4,\frac{1}{2}$
A normal vector to the plane is $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}$
$=\hat{i}\left(2 \alpha - 12\right)-\hat{j}\left(- 4\right)+\hat{k}\left(8 - 3 \alpha \right)$
If $\alpha =4$ , then a normal vector to one of the plane is $\overset{ \rightarrow }{n_{1}}=-4\hat{i}+4\hat{j}-4\hat{k}=-4\left(\hat{i} - \hat{j} + \hat{k}\right)$
If $\alpha =\frac{1}{2}$ , then a normal vector to the other plane is $\overset{ \rightarrow }{n_{2}}=-11\hat{i}+4\hat{j}+\frac{13}{2}\hat{k}$
$cos \theta =\frac{\overset{ \rightarrow }{n_{1}} \cdot \overset{ \rightarrow }{n_{2}}}{\left|\overset{ \rightarrow }{n_{1}}\right| \left|\overset{ \rightarrow }{n_{2}}\right|}=\frac{- 11 - 4 + \frac{13}{2}}{\sqrt{3} \sqrt{121 + 16 + \frac{169}{4}}}$
$\Rightarrow cos \theta = \frac{- 17}{\sqrt{3} \sqrt{717}}$
$\Rightarrow cos^{2} \theta = \frac{289}{2151}$