Question

In $R^3$, let $P_1$ be the plane which contains the line $L_1:\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}-\hat{k})$ and $P_2$ be the another plane which contains the line $L_1$ and a point with position vector $\hat{j}$. If the vector $(\hat{i}+\hat{j})$ is normal to $P_1$, then which of the following is (are) true?

• A. The equation of $P_1$ is $x+y=2$
• B. The equation of $P_2$ is $\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+\hat{k})=2$
• C. The acute angle between $P_1$ and $P_2$ is ${\cot }^{-1}\left(\sqrt{3}\right)$
• D. The angle between the plane $P_2$ and the line $L_1$ is ${\tan }^{-1}\left(\sqrt{3}\right)$

Answer 1

Answer: (A), (C)

The correct options are
A The equation of $P_1$ is $x+y=2$
C The acute angle between $P_1$ and $P_2$ is ${\cot }^{-1}\left(\sqrt{3}\right)$

Plane $P_1$ contains the line $\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}-\hat{k})$, hence contains the point $\hat{i}+\hat{j}+\hat{k}$ and is normal to vector $(\hat{i}+\hat{j}).$

Hence, equation of plane $P_1$ is,
$(\overrightarrow{r}-(\hat{i}+\hat{j}+\hat{k}\left)\right)\cdot (\hat{i}+\hat{j})=0$
$\Rightarrow x+y=2$

Plane $P_2$ contains the line $\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}-\hat{k})$ and point $\hat{j}.$
Hence, equation of plane $P_2$ is,
$\left|\begin{array}{ccc}x-0& y-1& z-0\\ 1-0& 1-1& 1-0\\ 1& -1& -1\end{array}\right|=0$

$\Rightarrow x+2y-z=2$

If $\theta$ is the acute angle between $P_1$ and $P_2$, then
$\cos \theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}=\left|\frac{(\hat{i}+\hat{j})\cdot (\hat{i}+2\hat{j}-\hat{k})}{\sqrt{2}\cdot \sqrt{6}}\right|$
$=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)={\cot }^{-1}\left(\sqrt{3}\right)$

As $L_1$ is contained in $P_2\Rightarrow \theta =0$