Question

Let $P_1$ denote the equation of a plane to which the vector $(\hat{i}+\hat{j})$ is normal and which contains the line $L$, whose equation is $\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}-\hat{k})$ , and $P_2$ denote the equation of the plane containing the line $L$ and a point with position vector $\hat{j}$. Which of the following holds good?

• 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 $P_1$ is ${\tan }^{-1}\sqrt{3}$.

Answer 1

Answer: (A), (C)

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

$(\hat{i}+\hat{j})$ is normal to $P_1$. Hence, the equation of $P_1$ can be written as $x+y=k$
Since, $P_1$ contains the line with the vector equation
$\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}-\hat{k})$, the point $(1,1,1)$ lies on $P_1$.
Hence,
$k=2$
Hence, $P_1:x+y=2$.
$P_2$ contains the line L and also the point $(0,1,0)$. The vector along the line L is $\hat{i}-\hat{j}-\hat{k}$
A vector along the plane through the point $(0,1,0)$ is $\hat{i}+\hat{k}$
Hence, the vector along the normal to the plane will be given by -
$\overline{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 1\\ 1& -1& -1\end{array}\right|$
$\overline{n_2}=\hat{i}+2\hat{j}-\hat{k}$
Hence, $P_2$ will have the equation $x+2y-z=k$
$(0,1,0)$ lies on the plane.
Hence, $P_2:x+2y-z=2$
The vector equation of the plane will be $\overline{r}.(\hat{i}+2\hat{j}-\hat{k})=2$
Let $\theta$ be the angle between $P_1$ and $P_2$.
$\cos \theta =\frac{1\times 1+2\times 1}{\sqrt{12}}=\frac{\sqrt{3}}{2}$
$\cot \theta =\sqrt{3}$
Hence, options A and C are correct.