Question

Let $P(1,3,5)$ and $Q(-2,1,4)$ be two points from which perpendiculars $PM$ and $QN$ are drawn to the $x-z$ plane. Find the angle that the line MN makes with the plane $x+y+z=5$.

Answer 1

Any plane can be written in the form $\overrightarrow{r}\cdot \overrightarrow{n}=d$

where $\overrightarrow{n}$ is normal vector and $\overrightarrow{r}$ is any general point on plane and $d$ is a constant.

Hence, normal vector of $x-z=0$ is $\hat{i}-\hat{k}$.

$\therefore \overrightarrow{PM}=\hat{i}+3\hat{j}+5\hat{k}+\lambda (\hat{i}-\hat{k})=(1+\lambda )\hat{i}+3\hat{j}+(5-\lambda )\hat{k}$

Now, as $M$ lies on $x-z=0$,

$(1+\lambda )-(5-\lambda )=0⟹\lambda =2$

$⟹M=(3,3,3)$

Also, $\overrightarrow{QN}=-2\hat{i}+\hat{j}+4\hat{k}+\alpha (\hat{i}-\hat{k})=(\alpha -2)\hat{i}+\hat{j}+(4-\alpha )\hat{k}$

Now, as $N$ lies on $x-z=0$,

$(\alpha -2)-(4-\alpha )=0⟹\alpha =3$

$⟹N=(1,1,1)$

$⟹\overrightarrow{MN}=\hat{i}+\hat{j}+\hat{k}+\beta (2\hat{i}+2\hat{j}+2\hat{k})$

Let direction vector of $\overrightarrow{MN}$ be $\overrightarrow{b}$.

$\therefore \overrightarrow{b}=2\hat{i}+2\hat{j}+2\hat{k}$

Now, normal of plane $x+y+z=5$ is $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

Hence, angle between normal of plane $x+y+z=5$ and $\overrightarrow{MN}={\cos }^{-1}\left|\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}\right|$

$={\cos }^{-1}\left|\frac{2+2+2}{\sqrt{3}\times 2\sqrt{3}}\right|$

$={\cos }^{-1}\left|\frac{6}{6}\right|$

$={\cos }^{-1}1$

$=0^{\circ }$

Hence, as the angle between normal vector and $\overrightarrow{MN}$ is $0^{\circ }$, the angle between the plane $x+y+z=5$ and $\overrightarrow{MN}$ is ${90}^{\circ }$.