Question

In $R^3$, let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_1:x+2y-z+1-=0$ and $P_2:2x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_1$. Which of the following points lie(s) on $M$?

• A. $(0,-\frac{5}{6},-\frac{2}{3})$
• B. $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$
• C. $(-\frac{5}{6},0,\frac{1}{6})$
• D. $(-\frac{1}{3},0,\frac{2}{3})$

Answer 1

Answer: (A), (B)

The correct options are
A $(0,-\frac{5}{6},-\frac{2}{3})$
B $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$

Line $L$ will be parallel to the line of intersection of $P_1$ and $P_2$

Let $a,b\text{and}c$ be the direction ratios of line $L$

$⟹a+2b–c=0$ and $2a–b+c=0$
$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 2& -1& 1\end{array}\right|=\hat{i}-3\hat{j}-5\hat{k}$

$⟹a:b:c::1:–3:–5$

Hence, the Equation of line L is $\frac{x-0}{1}=\frac{y-0}{-3}=\frac{z-0}{-5}$ as it passes through the origin.

The foot of perpendicular from origin to plane $P_1$ is $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$

$\therefore$ Equation of projection of line $L$ on plane $P_1$ $\left(M\right)$ is

$\frac{x+\frac{1}{6}}{1}=\frac{y+\frac{1}{3}}{-3}=\frac{z-\frac{1}{6}}{-5}=c$

From the options given, only $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$

and $(0,-\frac{5}{6},-\frac{2}{3})$ satisfy the the equation of line of projection $M$