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

Answer 1

Answer: (A), (B)

$P_1:x+2y-z+1=0$
$\&P_2:2x-y+z-1=0$
Direction Ratios of common line $\left(1,-3,-5\right)\Rightarrow \hat{i}-3\hat{j}-5\hat{k}$
$L:\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=t$
Let M$\left(\alpha ,\beta ,\gamma \right)$ is feet of perpendicular from $\left(t,-3t,-5t\right)$ on $P_1$
$\frac{\alpha -t}{1}=\frac{\beta +3t}{2}=\frac{\gamma +5t}{-1}=-\left(\frac{t-6t+5t+1}{6}\right)$
$\alpha =t-\frac{1}{6}\beta =-3t-\frac{1}{3}\gamma =-5t+\frac{1}{6}$
Only option $\left(A\right)\&\left(B\right)$ satisfies.