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: (B)

$\left( -\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$

Answer 2

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