In $\mathbb{R}^{3}$, 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 $M$?

In R³, 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 P₁: x + 2y - z + 1 = 0 and P₂: 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 P₁. Which of the following points lie(s) on M?

$\text{Given: } L \text{ passes through the origin}$

$P_1: x+2y-z+1=0 \quad P_2: 2x-y+z-1=0$

$\text{every point of } L \text{ is equidistant from } P_1, P_2$

$M = \text{locus of feet of } \perp \text{ from points of } L \text{ onto } P_1$

$\text{Which of (1),(2),(3),(4) lie on } M?$

$P_1: \vec n_1=(1,2,-1),\quad |\vec n_1|=\sqrt{6}$

$P_2: \vec n_2=(2,-1,1),\quad |\vec n_2|=\sqrt{6}$

$C(0,0,1) \text{ lies on both } P_1 \text{ and } P_2$

$L: \vec r = t\,\vec d, \quad \text{through } O(0,0,0)$

$\text{Plan: (i) constant distance } \Rightarrow \vec d \perp \vec n_1,\ \vec d \perp \vec n_2$

$\text{(ii) } \vec d = \vec n_1 \times \vec n_2$

$\text{(iii) foot formula } \Rightarrow M, \text{ then test options}$

$\text{dist}(t\vec d, P_1) = \dfrac{|(\vec n_1 \cdot \vec d)\,t + 1|}{\sqrt{6}}$

$\text{constant in } t \ \Rightarrow\ \vec n_1 \cdot \vec d = 0,\ \ \vec n_2 \cdot \vec d = 0$

$\vec d = \vec n_1 \times \vec n_2 = (1,-3,-5)$

$L: (x,y,z) = (t,\,-3t,\,-5t)$

$F = A - \dfrac{\vec n_1 \cdot A + 1}{|\vec n_1|^2}\,\vec n_1, \quad A \in L$

$|\vec n_1|^2 = 6$

$\vec n_1 \cdot A = t(\vec n_1 \cdot \vec d) = 0$

$F = A - \tfrac{1}{6}\,\vec n_1$

$F(t) = \left(t-\tfrac{1}{6},\ -3t-\tfrac{1}{3},\ -5t+\tfrac{1}{6}\right)$

$M \parallel L, \quad M \subset P_1$

$t = x+\tfrac{1}{6} \ \Rightarrow\ y=-3x-\tfrac{5}{6},\quad z=-5x-\tfrac{2}{3}$

$(1)\ x=0:\ y=-\tfrac{5}{6},\ z=-\tfrac{2}{3}\ \checkmark$

$(2)\ x=-\tfrac{1}{6}:\ y=-\tfrac{1}{3},\ z=\tfrac{1}{6}\ \checkmark$

$(3)\ x=-\tfrac{5}{6}:\ y=\tfrac{5}{3}\neq 0\ \ \text{✗}$

$(4)\ x=-\tfrac{1}{3}:\ y=\tfrac{1}{6}\neq 0\ \ \text{✗}$

$\text{Answer: Options } (1) \text{ and } (2)$