Let the line $L$ be the projection of the line
\[ \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2} \]
in the plane $x - 2y - z = 3$. If $d$ is the distance of the point $(0, 0, 6)$ from $L$, then $d^2$ is equal to _________.
Show Hint
If the dot product of the vector from a point to a line and the line's direction is zero, that point on the line is the foot of the perpendicular.
Step 1: Understanding the Concept:
The projection line $L$ lies in the given plane. We find two points on the projection: the intersection of the original line with the plane, and the foot of the perpendicular from another point on the line to the plane. Step 2: Detailed Explanation: 1. Intersection Point ($P$):
Any point on line is $(1+2\lambda, 3+\lambda, 4+2\lambda)$. Sub into plane:
$(1+2\lambda) - 2(3+\lambda) - (4+2\lambda) = 3 \implies 1+2\lambda-6-2\lambda-4-2\lambda = 3 \implies -9-2\lambda=3 \implies \lambda = -6$.
$P = (1-12, 3-6, 4-12) = (-11, -3, -8)$. 2. Foot of perpendicular ($Q$) from $A(1, 3, 4)$ to plane:
$\frac{x-1}{1} = \frac{y-3}{-2} = \frac{z-4}{-1} = -\frac{1-6-4-3}{1+4+1} = \frac{12}{6} = 2$.
$Q = (3, -1, 2)$. 3. Line $L$ through $P, Q$: Direction $\vec{PQ} = (14, 2, 10) \parallel (7, 1, 5)$.
Point $(0, 0, 6)$ is $S$. We need distance from $S$ to $L$:
$\vec{SQ} = (3-0, -1-0, 2-6) = (3, -1, -4)$.
$d^2 = |\vec{SQ}|^2 - \frac{(\vec{SQ} \cdot \vec{v})^2}{|\vec{v}|^2}$ where $\vec{v} = (7, 1, 5)$.
$\vec{SQ} \cdot \vec{v} = 21 - 1 - 20 = 0$.
Thus, the point $Q$ is the foot of perpendicular from $S$ to $L$.
$d^2 = |\vec{SQ}|^2 = 3^2 + (-1)^2 + (-4)^2 = 9 + 1 + 16 = 26$. Step 3: Final Answer:
The value of $d^2$ is 26.
