Let $A$ be the point $(1, 2, 3)$ and $B$ be a point on the line $\frac{x-1}{2} = \frac{y+1}{2} = \frac{z-3}{2} = k$. The value of $k$ such that line $AB$ is perpendicular to the plane $4x + 9y - 18z = 6$ is
Step-by-Step Solution
Key Concept: For a perpendicular from a point to a line, the connecting vector must be parallel to the plane's normal when the line lies in a plane.
Point $B$ on line $(2k + 1, 3k - 1, 4k + 5)$ has direction vector $\vec{AB} = (2k + 1, 3k - 1, 4k + 5)$ parallel to $(2k+1, 3k-1, 4k+5)$. For perpendicularity to plane, $\vec{AB}$ must be parallel to normal $(-2k+1, 3k-3, 4k+2k) = (-2k+1, 3k-3, 2k)$. Equating proportions: $\frac{2k+1}{-2k+1} = \frac{3k-1}{3k-3} = \frac{4k+5}{2k}$. Solving yields $k = \frac{2}{3}$. Distance from $(2,1,-1)$ is $\frac{|4(2)+2(1)+12(-1)-105|}{\sqrt{4^2+2^2+12^2}} = \frac{|-99|}{\sqrt{164}} = \frac{33}{\sqrt{41}}$.
Correct Answer: 2