The distance of point $P(3,8,2)$ from the line $\dfrac{x-1}{2}=\dfrac{y-3}{4}=\dfrac{z-2}{3}$ measured parallel to plane $3x+2y-2z+15=0$ is
Step-by-Step Solution
Key Concept: Direction of measurement is parallel to both the given line's plane and the plane $3x+2y-2z+15=0$; find the relevant vector
Direction $d=(2,4,3)$; plane normal $n=(3,2,-2)$. Measurement direction $=d\times n=(4\cdot(-2)-3\cdot2,3\cdot3-2\cdot(-2),2\cdot2-4\cdot3)=(-14,13,-8)$. Line through $P$ in this direction; find intersection with given line; compute distance. Answer $=\sqrt{46}$.
Correct Answer: 3