The distance of the line $\dfrac{x-2}{2}=\dfrac{y-6}{3}=\dfrac{z-3}{4}$ from the point $(1,4,0)$ along the line $\dfrac{x}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{3}$ is:
Step-by-Step Solution
Key Concept: Write the line through $(1,4,0)$ parallel to direction $(1,2,3)$; find its intersection with the given line; the required distance is from $(1,4,0)$ to that intersection point.
Line through $(1,4,0)$ with direction $(1,2,3)$: points $(\lambda+1,2\lambda+4,3\lambda)$.
Point on first line: $(2\mu+2,3\mu+6,4\mu+3)$.
Equating: $\lambda+1=2\mu+2$, $2\lambda+4=3\mu+6$, $3\lambda=4\mu+3$.
From first two: $\lambda=1$, $\mu=0$. Verify third: $3=3$ ✓.
Intersection point: $(2,6,3)$.
Distance from $(1,4,0)$ to $(2,6,3)$: $\sqrt{1+4+9}=\sqrt{14}$.
Correct Answer: 3