Let $P$ be the point of intersection of the lines $\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{1}$ and $\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{2}$. Then the shortest distance of $P$ from the line $4x=2y=z$ is:
Step-by-Step Solution
Key Concept: Find $P$: from $L_1$, point $=(\lambda+2,5\lambda+4,\lambda+2)$; from $L_2$, point $=(2\mu+3,3\mu+2,2\mu+3)$. Solve: $\lambda=-1,\mu=-1\Rightarrow P=(1,-1,1)$. Line $L_3$: $4x=2y=z\Rightarrow\frac{x}{1}=\frac{y}{2}=\frac{z}{4}$.
$P=(1,-1,1)$. Foot on $L_3$: $(1/7,2/7,4/7)$. $PQ=3\sqrt{14}/7$.
Correct Answer: 2