Let $L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $L_2:\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?
$\left(\dfrac{14}{3},-3,\dfrac{22}{3}\right)$
$\left(-\dfrac{5}{3},-7,1\right)$
$\left(2,3,\dfrac{1}{3}\right)$
$\left(\dfrac{8}{3},-1,\dfrac{1}{3}\right)$
Step-by-Step Solution
Key Concept: Find the feet $P$ on $L_1$ and $Q$ on $L_2$ of the common perpendicular, determine the direction of $PQ$, then verify which given option lies on the line through $P$ (or $Q$) in that direction.
$P=(2\lambda+1,3\lambda+2,4\lambda+3)$ on $L_1$, $Q=(3\mu+2,4\mu+4,5\mu+5)$ on $L_2$.
$\overrightarrow{PQ}\cdot(2,3,4)=0$ and $\overrightarrow{PQ}\cdot(3,4,5)=0$.
Solving: $\lambda=\tfrac{1}{3}$, $\mu=-\tfrac{1}{6}$. $P=\left(\tfrac{5}{3},3,\tfrac{13}{3}\right)$, $Q=\left(\tfrac{3}{2},\tfrac{10}{3},\tfrac{25}{6}\right)$.
Direction $PQ\propto(1,-2,1)$. Line: $\dfrac{x-\tfrac{5}{3}}{1}=\dfrac{y-3}{-2}=\dfrac{z-\tfrac{13}{3}}{1}=t$.
At $t=3$: $\left(\tfrac{14}{3},-3,\tfrac{22}{3}\right)$ ✓ (option 1).
Correct Answer: 1