Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

Question

Aicharya · Accepted Answer

$\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ & L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\end{aligned}$

$\begin{aligned}
& P(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & Q(3 \mu+2,4 \mu+4,5 \mu+5)
\end{aligned}$
Dr's of $P Q < 2 \lambda-3 \mu-1,3 \lambda-4 \mu-2$, $4 \lambda-5 \mu-2>$
$\begin{aligned}
& P Q=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5
\end{array}\right|=-\hat{i}+2 \hat{j}-\hat{k} \\ & \Rightarrow \frac{2 \lambda-3 \mu-1}{-1}=\frac{3 \lambda-4 \mu-2}{2}=\frac{4 \lambda-5 \mu-2}{-1} \\ & \Rightarrow \lambda=\frac{1}{3} \mu=\frac{-1}{6} \\ & \Rightarrow P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)
\end{aligned}$
Dr's $P Q\langle 1,-2,1\rangle$
$\therefore \quad \text { Line }$
$\frac{y-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{y-\frac{13}{3}}{1}$ ,

Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2:…

Solution