Let L_1:(x-1)/(2)=(y-2)/(3)=(z-3)/(4) and L_2:(x-2)/(3)=(y-4)/(4)=(z-5)/(5) be two lines.

Question

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

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 $P Q<2 \lambda-3 \mu-1,3 \lambda-4 \mu-2$,

$$ \begin{aligned} & 4 \lambda-5 \mu-2> \ & P Q=\left|\[\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{array}\end{aligned}$$\right|=-\hat{i}+2 \hat{j}-\hat{k}  \]

$$ \begin{aligned} & \Rightarrow \quad \frac{2 \lambda-3 \mu-1}{-1}=\frac{3 \lambda-4 \mu-2}{2}=\frac{4 \lambda-5 \mu-2}{-1} \ & \Rightarrow \quad \lambda=\frac{1}{3} \mu=\frac{-1}{6} \ & \Rightarrow \quad P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right) \end{aligned} $$

Dr's $P Q\langle 1,-2,1\rangle$

$\therefore$ Line

$$ \frac{y-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{y-\frac{13}{3}}{1} $$

Solutions · 1