Question

Find the shortest distance between the lines $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ and $3x-y-2z+4=0=2x+y+z+1$.

Answer 1

The equation of the plane containing the line $3x-y-2z+4=0=2x+y+z+1$ is

$$\begin{gathered} \left(3x-y-2z+4\right)+\lambda \left(2x+y+z+1\right)=0 \\ \mathrm{Or}\left(3+2\lambda \right)x+\left(\lambda -1\right)y+\left(\lambda -2\right)z+\left(\lambda +4\right)=0.....\left(1\right) \end{gathered}$$

If it is parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$, then

$$\begin{gathered} 2\left(3+2\lambda \right)+4\left(\lambda -1\right)+\left(\lambda -2\right)=0 \\ \Rightarrow 9\lambda =0 \\ \Rightarrow \lambda =0 \end{gathered}$$

Putting $\lambda =0$ in (1), we get

$3x-y-2z+4=0.....\left(2\right)$

This is the equation of the plane containing the second line and parallel to the first line.

Now, the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ passes through (1, 3, −2).

∴ Shortest distance between the given lines

= Length of the perpendicular from (1, 3, −2) to the plane $3x-y-2z+4=0$

$$\begin{gathered} =\left|\frac{3\times 1-3-2\times \left(-2\right)+4}{\sqrt{3^2+{\left(-1\right)}^2+{\left(-2\right)}^2}}\right| \\ =\left|\frac{3-3+4+4}{\sqrt{9+1+4}}\right| \\ =\frac{8}{\sqrt{14}}\mathrm{units} \end{gathered}$$