Question

Show that the lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}\mathrm{and}\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$ do not intersect.

Answer 1

The coordinates of any point on the first line are given by

$$\begin{gathered} \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda \\ \Rightarrow x=3\lambda +1 \\ y=2\lambda -1 \\ z=5\lambda +1 \end{gathered}$$

The coordinates of a general point on the first line are $\left(3\lambda +1,2\lambda -1,5\lambda +1\right)$.

The coordinates of any point on the second line are given by

$$\begin{gathered} \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu \\ \Rightarrow x=4\mu -2 \\ y=3\mu +1 \\ z=-2\mu -1 \end{gathered}$$

The coordinates of a general point on the second line are $\left(4\mu -2,3\mu +1,-2\mu -1\right)$.

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$$\begin{gathered} 3\lambda +1=4\mu -2,2\lambda -1=3\mu +1,5\lambda +1=-2\mu -1 \\ \Rightarrow 3\lambda -4\mu =-3...\left(1\right) \\ 2\lambda -3\mu =2...\left(2\right) \\ 5\lambda +2\mu =-2...\left(3\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \lambda =-17 \\ \mu =-12 \\ \mathrm{Substituting}\lambda =-17\mathrm{and}\mu =-12\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=3\lambda +2\mu \\ =3\left(-17\right)+2\left(-12\right) \\ =-75 \\ \ne -2 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{do}\mathrm{not}\mathrm{intersect}. \end{gathered}$$