Question

Show that the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\mathrm{and}\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$ intersect. Find their point of intersection.

Answer 1

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

$$\begin{gathered} \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda \\ \Rightarrow x=3\lambda -1 \\ y=5\lambda -3 \\ z=7\lambda -5 \end{gathered}$$

The coordinates of a general point on the first line are $\left(3\lambda -1,5\lambda -3,7\lambda -5\right)$.

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

$$\begin{gathered} \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu \\ \Rightarrow x=\mu +2 \\ y=3\mu +4 \\ z=5\mu +6 \end{gathered}$$

The coordinates of a general point on the second line are $\left(\mu +2,3\mu +4,5\mu +6\right)$.

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

$$\begin{gathered} 3\lambda -1=\mu +2,5\lambda -3=3\mu +4,7\lambda -5=5\mu +6 \\ \Rightarrow 3\lambda -\mu =3...\left(1\right) \\ 5\lambda -3\mu =7...\left(2\right) \\ 7\lambda -5\mu =11...\left(3\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \lambda =\frac{1}{2} \\ \mu =-\frac{3}{2} \\ \mathrm{Substituting}\lambda =\frac{1}{2}\mathrm{and}\mu =-\frac{3}{2}\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=7\lambda -5\mu \\ =7\left(\frac{1}{2}\right)-5\left(-\frac{3}{2}\right) \\ =11 \\ =\mathrm{RHS} \\ \mathrm{Since}\mathrm{\lambda }=\frac{1}{2}\mathrm{and}\mu =-\frac{3}{2}\mathrm{satisfy}\left(3\right),\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{intersect}. \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\lambda \mathrm{in}\mathrm{the}\mathrm{general}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line},\mathrm{we}\mathrm{get} \\ x=\frac{1}{2} \\ y=-\frac{1}{2} \\ z=-\frac{3}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{intersect}\mathrm{at}\mathrm{point}\left(\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\right). \end{gathered}$$