Question

Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.

Answer 1

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

$$\begin{gathered} \frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}=\lambda \\ \Rightarrow x=4\lambda \\ y=6\lambda -1 \\ z=2\lambda -1 \end{gathered}$$

The coordinates of a general point on AB are $\left(4\lambda ,6\lambda -1,2\lambda -1\right)$.

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

$$\begin{gathered} \frac{x-3}{3+4}=\frac{y-9}{9-4}=\frac{z-4}{4-4}=\mu \\ \Rightarrow x=7\mu +3 \\ y=5\mu +9 \\ z=4 \end{gathered}$$

The coordinates of a general point on CD are $\left(7\mu +3,5\mu +9,4\right)$.
If the lines AB and CD intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$$\begin{gathered} 4\lambda =7\mu +3,6\lambda -1=5\mu +9,2\lambda -1=4 \\ \Rightarrow 4\lambda -7\mu =3...\left(1\right) \\ 6\lambda -5\mu =10...\left(2\right) \\ \lambda =\frac{5}{2}...\left(3\right) \\ \mathrm{Solving}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \lambda =\frac{5}{2} \\ \mu =1 \\ \mathrm{Substituting}\lambda =\frac{5}{2}\mathrm{and}\mu =1\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=4\lambda -7\mu \\ =4\left(\frac{5}{2}\right)-7\left(1\right) \\ =3 \\ =\mathrm{RHS} \\ \mathrm{Since}\mathrm{\lambda }=\frac{5}{2}\mathrm{and}\mu =1\mathrm{satisfy}\left(3\right),\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{intersect}. \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\lambda \mathrm{in}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{a}\mathrm{general}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{line}AB,\mathrm{we}\mathrm{get} \\ x=10 \\ y=14 \\ z=4 \\ \mathrm{Hence},AB\mathrm{and}CD\mathrm{intersect}\mathrm{at}\mathrm{point}\left(10,14,4\right). \end{gathered}$$