# Show that the lines $\large \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z}{3}$ and $\large\frac{x-2}{1}=\frac{y-1}{2}=\frac{-z-1}{1}$ intersect and find their point of intersection.

Toolbox:
• The lines $\overrightarrow r=\overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r=\overrightarrow a_2+s\overrightarrow u$ intersect if $[(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]=0$
Step 1
$\large\frac{x-1}{1} = \large\frac{y+1}{-1} = \large\frac{z}{3}$ (i)
$\large\frac{x-2}{1} = \large\frac{y-1}{2} = \large\frac{-z-1}{1}$ (ii)
(ii) can be rewritten as $\large\frac{x-2}{1} = \large\frac{y-1}{2} = \large\frac{z+1}{-1}$
(i) passes through $(1, -1, 0) [ \overrightarrow a_1=\overrightarrow i-\overrightarrow j]$ and is parallel to $\overrightarrow u=\overrightarrow i-\overrightarrow j+3\overrightarrow k$
(ii) passes through $(2, 1, -1) [a_2=2\overrightarrow i+\overrightarrow j-\overrightarrow k]$ and is parallel to $\overrightarrow v=\overrightarrow i+2\overrightarrow j-\overrightarrow k$
Step 2
$\overrightarrow a_2-\overrightarrow a_1 = \overrightarrow i+2\overrightarrow j-\overrightarrow k$
Consider $[ (\overrightarrow a_2-\overrightarrow a_1)\: \: \overrightarrow u\: \overrightarrow v] = \begin{vmatrix} 1 & 2 & -1 \\ 1 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix} = 1(1-6)-2(-1-3)-1(2+1)$
$= -5+8-3=0$
$[(\overrightarrow a_2-\overrightarrow a_1)\:\: \overrightarrow u\: \overrightarrow v]=0 \Rightarrow$ the lines intersect.
To find the point of intersection, let $\large\frac{x-1}{1}=\large\frac{y+1}{-1}=\large\frac{z}{3}=\lambda$ at the point of intersection.
$\Rightarrow x = \lambda+1, \: y=-\lambda-1, \: z=3\lambda$
$\therefore ( \lambda+1, -\lambda-1, 3\lambda)$ lies on (ii). Subset in (ii).
$\large\frac{\lambda+1-2}{1} = \large\frac{-\lambda-1-1}{2} = \large\frac{3\lambda+1}{-1}$
$\lambda-1=\large\frac{-\lambda-2}{2}\Rightarrow 2\lambda-2=-\lambda-2\: or\: \lambda=0$
$\therefore$ the point of intersection is $(1, -1, 0)$

answered Jun 12, 2013
edited Jun 24, 2013