Question

Show that the lines $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}$ and $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ intersect. Find the coordinates of their point of intersection.

Answer 1

$\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}=\lambda \Rightarrow x=\lambda +4,y=-4\lambda -3,z=7\lambda -\mathrm{1.....}\left(i\right)\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\mu \Rightarrow x=2\mu +1,y=-3\mu -1,z=8\mu -\mathrm{10....}\left(ii\right)$

Comapring $x$ from $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow \lambda +4=2\mu +1\Rightarrow 2\mu -\lambda =\mathrm{3.......}\left(iii\right)$

Comparing $y$ from $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow -4\lambda -3=-3\mu -1\Rightarrow 3\mu -4\lambda =\mathrm{2.....}\left(iv\right)$

Solving $\left(iii\right)$ and $\left(iv\right)$

$\Rightarrow \lambda =1,\mu =2$

Substituing $\lambda$ and in $\left(i\right)$

$\Rightarrow x=1+4,y=-4-3,z=7-1\Rightarrow x=5,y=-7,z=6$

So the point of intersection is $(5,-7,6)$