Question

Find the point of intersection of the lines $\overline{AB}$ and $\overline{CD}$ where $A=(7,-6,1),B=(17,-18,-3),C=(1,4,-5)$ and $D=(3,-4,11)$.

Answer 1

enqn of line $\overrightarrow{AB}$
$\frac{x-7}{17-7}=\frac{y+6}{-18+6}=\frac{z-1}{-3-1}$
$\frac{x-7}{10}=\frac{y+6}{-12}=\frac{z-1}{-4}=k$
$x=10k+7$
$y=-12k-6$
$z=-4k+1$
eqn of line $\overrightarrow{CO}$
$\frac{x-1}{3-1}=\frac{y-y}{-4-4}=\frac{z+5}{11+5}$
$\frac{x-1}{2}=\frac{y-4}{-8}=\frac{z+5}{16}=\mu$
$x=2-4+1$
$y=-5\mu +4$
$z=18\mu -5$
Proof of intersection,
$x=10k+7=2\mu +1$
$y=-12k-=-3\mu +4$
$z=-4k+1=16\mu -5$
solving any two
$-12k-6=-8\mu +4$
$-19k+3=45\mu -15$
$+$ $-$ $-$ $+$
___________________________
$-9=-56\mu +15$ $\Rightarrow 56\mu =26$