Write in canonical form the line 2x-3y-3z-9=0, x-2y+z+3=0

Write in canonical form the line $2x-3y-3z-9=0,\ x-2y+z+3=0$.

Explanation of Method

Converting a line given by two planes into canonical form.

\[ \vec d=\{2,-3,-3\}\times\{1,-2,1\}=\{-9,-5,-1\}\parallel\{9,5,1\}. \]

With $z=0$: $2x-3y=9$ and $x-2y=-3$, so $M_0(27,15,0)$. This gives

\[\frac{x-27}{9}=\frac{y-15}{5}=\frac z1\]

A simpler point is obtained with $x=0$: then $-3y-3z=9$ and $-2y+z=-3$, so $y=0$, $z=-3$. The same line is

\[\frac x9=\frac y5=\frac{z+3}{1}\]

Check: $(0,0,-3)$ gives $9-9=0$ and $-3+3=0$. $(27,15,0)$ corresponds to $t=3$, since $(0,0,-3)+3\{9,5,1\}=(27,15,0)$.

$(0,0,-3)$

The line of intersection (black) with direction $\vec n_1\times\vec n_2\parallel\{9,5,1\}$ through $(0,0,-3)$.

\[\boxed{\,\frac{x-27}{9}=\frac{y-15}{5}=\frac z1\,}\]

\[\boxed{\,\frac x9=\frac y5=\frac{z+3}{1}\,}\]