## Task Condition

Write the canonical equations of the line.

$$
\begin{aligned}
& 2 x+3 y+z+6=0 \\
& x-3 y-2 z+3=0
\end{aligned}
$$
## Solution

Canonical equations of a line:

$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$

where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}_{\text {- its direction }}$ vector.

Since the line belongs to both planes simultaneously, its direction vector $\vec{s}$ is orthogonal to the normal vectors of both planes. The normal vectors of the planes are:

$\overrightarrow{n_{1}}=\{2 ; 3 ; 1\}$

$\overrightarrow{n_{2}}=\{1 ;-3 ;-2\}$

Find the direction vector $\vec{s}:$

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
2 & 3 & 1 \\
1 & -3 & -2
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
3 & 1 \\
-3 & -2
\end{array}\right|-j \cdot\left|\begin{array}{cc}
2 & 1 \\
1 & -2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
2 & 3 \\
1 & -3
\end{array}\right|= \\
& =-3 i+5 j-9 k=\{-3 ; 5 ;-9\}
\end{aligned}
$$

Find some point on the line $\left(x_{0} ; y_{0} ; z_{0}\right)$. Let $z_{0}=0$, then

$$
\begin{aligned}
& \left\{\begin{array}{l}
2 x_{0}+3 y_{0}+0+6=0 \\
x_{0}-3 y_{0}-2 \cdot 0+3=0
\end{array}\right. \\
& \left\{\begin{array}{l}
2 x_{0}+3 y_{0}+6=0 \\
x_{0}-3 y_{0}+3=0
\end{array}\right. \\
& \left\{\begin{array}{l}
2 x_{0}+3 y_{0}+6=0 \\
3 x_{0}+9=0
\end{array}\right. \\
& \left\{\begin{array}{l}
2 x_{0}+3 y_{0}+6=0 \\
x_{0}=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
2 \cdot(-3)+3 y_{0}+6=0 \\
x_{0}=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=0 \\
x_{0}=-3
\end{array}\right.
\end{aligned}
$$

Therefore, the point $(-3 ; 0 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

$\frac{x+3}{-3}=\frac{y}{5}=\frac{z}{-9}$

## Problem Kuznetsov Analytic Geometry 13-5