## Task Condition

Write the canonical equations of the line.

$x-3 y+2 z+2=0$

$x+3 y+z+14=0$
## Solution

Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$

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

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

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

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
1 & -3 & 2 \\
1 & 3 & 1
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
-3 & 2 \\
3 & 1
\end{array}\right|-j \cdot\left|\begin{array}{cc}
1 & 2 \\
1 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
1 & -3 \\
1 & 3
\end{array}\right|= \\
& =-9 i+j+6 k=\{-9 ; 1 ; 6\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}-3 y_{0}+2 \cdot 0+2=0 \\
x_{0}+3 y_{0}+0+14=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-3 y_{0}+2=0 \\
x_{0}+3 y_{0}+14=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-3 y_{0}+2=0 \\
2 x_{0}+16=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-3 y_{0}+2=0 \\
x_{0}=-8
\end{array}\right. \\
& \left\{\begin{array}{l}
-8-3 y_{0}+2=0 \\
x_{0}=-8
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=-2 \\
x_{0}=-8
\end{array}\right.
\end{aligned}
$$

Therefore, the point $(-8 ;-2 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:
$\frac{x+8}{-9}=\frac{y+2}{1}=\frac{z}{6}$

## Problem Kuznetsov Analytic Geometry 13-2