## Task Condition

Write the canonical equations of the line.

$$
\begin{aligned}
& x+5 y-z+11=0 \\
& x-y+2 z-1=0
\end{aligned}
$$
## Solution

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

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

$\overrightarrow{n_{1}}=\{1 ; 5 ;-1\}$

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

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
1 & 5 & -1 \\
1 & -1 & 2
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
5 & -1 \\
-1 & 2
\end{array}\right|-j \cdot\left|\begin{array}{cc}
1 & -1 \\
1 & 2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
1 & 5 \\
1 & -1
\end{array}\right|= \\
& =9 i-3 j-6 k=\{9 ;-3 ;-6\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}+5 y_{0}-0+11=0 \\
x_{0}-y_{0}+2 \cdot 0-1=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}+5 y_{0}+11=0 \\
x_{0}-y_{0}-1=0
\end{array}\right. \\
& \left\{\begin{array}{l}
6 y_{0}+12=0 \\
x_{0}-y_{0}-1=0
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=-2 \\
x_{0}-y_{0}-1=0
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=-2 \\
x_{0}-(-2)-1=0
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=-2 \\
x_{0}=-1
\end{array}\right.
\end{aligned}
$$

Thus, the point $(-1 ;-2 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

$$
\frac{x+1}{9}=\frac{y+2}{-3}=\frac{z}{-6}
$$

## Problem Kuznetsov Analytic Geometry 13-22