## Task Condition

Write the canonical equations of the line.

$5 x+y+2 z+4=0$

$x-y-3 z+2=0$
## 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 are:

$\overrightarrow{n_{1}}=\{5 ; 1 ; 2\}$

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

Let's find the direction vector $\vec{s}:$

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
5 & 1 & 2 \\
1 & -1 & -3
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
1 & 2 \\
-1 & -3
\end{array}\right|-j \cdot\left|\begin{array}{cc}
5 & 2 \\
1 & -3
\end{array}\right|+k \cdot\left|\begin{array}{cc}
5 & 1 \\
1 & -1
\end{array}\right|= \\
& =-i+17 j-6 k=\{-1 ; 17 ;-6\}
\end{aligned}
$$

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

$$
\left\{\begin{array}{l}
5 x_{0}+y_{0}+2 \cdot 0+4=0 \\
x_{0}-y_{0}-3 \cdot 0+2=0
\end{array}\right.
$$

$\left\{\begin{array}{l}5 x_{0}+y_{0}+4=0 \\ x_{0}-y_{0}+2=0\end{array}\right.$

$\left\{\begin{array}{l}5 x_{0}+y_{0}+4=0 \\ 6 x_{0}+6=0\end{array}\right.$

$\left\{\begin{array}{l}5 x_{0}+y_{0}+4=0 \\ x_{0}=-1\end{array}\right.$

$\left\{\begin{array}{l}5 \cdot(-1)+y_{0}+4=0 \\ x_{0}=-1\end{array}\right.$

$\left\{\begin{array}{l}y_{0}=1 \\ x_{0}=-1\end{array}\right.$

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

We obtain the canonical equations of the line:

$\frac{x+1}{-1}=\frac{y-1}{17}=\frac{z}{-6}$

## Problem Kuznetsov Analytic Geometry 13-18