## Task Condition

Write the canonical equations of the line.

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

$2 x-y+z+6=0$
## Solution

Canonical equations of a line:

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

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

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

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

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
2 & 1 & -3 \\
2 & -1 & 1
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
1 & -3 \\
-1 & 1
\end{array}\right|-j \cdot\left|\begin{array}{cc}
2 & -3 \\
2 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
2 & 1 \\
2 & -1
\end{array}\right|= \\
& =-2 i-8 j-4 k=\{-2 ;-8 ;-4\}
\end{aligned}
$$

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

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
2 x_{0}+y_{0}-2=0 \\
2 x_{0}-y_{0}+6=0
\end{array}\right. \\
& \left\{\begin{array}{l}
2 x_{0}+y_{0}-2=0 \\
4 x_{0}+4=0
\end{array}\right. \\
& \left\{\begin{array}{l}
2 x_{0}+y_{0}-2=0 \\
x_{0}=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
2 \cdot(-1)+y_{0}-2=0 \\
x_{0}=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=4 \\
x_{0}=-1
\end{array}\right.
\end{aligned}
$$

Thus, the point $(-1 ; 4 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

$\frac{x+1}{-2}=\frac{y-4}{-8}=\frac{z}{-4}$

## Problem Kuznetsov Analytic Geometry 13-20