## Task Condition

Write the canonical equations of the line.

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

$2 x-y-3 z+6=0$
## Solution

Canonical equations of a line:

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

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

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

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

$\vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}i & j & k \\ 2 & 1 & 1 \\ 2 & -1 & -3\end{array}\right|=$

$=i \cdot\left|\begin{array}{cc}1 & 1 \\ -1 & -3\end{array}\right|-j \cdot\left|\begin{array}{cc}2 & 1 \\ 2 & -3\end{array}\right|+k \cdot\left|\begin{array}{cc}2 & 1 \\ 2 & -1\end{array}\right|=$

$$
=-2 i+8 j-4 k=\{-2 ; 8 ;-4\}
$$

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

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

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

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

$\left\{\begin{array}{l}2 x_{0}+y_{0}-2=0 \\ x_{0}=-1\end{array}\right.$

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

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

Thus, the point $(-1 ; 4 ; 0)_{\text { belongs to the line. }}$.

We obtain the canonical equations of the line:

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

## Problem Kuznetsov Analytic Geometry 13-1