## Problem Statement

Write the canonical equations of the line.

$$
\begin{aligned}
& 4 x+y-3 z+2=0 \\
& 2 x-y+z-8=0
\end{aligned}
$$
## 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:

$\overrightarrow{n_{1}}=\{4 ; 1 ;-3\}$

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

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
4 & 1 & -3 \\
2 & -1 & 1
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
1 & -3 \\
-1 & 1
\end{array}\right|-j \cdot\left|\begin{array}{cc}
4 & -3 \\
2 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right|= \\
& =-2 i-10 j-6 k=\{-2 ;-10 ;-6\}
\end{aligned}
$$

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

$\left\{\begin{array}{l}4 x_{0}+y_{0}-3 \cdot 0+2=0 \\ 2 x_{0}-y_{0}+0-8=0\end{array}\right.$

$\left\{\begin{array}{l}4 x_{0}+y_{0}+2=0 \\ 2 x_{0}-y_{0}-8=0\end{array}\right.$

$\left\{\begin{array}{l}4 x_{0}+y_{0}+2=0 \\ 6 x_{0}-6=0\end{array}\right.$

$\left\{\begin{array}{l}4 x_{0}+y_{0}+2=0 \\ x_{0}=1\end{array}\right.$

$\left\{\begin{array}{l}4 \cdot 1+y_{0}+2=0 \\ x_{0}=1\end{array}\right.$

$\left\{\begin{array}{l}y_{0}=-6 \\ x_{0}=1\end{array}\right.$

Thus, the point $(1 ;-6 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

$\frac{x-1}{-2}=\frac{y+6}{-10}=\frac{z}{-6}$

## Problem Kuznetsov Analytic Geometry 13-11