Condition of the problem

Write the canonical equations of the line.

\[
\begin{aligned}
& 3 x+y-z-6=0 \\
& 3 x-y+2 z=0
\end{aligned}
\]
## Solution

Canonical equations of a line:

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

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

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

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

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

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

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
3 x_{0}+y_{0}-0-6=0 \\
3 x_{0}-y_{0}+2 \cdot 0=0
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x_{0}+y_{0}-6=0 \\
3 x_{0}-y_{0}=0
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x_{0}+y_{0}-6=0 \\
-2 y_{0}+6=0
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x_{0}+y_{0}-6=0 \\
y_{0}=3
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x_{0}+3-6=0 \\
y_{0}=3
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}=1 \\
y_{0}=3
\end{array}\right.
\end{aligned}
$$

Thus, the point $(1 ; 3 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

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

## Problem Kuznetsov Analytic Geometry 13-6