## problem statement

Write the canonical equations of the line.

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

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

Find the direction vector $\vec{s}:$

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
1 & -2 & 1 \\
2 & 2 & -1
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
-2 & 1 \\
2 & -1
\end{array}\right|-j \cdot\left|\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
1 & -2 \\
2 & 2
\end{array}\right|= \\
& =0 \cdot i+3 j+6 k=\{0 ; 3 ; 6\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}-2 y_{0}+0-4=0 \\
2 x_{0}+2 y_{0}-0-8=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-2 y_{0}-4=0 \\
2 x_{0}+2 y_{0}-8=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-2 y_{0}-4=0 \\
3 x_{0}-12=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-2 y_{0}-4=0 \\
x_{0}=4
\end{array}\right. \\
& \left\{\begin{array}{l}
4-2 y_{0}-4=0 \\
x_{0}=4
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=0 \\
x_{0}=4
\end{array}\right.
\end{aligned}
$$

Therefore, the point $(4 ; 0 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:
$\frac{x-4}{0}=\frac{y}{3}=\frac{z}{6}$

Problem Kuznetsov Analytic Geometry 13-3