## problem statement

Write the canonical equations of the line.

$x-y-z-2=0$

$x-2 y+z+4=0$
## 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:

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

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

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

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

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}-y_{0}-0-2=0 \\
x_{0}-2 y_{0}+0+4=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-y_{0}-2=0 \\
x_{0}-2 y_{0}+4=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-y_{0}-2=0 \\
-y_{0}+6=0
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-y_{0}-2=0 \\
y_{0}=6
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}-6-2=0 \\
y_{0}=6
\end{array}\right. \\
& \left\{\begin{array}{l}
x_{0}=8 \\
y_{0}=6
\end{array}\right.
\end{aligned}
$$

Thus, the point $(8 ; 6 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

$$
\frac{x-8}{-3}=\frac{y-6}{-2}=\frac{z}{-1}
$$

## Problem Kuznetsov Analytic Geometry 13-10