## Problem Statement

Write the canonical equations of the line.

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

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

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

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

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

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

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

Thus, the point $(0 ; 2 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

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

## Problem Kuznetsov Analytic Geometry 13-4