## Task Condition

Write the canonical equations of the line.

$8 x-y-3 z-1=0$

$x+y+z+10=0$
## Solution

Canonical equations of a line:

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

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

$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{8 ;-1 ;-3\} \\
& \overrightarrow{n_{2}}=\{1 ; 1 ; 1\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
8 & -1 & -3 \\
1 & 1 & 1
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
-1 & -3 \\
1 & 1
\end{array}\right|-j \cdot\left|\begin{array}{cc}
8 & -3 \\
1 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
8 & -1 \\
1 & 1
\end{array}\right|=
\end{aligned}
$$

$$
=2 i-11 j+9 k=\{2 ;-11 ; 9\}
$$

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

$$
\left\{\begin{array}{l}
8 x_{0}-y_{0}-3 \cdot 0-1=0 \\
x_{0}+y_{0}+0+10=0
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
8 x_{0}-y_{0}-1=0 \\
x_{0}+y_{0}+10=0
\end{array}\right. \\
& \left\{\begin{array}{l}
8 x_{0}-y_{0}-1=0 \\
9 x_{0}+9=0
\end{array}\right. \\
& \left\{\begin{array}{l}
8 x_{0}-y_{0}-1=0 \\
x_{0}=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
8 \cdot(-1)-y_{0}-1=0 \\
x_{0}=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=-9 \\
x_{0}=-1
\end{array}\right.
\end{aligned}
$$

Therefore, the point $(-1 ;-9 ; 0)$ belongs to the line. We obtain the canonical equations of the line:

$\frac{x+1}{2}=\frac{y+9}{-11}=\frac{z}{9}$

Problem Kuznetsov Analytic Geometry 13-14