## Task Condition

Write the canonical equations of the line.

$6 x-5 y-4 z+8=0$

$6 x+5 y+3 z+4=0$
## Solution

Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$

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

$\overrightarrow{n_{1}}=\{6 ;-5 ;-4\}$

$\overrightarrow{n_{2}}=\{6 ; 5 ; 3\}$

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
6 & -5 & -4 \\
6 & 5 & 3
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
-5 & -4 \\
5 & 3
\end{array}\right|-j \cdot\left|\begin{array}{cc}
6 & -4 \\
6 & 3
\end{array}\right|+k \cdot\left|\begin{array}{cc}
6 & -5 \\
6 & 5
\end{array}\right|= \\
& =5 i-42 j+60 k=\{5 ;-42 ; 60\}
\end{aligned}
$$

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

$\left\{\begin{array}{l}6 x_{0}-5 y_{0}-4 \cdot 0+8=0 \\ 6 x_{0}+5 y_{0}+3 \cdot 0+4=0\end{array}\right.$

$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+8=0 \\ 6 x_{0}+5 y_{0}+4=0\end{array}\right.$

$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+8=0 \\ 12 x_{0}+12=0\end{array}\right.$

$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+8=0 \\ x_{0}=-1\end{array}\right.$

$\left\{\begin{array}{l}6 \cdot(-1)-5 y_{0}+8=0 \\ x_{0}=-1\end{array}\right.$

$\left\{\begin{array}{l}y_{0}=\frac{2}{5} \\ x_{0}=-1\end{array}\right.$

Therefore, the point $\left(-1 ; \frac{2}{5} ; 0\right)$ belongs to the line.

We obtain the canonical equations of the line:
$\frac{x+1}{5}=\frac{y-\frac{2}{5}}{-42}=\frac{z}{60}$

Problem Kuznetsov Analytic Geometry 13-15