## problem statement

Write the canonical equations of the line.

$6 x-5 y+3 z+8=0$

$6 x+5 y-4 z+4=0$
## 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:

$\overrightarrow{n_{1}}=\{6 ;-5 ; 3\}$

$\overrightarrow{n_{2}}=\{6 ; 5 ;-4\}$

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

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

$=5 i+42 j+60 k=\{5 ; 42 ; 60\}$

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

$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+3 \cdot 0+8=0 \\ 6 x_{0}+5 y_{0}-4 \cdot 0+4=0\end{array}\right.$
$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+8=0 \\ 6 x_{0}+5 y_{0}+4=0\end{array}\right.$
$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+8=0 \\ 12 x_{0}+12=0\end{array}\right.$
$\left\{\begin{array}{l}6 x_{0}-5 y_{0}+8=0 \\ x_{0}=-1\end{array}\right.$
$\left\{\begin{array}{l}6 \cdot(-1)-5 y_{0}+8=0 \\ x_{0}=-1\end{array}\right.$
$\left\{\begin{array}{l}y_{0}=\frac{2}{5} \\ x_{0}=-1\end{array}\right.$

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 13-30