## Task Condition

Write the canonical equations of the line.

$3 x+3 y-2 z-1=0$

$2 x-3 y+z+6=0$
## 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:

$\overrightarrow{n_{1}}=\{3 ; 3 ;-2\}$

$\overrightarrow{n_{2}}=\{2 ;-3 ; 1\}$

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
3 & 3 & -2 \\
2 & -3 & 1
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
3 & -2 \\
-3 & 1
\end{array}\right|-j \cdot\left|\begin{array}{cc}
3 & -2 \\
2 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
3 & 3 \\
2 & -3
\end{array}\right|= \\
& =-3 i-7 j-15 k=\{-3 ;-7 ;-15\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
3 x_{0}+3 y_{0}-2 \cdot 0-1=0 \\
2 x_{0}-3 y_{0}+0+6=0
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x_{0}+3 y_{0}-1=0 \\
5 x_{0}+5=0
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x_{0}+3 y_{0}-1=0 \\
x_{0}=-1
\end{array}\right.
\end{aligned}
$$

$\left\{\begin{array}{l}3 y_{0}=4 \\ x_{0}=-1\end{array}\right.$

$\left\{\begin{array}{l}x_{0}=-1 \\ y_{0}=\frac{4}{3}\end{array}\right.$

Thus, the point $\left(-1 ; \frac{4}{3} ; 0\right)$ belongs to the line. We obtain the canonical equations of the line:

$\frac{x+1}{-3}=\frac{y-\frac{4}{3}}{-7}=\frac{z}{-15}$

## Problem Kuznetsov Analytic Geometry 13-12