## Task Condition

Write the canonical equations of the line.

$x+5 y-z-5=0$

$2 x-5 y+2 z+5=0$
## Solution

Canonical equations of a line:

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

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

$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{1 ; 5 ;-1\} \\
& \overrightarrow{n_{2}}=\{2 ;-5 ; 2\}
\end{aligned}
$$

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

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

$=i \cdot\left|\begin{array}{cc}5 & -1 \\ -5 & 2\end{array}\right|-j \cdot\left|\begin{array}{cc}1 & -1 \\ 2 & 2\end{array}\right|+k \cdot\left|\begin{array}{cc}1 & 5 \\ 2 & -5\end{array}\right|=$

$=5 i-4 j-15 k=\{5 ;-4 ;-15\}$

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

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

Thus, the point $(0 ; 1 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:

$\frac{x}{5}=\frac{y-1}{-4}=\frac{z}{-15}$

Problem Kuznetsov Analytic Geometry 13-16