## Task Condition

Write the canonical equations of the line.

$6 x-7 y-z-2=0$

$x+7 y-4 z-5=0$
## Solution

Canonical equations of a line:

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

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

$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{6 ;-7 ;-1\} \\
& \overrightarrow{n_{2}}=\{1 ; 7 ;-4\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}
i & j & k \\
6 & -7 & -1 \\
1 & 7 & -4
\end{array}\right|= \\
& =i \cdot\left|\begin{array}{cc}
-7 & -1 \\
7 & -4
\end{array}\right|-j \cdot\left|\begin{array}{cc}
6 & -1 \\
1 & -4
\end{array}\right|+k \cdot\left|\begin{array}{cc}
6 & -7 \\
1 & 7
\end{array}\right|=
\end{aligned}
$$

$$
=35 i+23 j+49 k=\{35 ; 23 ; 49\}
$$

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

$$
\left\{\begin{array}{l}
6 x_{0}-7 y_{0}-0-2=0 \\
x_{0}+7 y_{0}-4 \cdot 0-5=0
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
6 x_{0}-7 y_{0}-2=0 \\
x_{0}+7 y_{0}-5=0
\end{array}\right. \\
& \left\{\begin{array}{l}
6 x_{0}-7 y_{0}-2=0 \\
7 x_{0}-7=0
\end{array}\right. \\
& \left\{\begin{array}{l}
6 x_{0}-7 y_{0}-2=0 \\
x_{0}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
6 \cdot 1-7 y_{0}-2=0 \\
x_{0}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{0}=\frac{4}{7} \\
x_{0}=1
\end{array}\right.
\end{aligned}
$$

Thus, the point $\left(1 ; \frac{4}{7} ; 0\right)$ belongs to the line. We obtain the canonical equations of the line:

$$
\frac{x-1}{35}=\frac{y-\frac{4}{7}}{23}=\frac{z}{49}
$$

Problem Kuznetsov Analytic Geometry 13-24