Problem 941

## 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:

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

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

$$\begin{aligned} & \vec{s}=\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc} i & j & k \\ 1 & -2 & 1 \\ 2 & 2 & -1 \end{array}\right|= \\ & =i \cdot\left|\begin{array}{cc} -2 & 1 \\ 2 & -1 \end{array}\right|-j \cdot\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right|+k \cdot\left|\begin{array}{cc} 1 & -2 \\ 2 & 2 \end{array}\right|= \\ & =0 \cdot i+3 j+6 k=\{0 ; 3 ; 6\} \end{aligned}$$

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

$$\begin{aligned} & \left\{\begin{array}{l} x_{0}-2 y_{0}+0-4=0 \\ 2 x_{0}+2 y_{0}-0-8=0 \end{array}\right. \\ & \left\{\begin{array}{l} x_{0}-2 y_{0}-4=0 \\ 2 x_{0}+2 y_{0}-8=0 \end{array}\right. \\ & \left\{\begin{array}{l} x_{0}-2 y_{0}-4=0 \\ 3 x_{0}-12=0 \end{array}\right. \\ & \left\{\begin{array}{l} x_{0}-2 y_{0}-4=0 \\ x_{0}=4 \end{array}\right. \\ & \left\{\begin{array}{l} 4-2 y_{0}-4=0 \\ x_{0}=4 \end{array}\right. \\ & \left\{\begin{array}{l} y_{0}=0 \\ x_{0}=4 \end{array}\right. \end{aligned}$$

Therefore, the point $(4 ; 0 ; 0)$ belongs to the line.

We obtain the canonical equations of the line:
$\frac{x-4}{0}=\frac{y}{3}=\frac{z}{6}$

Problem Kuznetsov Analytic Geometry 13-3