## Problem Statement

Find the point of intersection of the line and the plane.

$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y}{0}=\frac{z+3}{2} \\
& 2 x-y+4 z=0
\end{aligned}
$$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y}{0}=\frac{z+3}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+t \\
y=0 \\
z=-3+2 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$2(1+t)-0+4(-3+2 t)=0$

$2+2 t+0-12+8 t=0$

$10 t-10=0$

$t=1$

Find the coordinates of the intersection point of the line and the plane:

$$
\left\{\begin{array}{l}
x=1+1=2 \\
y=0 \\
z=-3+2 \cdot 1=-1
\end{array}\right.
$$

We get:

$(2 ; 0 ;-1)$

## Problem Kuznetsov Analytic Geometry 14-4