Condition of the problem

Find the point of intersection of the line and the plane.
$\frac{x-3}{-1}=\frac{y-4}{5}=\frac{z-4}{2}$

$7 x+y+4 z-47=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-3}{-1}=\frac{y-4}{5}=\frac{z-4}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3-t \\
y=4+5 t \\
z=4+2 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$7(3-t)+(4+5 t)+4(4+2 t)-47=0$

$21-7 t+4+5 t+16+8 t-47=0$

$6 t-6=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=3-1=2 \\
y=4+5 \cdot 1=9 \\
z=4+2 \cdot 1=6
\end{array}\right.
$$

We get:

$(2 ; 9 ; 6)$

## Problem Kuznetsov Analytic Geometry 14-16