## problem statement

Write the equation of the plane passing through point $A$ and perpendicular to the vector $\overrightarrow{B C}$.

$A(7 ; -5 ; 0)$

$B(8 ; 3 ; -1)$

$C(8 ; 5 ; 1)$
## Solution

Let's find the vector $\overrightarrow{B C}$:

$\overrightarrow{B C}=\{8-8 ; 5-3 ; 1-(-1)\}=\{0 ; 2 ; 2\}$

Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:

$0 \cdot(x-7)+2 \cdot(y-(-5))+2 \cdot(z-0)=0$

## Simplify:

$$
\begin{aligned}
& 2 y+10+2 z-0=0 \\
& 2 y+2 z+10=0 \\
& y+z+5=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-29