## Problem Statement

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

$A(5 ; 3 ;-1)$

$B(0 ; 0 ;-3)$

$C(5 ;-1 ; 0)$
## Solution

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

$$
\overrightarrow{BC}=\{5-0 ;-1-0 ; 0-(-3)\}=\{5 ;-1 ; 3\}
$$

Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$5 \cdot(x-5)-(y-3)+3 \cdot(z-(-1))=0$

$5 x-25-y+3+3 z+3=0$

$5 x-y+3 z-19=0$

## Problem Kuznetsov Analytic Geometry 9-27