## Problem Statement

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

$A(1 ; 9 ;-4)$

$B(5 ; 7 ; 1)$

$C(3 ; 5 ; 0)$
## Solution

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

$\overrightarrow{BC}=\{3-5 ; 5-7 ; 0-1\}=\{-2 ;-2 ;-1\}$

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:

$-2 \cdot(x-1)-2 \cdot(y-9)-(z-(-4))=0$

$-2 x+2-2 y+18-z+(-4)=0$

$-2 x-2 y-z+16=0$

## Problem Kuznetsov Analytic Geometry 9-9