## Problem Statement

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

$A(1; -1; 5)$

$B(0; 7; 8)$

$C(-1; 3; 8)$
## Solution

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

$\overrightarrow{BC}=\{-1-0 ; 3-7 ; 8-8\}=\{-1 ;-4 ; 0\}$

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:

$-(x-1)-4 \cdot(y-(-1))+0 \cdot(z-5)=0$

$-x+1-4 y-4=0$

$-x-4 y-3=0$

$x+4 y+3=0$

## Problem Kuznetsov Analytic Geometry 9-15