## Problem Statement

Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.

$A(-2 ; 4 ;-6), B(0 ; 2 ;-4), C(-6 ; 8 ;-10)$
## Solution

Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:

$\overrightarrow{A B}=(0-(-2) ; 2-4 ;-4-(-6))=(2 ;-2 ; 2)$

$\overrightarrow{A C}=(-6-(-2) ; 8-4 ;-10-(-6))=(-4 ; 4 ;-4)$

We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:

$\cos (\overrightarrow{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{2 \cdot(-4)+(-2) \cdot 4+2 \cdot(-4)}{\sqrt{2^{2}+(-2)^{2}+2^{2}} \cdot \sqrt{(-4)^{2}+4^{2}+(-4)^{2}}}=$

$$
=\frac{-8-8-8}{\sqrt{4+4+4} \cdot \sqrt{16+16+16}}=\frac{-24}{2 \sqrt{3} \cdot 4 \sqrt{3}}=\frac{-24}{24}=-1
$$

Thus, the cosine of the angle:

$\cos (\overrightarrow{A B, A C})=-1$

and consequently the angle

$\widehat{A B,} \overrightarrow{A C}=\pi$

## Problem Kuznetsov Analytic Geometry 4-31