## 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 $\overrightarrow{A C}$:

$$
\begin{aligned}
& \overrightarrow{A B}=(0-2 ;-2-(-4) ; 4-6)=(-2 ; 2 ;-2) \\
& \overrightarrow{A C}=(6-2 ;-8-(-4) ; 10-6)=(4 ;-4 ; 4)
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos (\overrightarrow{A B, A \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{(-2) \cdot 4+2 \cdot(-4)+(-2) \cdot 4}{\sqrt{(-2)^{2}+2^{2}+(-2)^{2}} \cdot \sqrt{4^{2}+(-4)^{2}+4^{2}}}= \\
& =\frac{-8-8-8}{\sqrt{4+4+4} \cdot \sqrt{16+16+16}}=\frac{-24}{\sqrt{12} \cdot \sqrt{48}}=-1
\end{aligned}
$$

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 4-8