## Problem Statement

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

$A(-1 ; 2 ;-3), B(0 ; 1 ;-2), C(-3 ; 4 ;-5)$
## Solution

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

$\overrightarrow{A B}=(0-(-1) ; 1-2 ;-2-(-3))=(1 ;-1 ; 1)$

$\overrightarrow{A C}=(-3-(-1) ; 4-2 ;-5-(-3))=(-2 ; 2 ;-2)$

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

$$
\begin{aligned}
& \cos (\overrightarrow{A B, \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{1 \cdot(-2)+(-1) \cdot 2+1 \cdot(-2)}{\sqrt{1^{2}+(-1)^{2}+1^{2}} \cdot \sqrt{(-2)^{2}+2^{2}+(-2)^{2}}}= \\
& =\frac{-2-2-2}{\sqrt{1+1+1} \cdot \sqrt{4+4+4}}=\frac{-6}{\sqrt{3} \cdot \sqrt{12}}=-1
\end{aligned}
$$

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 4-24