## 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 $\overrightarrow{A C}$:

$$
\begin{aligned}
& \cos (\overrightarrow{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{(-1) \cdot 2+1 \cdot(-2)+(-1) \cdot 2}{\sqrt{(-1)^{2}+1^{2}+(-1)^{2}} \cdot \sqrt{2^{2}+(-2)^{2}+2^{2}}}= \\
& =\frac{-2-2-2}{\sqrt{1+1+1} \cdot \sqrt{4+4+4}}=\frac{-6}{\sqrt{3} \cdot \sqrt{12}}=\frac{-6}{\sqrt{36}}=-1
\end{aligned}
$$

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 4-1