## Problem Statement

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

$A(5 ; 3 ;-1), B(5 ; 2 ; 0), C(6 ; 4 ;-1)$
## Solution

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

$\overrightarrow{A B}=(5-5 ; 2-3 ; 0-(-1))=(0 ;-1 ; 1)$

$\overrightarrow{A C}=(6-5 ; 4-3 ;-1-(-1))=(1 ; 1 ; 0)$

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

$\cos (\overrightarrow{A B}, \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$

$$
\begin{aligned}
& =\frac{0 \cdot 1+(-1) \cdot 1+1 \cdot 0}{\sqrt{0^{2}+(-1)^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+0^{2}}}= \\
& =\frac{0-1+0}{\sqrt{0+1+1} \cdot \sqrt{1+1+0}}=\frac{-1}{\sqrt{2} \cdot \sqrt{2}}=-\frac{1}{2}
\end{aligned}
$$

Thus, the cosine of the angle:

$\cos (\overrightarrow{A B}, \widehat{A C})=-\frac{1}{2}$

and consequently the angle

$\widehat{A B,} \overrightarrow{A C}=120^{\circ}$

## Problem Kuznetsov Analytic Geometry 4-6