## Problem Statement

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

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

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

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

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

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

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

$$
\begin{aligned}
& =\frac{2 \cdot 1+(-2) \cdot(-2)+(-1) \cdot(-2)}{\sqrt{2^{2}+(-2)^{2}+(-1)^{2}} \cdot \sqrt{1^{2}+(-2)^{2}+(-2)^{2}}}= \\
& =\frac{2+4+2}{\sqrt{4+4+1} \cdot \sqrt{1+4+4}}=\frac{8}{\sqrt{9} \cdot \sqrt{9}}=\frac{8}{9}
\end{aligned}
$$

Thus, the cosine of the angle:

$\cos (\overrightarrow{A B}, \overrightarrow{A C})=\frac{8}{9}$

and consequently the angle

$\overrightarrow{A B}, \overrightarrow{A C}=27^{\circ} 15' 58''$

## Problem Kuznetsov Analytic Geometry 4-26