## Problem Statement

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

$$
A(-3 ; -7 ; -5), B(0 ; -1 ; -2), C(2 ; 3 ; 0)
$$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A B}=(0-(-3) ;-1-(-7) ;-2-(-5))=(3 ; 6 ; 3) \\
& \overrightarrow{A C}=(2-(-3) ; 3-(-7) ; 0-(-5))=(5 ; 10 ; 5)
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos (\overrightarrow{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{3 \cdot 5+6 \cdot 10+3 \cdot 5}{\sqrt{3^{2}+6^{2}+3^{2}} \cdot \sqrt{5^{2}+10^{2}+5^{2}}}= \\
& =\frac{15+60+15}{\sqrt{9+36+9} \cdot \sqrt{25+100+25}}=\frac{90}{\sqrt{54} \cdot \sqrt{150}}=\frac{90}{\sqrt{8100}}=1
\end{aligned}
$$

Thus, the cosine of the angle:

$\cos (\overrightarrow{A B, A C})=1$

and consequently the angle

$\widehat{A B,} \overrightarrow{A C}=0$

## Problem Kuznetsov Analytic Geometry 4-7