## Problem Statement

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

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

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

$\overrightarrow{A B}=(-1-2 ;-3-3 ;-1-2)=(-3 ;-6 ;-3)$

$\overrightarrow{A C}=(-3-2 ;-7-3 ;-3-2)=(-5 ;-10 ;-5)$

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

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

$=\frac{(-3) \cdot(-5)+(-6) \cdot(-10)+(-3) \cdot(-5)}{\sqrt{(-3)^{2}+(-6)^{2}+(-3)^{2}} \cdot \sqrt{(-5)^{2}+(-10)^{2}+(-5)^{2}}}=$

$=\frac{15+60+15}{\sqrt{9+36+9} \cdot \sqrt{25+100+25}}=\frac{90}{\sqrt{54} \cdot \sqrt{150}}=1$

Thus, the cosine of the angle:

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

and consequently the angle

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

## Problem Kuznetsov Analytic Geometry 4-22