## Problem Statement

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

$A(2 ; 2 ; 7), B(0 ; 0 ; 6), C(-2 ; 5 ; 7)$
## Solution

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

$\overrightarrow{A B}=(0-2 ; 0-2 ; 6-7)=(-2 ;-2 ;-1)$

$\overrightarrow{A C}=(-2-2 ; 5-2 ; 7-7)=(-4 ; 3 ; 0)$

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

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

$=\frac{(-2) \cdot(-4)+(-2) \cdot 3+(-1) \cdot 0}{\sqrt{(-2)^{2}+(-2)^{2}+(-1)^{2}} \cdot \sqrt{(-4)^{2}+3^{2}+0^{2}}}=$

$=\frac{8+(-6)+0}{\sqrt{4+4+1} \cdot \sqrt{16+9+0}}=\frac{2}{\sqrt{9} \cdot \sqrt{25}}=\frac{2}{15}$

Thus, the cosine of the angle:

$\cos (\overrightarrow{A B, A C})=\frac{2}{15}$

and consequently the angle

$\widehat{A B,} \overrightarrow{A C}=82^{0} 20^{\prime} 15^{\prime \prime}$

## Problem Kuznetsov Analytical Geometry 4-23