## Problem Statement

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

$A(2, -8, -1), B(4, -6, 0), C(-2, -5, -1)$
## Solution

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

$\overrightarrow{A B}=(4-2 ;-6-(-8) ; 0-(-1))=(2 ; 2 ; 1)$

$\overrightarrow{A C}=(-2-2 ;-5-(-8) ;-1-(-1))=(-4 ; 3 ; 0)$

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

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

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

$=\frac{-8+6+0}{\sqrt{4+4+1} \cdot \sqrt{16+9+0}}=\frac{-2}{\sqrt{9} \cdot \sqrt{25}}=-\frac{2}{3 \cdot 5}=-\frac{2}{15}$

Thus, the cosine of the angle:

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

and consequently the angle

$\widehat{A B,} \overrightarrow{A C}=97^{0} 39^{\prime} 44^{\prime \prime}$

## Problem Kuznetsov Analytical Geometry 4-15