## Problem Statement

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

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

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

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

$\overrightarrow{A C}=(-8-(-1) ;-2-(-2) ; 2-1)=(-7 ; 0 ; 1)$

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

$\cos (\overrightarrow{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{-3 \cdot(-7)+0 \cdot 0+4 \cdot 1}{\sqrt{(-3)^{2}+0^{2}+4^{2}} \cdot \sqrt{(-7)^{2}+0^{2}+1^{2}}}=$
$=\frac{21+0+4}{\sqrt{9+0+16} \cdot \sqrt{49+0+1}}=\frac{25}{\sqrt{25} \cdot \sqrt{50}}=\frac{1}{\sqrt{2}}$

Thus, the cosine of the angle is:

$\cos (\overrightarrow{A B}, \overrightarrow{A C})=\frac{1}{\sqrt{2}}$

and consequently, the angle

$\overrightarrow{A B,} \overrightarrow{A C}=\frac{\pi}{4}$

## Problem Kuznetsov Analytic Geometry 4-12