## Problem Statement

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

$$
A(-4 ; -2 ; 0), B(-1 ; -2 ; 4), C(3 ; -2 ; 1)
$$
## Solution

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

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

$\overrightarrow{A C}=(3-(-4) ;-2-(-2) ; 1-0)=(7 ; 0 ; 1)$

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 4-5