## Problem Statement

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

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

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

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

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

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

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

$=\frac{3 \cdot 4+0 \cdot 0+4 \cdot 3}{\sqrt{3^{2}+0^{2}+4^{2}} \cdot \sqrt{4^{2}+0^{2}+3^{2}}}=$
$=\frac{12+0+12}{\sqrt{9+0+16} \cdot \sqrt{16+0+9}}=\frac{24}{\sqrt{25} \cdot \sqrt{25}}=\frac{24}{25}=0.96$

Thus, the cosine of the angle is:

$\cos (\overrightarrow{A B,} \overrightarrow{A C})=0.96$

and consequently the angle

$\overrightarrow{A B}, \overrightarrow{A C}=16^{\circ} 15^{\prime} 37^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 4-9