## Problem Statement

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

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

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

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

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

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

$=\frac{8 \cdot 6+0 \cdot 0+6 \cdot 8}{\sqrt{8^{2}+0^{2}+6^{2}} \cdot \sqrt{6^{2}+0^{2}+8^{2}}}=$

$=\frac{48+0+48}{\sqrt{64+0+36} \cdot \sqrt{36+0+64}}=\frac{96}{\sqrt{100} \cdot \sqrt{100}}=0.96$

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 4-17