## Problem Statement

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

$A(-4 ; 0 ; 4), B(-1 ; 6 ; 7), C(1 ; 10 ; 9)$
## Solution

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

$\overrightarrow{A B}=(-1-(-4) ; 6-0 ; 7-4)=(3 ; 6 ; 3)$

$\overrightarrow{A C}=(1-(-4) ; 10-0 ; 9-4)=(5 ; 10 ; 5)$

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

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

$=\frac{3 \cdot 5+6 \cdot 10+3 \cdot 5}{\sqrt{3^{2}+6^{2}+3^{2}} \cdot \sqrt{5^{2}+10^{2}+5^{2}}}=$

$=\frac{15+60+15}{\sqrt{9+36+9} \cdot \sqrt{25+100+25}}=\frac{90}{\sqrt{54} \cdot \sqrt{150}}=1$

Thus, the cosine of the angle:

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

and consequently the angle

$\overrightarrow{A B}, \overrightarrow{A C}=0$

Problem Kuznetsov Analytic Geometry 4-30