## Problem Statement

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

$A(0 ; 0 ; 4), B(-3 ;-6 ; 1), C(-5 ;-10 ;-1)$
## Solution

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

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

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

We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ 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:

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

and consequently the angle

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

## Problem Kuznetsov Analytic Geometry 4-14