## Problem Statement

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

$A(3, -6, 9), B(0, -3, 6), C(9, -12, 15)$
## Solution

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

$\overrightarrow{A B}=(0-3 ;-3-(-6) ; 6-9)=(-3 ; 3 ;-3)$

$\overrightarrow{A C}=(9-3 ;-12-(-6) ; 15-9)=(6 ;-6 ; 6)$

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

$$
\begin{aligned}
& \cos (\overrightarrow{A B, \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{(-3) \cdot 6+3 \cdot(-6)+(-3) \cdot 6}{\sqrt{(-3)^{2}+3^{2}+(-3)^{2}} \cdot \sqrt{6^{2}+(-6)^{2}+6^{2}}}= \\
& =\frac{-18-18-18}{\sqrt{9+9+9} \cdot \sqrt{36+36+36}}=\frac{-54}{\sqrt{27} \cdot \sqrt{108}}=-\frac{54}{\sqrt{2916}}=-1
\end{aligned}
$$

Thus, the cosine of the angle:

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

and consequently the angle

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

## Problem Kuznetsov Analytic Geometry 4-16