## Problem Statement

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

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

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

$$
\begin{aligned}
& \overrightarrow{A B}=(-12-0 ;-3-(-3) ;-3-6)=(-12 ; 0 ;-9) \\
& \overrightarrow{A C}=(-9-0 ;-3-(-3) ;-6-6)=(-9 ; 0 ;-12)
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos (\overrightarrow{A B}, \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{(-12) \cdot(-9)+0 \cdot 0+(-9) \cdot(-12)}{\sqrt{(-12)^{2}+0^{2}+(-9)^{2}} \cdot \sqrt{(-9)^{2}+0^{2}+(-12)^{2}}}= \\
& =\frac{108+0+108}{\sqrt{144+0+81} \cdot \sqrt{81+0+144}}=\frac{216}{\sqrt{225} \cdot \sqrt{225}}=\frac{216}{225}=0.96
\end{aligned}
$$

Thus, the cosine of the angle:

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

and consequently the angle

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

## Problem Kuznetsov Analytical Geometry 4-2