## problem statement

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

$A(-2 ; 1 ; 1), B(2 ; 3 ;-2), C(0 ; 0 ; 3)$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-2) ; 3-1 ;-2-1)=(4 ; 2 ;-3) \\
& \overrightarrow{A C}=(0-(-2) ; 0-1 ; 3-1)=(2 ;-1 ; 2)
\end{aligned}
$$

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

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

$=\frac{4 \cdot 2+2 \cdot(-1)+(-3) \cdot 2}{\sqrt{4^{2}+2^{2}+(-3)^{2}} \cdot \sqrt{2^{2}+(-1)^{2}+2^{2}}}=$

$=\frac{8-2-6}{\sqrt{16+4+9} \cdot \sqrt{4+1+4}}=\frac{0}{\sqrt{29} \cdot \sqrt{9}}=0$

Thus, the cosine of the angle:

$\cos (\overrightarrow{A B,} \overrightarrow{A C})=0$

and consequently the angle

$\widehat{A B,} \overrightarrow{A C}=\frac{\pi}{2}$

Problem Kuznetsov Analytic Geometry 4-27