## problem statement

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

$A(-4 ; 3 ; 0), B(0 ; 1 ; 3), C(-2 ; 4 ;-2)$
## Solution

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

$\overrightarrow{A B}=(0-(-4) ; 1-3 ; 3-0)=(4 ;-2 ; 3)$

$\overrightarrow{A C}=(-2-(-4) ; 4-3 ;-2-0)=(2 ; 1 ;-2)$

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

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

$=\frac{4 \cdot 2+(-2) \cdot 1+3 \cdot(-2)}{\sqrt{4^{2}+(-2)^{2}+3^{2}} \cdot \sqrt{2^{2}+1^{2}+(-2)^{2}}}=$
$=\frac{8-2-6}{\sqrt{16+4+9} \cdot \sqrt{4+1+4}}=\frac{0}{\sqrt{29} \cdot \sqrt{9}}=0$

Thus, the cosine of the angle:

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

and consequently the angle

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

Problem Kuznetsov Analytic Geometry 4-19