## problem statement

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

$A(6 ; 2 ;-3), B(6 ; 3 ;-2), C(7 ; 3 ;-3)$
## Solution

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

$\overrightarrow{A B}=(6-6 ; 3-2 ;-2-(-3))=(0 ; 1 ; 1)$

$\overrightarrow{A C}=(7-6 ; 3-2 ;-3-(-3))=(1 ; 1 ; 0)$

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

$\cos (\overrightarrow{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{0 \cdot 1+1 \cdot 1+1 \cdot 0}{\sqrt{0^{2}+1^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+0^{2}}}=$
$=\frac{0+1+0}{\sqrt{0+1+1} \cdot \sqrt{1+1+0}}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}$

Thus, the cosine of the angle:

$\cos (\widehat{A B, A C})=\frac{1}{2}$

and consequently the angle

$\overrightarrow{A B,} \overrightarrow{A C}=\frac{\pi}{3}$

## Problem Kuznetsov Analytic Geometry 4-13