## Problem Statement

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

$A(2 ; 1 ;-1), B(6 ;-1 ;-4), C(4 ; 2 ; 1)$
## Solution

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

$\overrightarrow{A B}=(6-2 ;-1-1 ;-4-(-1))=(4 ;-2 ;-3)$

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

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

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

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

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

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 4-11