## Problem Statement

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

$A(-1 ; 2 ;-3), B(3 ; 4 ;-6), C(1 ; 1 ;-1)$
## Solution

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

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

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

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

$$
\begin{aligned}
& \cos (\overrightarrow{A B, A \overrightarrow{A C}})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}= \\
& =\frac{4 \cdot 2+2 \cdot(-1)+(-3) \cdot 2}{\sqrt{4^{2}+2^{2}+(-3)^{2}} \cdot \sqrt{2^{2}+(-1)^{2}+2^{2}}}= \\
& =\frac{8-2-6}{\sqrt{16+4+9} \cdot \sqrt{4+1+4}}=\frac{0}{\sqrt{29} \cdot \sqrt{9}}=0
\end{aligned}
$$

Thus, the cosine of the angle:

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

and consequently the angle

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

## Problem Kuznetsov Analytic Geometry 4-4