## Problem Statement

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

$A(7 ; 0 ; 2), B(7 ; 1 ; 3), C(8 ;-1 ; 2)$
## Solution

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

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

$\overrightarrow{A C}=(8-7 ;-1-0 ; 2-2)=(1 ;-1 ; 0)$

We find the cosine of the angle $\phi$ between the 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 is:

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

and consequently the angle is:

$\widehat{A B,} \overrightarrow{A C}=120^{\circ}$

Problem Kuznetsov Analytical Geometry 4-21