## Problem Statement

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

$A(0 ; 3 ;-6), B(9 ; 3 ; 6), C(12 ; 3 ; 3)$
## Solution

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

$\overrightarrow{A B}=(9-0 ; 3-3 ; 6-(-6))=(9 ; 0 ; 12)$

$\overrightarrow{A C}=(12-0 ; 3-3 ; 3-(-6))=(12 ; 0 ; 9)$

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

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

$=\frac{9 \cdot 12+0 \cdot 0+12 \cdot 9}{\sqrt{9^{2}+0^{2}+12^{2}} \cdot \sqrt{12^{2}+0^{2}+9^{2}}}=$

$=\frac{108+0+108}{\sqrt{81+0+144} \cdot \sqrt{144+0+81}}=\frac{216}{\sqrt{225} \cdot \sqrt{225}}=\frac{216}{225}=0.96$

Thus, the cosine of the angle is:

$\cos (\overrightarrow{A B}, \overrightarrow{A C})=0.96$

and consequently the angle is:

$\widehat{A B,} \overrightarrow{A C}=16^{\circ} 15^{\prime} 37^{\prime \prime}$

## Problem Kuznetsov Analytical Geometry 4-25