## Problem Statement

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

$A(1, -1, 0), B(-2, -1, 4), C(8, -1, -1)$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A B}=(-2-1 ;-1-(-1) ; 4-0)=(-3 ; 0 ; 4) \\
& \overrightarrow{A C}=(8-1 ;-1-(-1) ;-1-0)=(7 ; 0 ;-1)
\end{aligned}
$$

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

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

$=\frac{(-3) \cdot 7+0 \cdot 0+4 \cdot(-1)}{\sqrt{(-3)^{2}+0^{2}+4^{2}} \cdot \sqrt{7^{2}+0^{2}+(-1)^{2}}}=$

$=\frac{-21+0-4}{\sqrt{9+0+16} \cdot \sqrt{49+0+1}}=\frac{-25}{\sqrt{25} \cdot \sqrt{50}}=-\frac{1}{\sqrt{2}}$

Thus, the cosine of the angle is:

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

and consequently, the angle is

$\widehat{A B,} \overrightarrow{A C}=135^{\circ}$

## Problem Kuznetsov Analytical Geometry 4-20