## Problem Statement

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

$A(1 ; 4 ;-1), B(-2 ; 4 ;-5), C(8 ; 4 ; 0)$
## Solution

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

$\overrightarrow{A B}=(-2-1 ; 4-4 ;-5-(-1))=(-3 ; 0 ;-4)$

$\overrightarrow{A C}=(8-1 ; 4-4 ; 0-(-1))=(7 ; 0 ; 1)$

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

$\overrightarrow{A B}, \overrightarrow{A C}=135^{0}$

## Problem Kuznetsov Analytic Geometry 4-28