# What is the angle between <-8,2,8> and < 6,-4,5>?

Aug 13, 2017

The angle is ${80.9}^{o}$(3sf)

#### Explanation:

The angle $\theta$ between two vectors $\boldsymbol{\vec{A}}$ and $\boldsymbol{\vec{B}}$ is related to the modulus (or magnitude) and scaler (or dot) product of $\boldsymbol{\vec{A}}$ and $\boldsymbol{\vec{B}}$ by the relationship:

$\boldsymbol{\vec{A} \cdot \vec{B}} = | | \boldsymbol{\vec{A}} | | \setminus | | \boldsymbol{\vec{B}} | | \setminus \cos \theta$

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle between $\boldsymbol{\vec{u}}$ and $\boldsymbol{\vec{v}}$ be $\theta$ and let:

$\boldsymbol{\vec{u}} = \left\langle- 8 , 2 , 8\right\rangle$
$\boldsymbol{\vec{v}} = \left\langle6 , - 4 , 5\right\rangle$

The vector norm is given by;

$| \boldsymbol{\vec{u}} | = | | \left\langle- 8 , 2 , - 8\right\rangle | | = \sqrt{64 + 4 + 64} = \sqrt{132}$
$| \boldsymbol{\vec{v}} | = | | \left\langle6 , - 4 , 5\right\rangle | | = \sqrt{36 + 16 + 25} = \sqrt{77}$

And the scaler (or "dot") product is:

$\boldsymbol{\vec{u} \cdot \vec{v}} = \left\langle- 8 , 2 , 8\right\rangle \boldsymbol{\cdot} \left\langle6 , - 4 , 5\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(- 8\right) \left(6\right) + \left(2\right) \left(- 4\right) + \left(8\right) \left(5\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 48 - 8 + 40$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 16$

And so using $\boldsymbol{\vec{A} \cdot \vec{B}} = | | \boldsymbol{\vec{A}} | | \setminus | | \boldsymbol{\vec{B}} | | \setminus \cos \theta$ we have:

$- 16 = \sqrt{132} \sqrt{77} \cos \theta$
$\therefore \cos \theta = - \frac{16}{\sqrt{10164}}$
$\text{ } = - \frac{8 \sqrt{21}}{231}$
$\text{ } = - 0.15870392 \ldots$
$\implies \theta = 99.1316 {\ldots}^{o}$

This is not acute, so the acute angle between the vectors is:

${\theta}_{\text{acute}} = {180}^{o} - 99.1316 {\ldots}^{o}$
$\text{ } = 80.8683 {\ldots}^{o}$