# If A = <3 ,-1 ,5 >, B = <5 ,2 ,-9 > and C=A-B, what is the angle between A and C?

Feb 10, 2017

38.43 ° (2dp)

#### Explanation:

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

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

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

So for this problem,

$\vec{A} = \left\langle3 , - 1 , 5\right\rangle$
$\vec{B} = \left\langle5 , 2 , - 9\right\rangle$

And so;

$\vec{C} = \vec{A} - \vec{B}$
$\setminus \setminus \setminus \setminus = \left\langle3 , - 1 , 5\right\rangle - \left\langle5 , 2 , - 9\right\rangle$
$\setminus \setminus \setminus \setminus = \left\langle- 2 , - 3 , 14\right\rangle$

The moduli are given by;

$| | \vec{A} | | = | | \left\langle3 , - 1 , 5\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(3\right)}^{2} + {\left(- 1\right)}^{2} + {\left(5\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{9 + 1 + 25}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{35}$

$| | \vec{C} | | = | | \left\langle- 2 , - 3 , 14\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(- 2\right)}^{2} + {\left(- 3\right)}^{2} + {\left(14\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{4 + 9 + 196}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{209}$

And the scaler product is:

$\vec{A} \cdot \vec{C} = | | \left\langle3 , - 1 , 5\right\rangle | | \cdot | | \left\langle- 2 , - 3 , 14\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(3\right) \left(- 2\right) + \left(- 1\right) \left(- 3\right) + \left(5\right) \left(14\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 6 + 3 + 70$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 67$

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

$67 = \sqrt{35} \cdot \sqrt{209} \cdot \cos \theta$
$\therefore \cos \theta = \frac{67}{\sqrt{7315}}$
$\therefore \cos \theta = 0.783371 \ldots$
 :. theta = 38.43 ° (2dp)

So the acute angle between the vectors is 38.43 ° (2dp)