# If A = <8 ,1 ,-5 >, B = <6 ,5 ,-8 > and C=A-B, what is the angle between A and C?

Feb 15, 2016

$\theta = {93.366}^{\circ}$

#### Explanation:

$\vec{A} = < 8 , 1 - 5 >$ => $| \vec{A} | = \sqrt{64 + 1 + 25} = \sqrt{90} = 3 \sqrt{10}$
$\vec{B} = < 6 , 5 , - 8 >$
$\vec{C} = \vec{A} - \vec{B} = < 8 - 6 , 1 - 5 , - 5 + 8 > = < 2 , - 4 , 3 >$
$\to \setminus \vec{C} | = \sqrt{4 + 16 + 9} = \sqrt{29}$

$\vec{A} \cdot \vec{C} = | \vec{A} | | \vec{C} | \cdot \cos \theta$
$\cos \theta = \frac{\vec{A} \cdot \vec{C}}{| \vec{A} | | \vec{C} |} = \frac{8 \cdot 2 + 1 \left(- 4\right) + \left(- 5\right) \left(3\right)}{3 \sqrt{10} \cdot \sqrt{29}} = \frac{16 - 4 - 15}{3 \sqrt{290}} = - \frac{3}{3 \sqrt{290}} = - \frac{1}{\sqrt{290}}$ $\to \theta = {93.366}^{\circ}$