# If A = <2 ,1 ,6 >, B = <5 ,1 ,3 > and C=A-B, what is the angle between A and C?

Jan 14, 2016

<$\theta = {63.8}^{\circ}$ ( 1 decimal place )

#### Explanation:

$\vec{C} = \vec{A} - \vec{B} = \left(2 , 1 , 6\right) - \left(5 , 1 , 3\right) = \left(- 3 , 0 , 3\right)$

To calculate the angle$\left(\theta\right) b e t w e e n \vec{A} \mathmr{and} \vec{C}$

use the following formula :

$\cos \theta = \frac{\vec{A} . \vec{C}}{| \vec{A} | | \vec{C} |}$

The 'dot product' $\vec{A} . \vec{C} = \left(2 , 1 , 6\right) . \left(- 3 , 0 , 3\right)$

= -6 + 0 + 18 = 12

and$| \vec{A} | = \sqrt{{2}^{2} + {1}^{2} + {6}^{2}} = \sqrt{4 + 1 + 36} = \sqrt{41}$

$| \vec{C} | = \sqrt{{\left(- 3\right)}^{2} + {0}^{2} + {3}^{2}} = \sqrt{9 + 0 + 9} = \sqrt{18}$

substituting these values into the formula :

$\cos \theta = \frac{12}{\sqrt{41} \times \sqrt{18}} = 0.4417 \ldots$

$\Rightarrow \theta = {63.8}^{\circ}$