# If A = <4 ,1 ,8 >, B = <6 ,7 ,3 > and C=A-B, what is the angle between A and C?

Feb 1, 2016

${\cos}^{- 1} \left(\frac{2 \sqrt{65}}{45}\right)$

#### Explanation:

The first step is to find $C$.

$C = < 4 , 1 , 8 > - < 6 , 7 , 3 >$

$= < - 2 , - 6 , 5 >$

To find the acute angle between $A$ and $C$, you could use either the dot product or the cross product. I prefer to use the dot product.

$A \cdot C = | A | | C | \cos \theta$

$A \cdot C = \left(4\right) \left(- 2\right) + \left(1\right) \left(- 6\right) + \left(8\right) \left(5\right) = 26$

$| A | = \sqrt{{4}^{2} + {1}^{2} + {8}^{2}} = 9$

$| C | = \sqrt{{\left(- 2\right)}^{2} + {\left(- 6\right)}^{2} + {5}^{2}} = \sqrt{65}$

Therefore,

$\cos \theta = \frac{26}{9 \sqrt{65}} = \frac{2 \sqrt{65}}{45}$

$\theta = {\cos}^{- 1} \left(\frac{2 \sqrt{65}}{45}\right)$