# If A = <1 ,6 ,-8 >, B = <-9 ,4 ,1 > and C=A-B, what is the angle between A and C?

Apr 28, 2018

$\theta = \arccos \left(\frac{94}{\sqrt{18685}}\right) \approx {46.55}^{\circ}$

#### Explanation:

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

The dot product divided by the magnitudes gives the cosine of the angle.

$\cos \theta = \frac{A \cdot C}{| A | | C |}$

$C = A - B = \left(1 , 6 , - 8\right) - \left(- 9 , 4 , 1\right) = \left(10 , 2 , - 9\right)$

$\cos \theta = \frac{\left(1 , 6 , - 8\right) \cdot \left(10 , 2 , - 9\right)}{| \left(\left(1 , 6 , - 8\right)\right) | | \left(\left(10 , 2 , - 9\right)\right) |}$

$\cos \theta = \frac{1 \left(10\right) + 6 \left(2\right) + - 8 \left(- 9\right)}{\sqrt{\left({1}^{2} + {6}^{2} + {8}^{2}\right) \left({10}^{2} + {2}^{2} + {9}^{2}\right)}}$

$\cos \theta = \frac{94}{\sqrt{18685}}$

$\theta = \arccos \left(\frac{94}{\sqrt{18685}}\right)$

$\theta \approx {46.55}^{\circ}$