# If A = <8 ,1 ,4 >, B = <6 ,5 ,-8 > and C=A-B, what is the angle between A and C?

Mar 29, 2018

First figure out the $C$.

$C = A - B = \left\langle 8 , 1 , 4\right\rangle - \left\langle 6 , 5 , - 8\right\rangle = \left\langle 8 - 6 , 1 - 5 , 4 - - 8\right\rangle = \left\langle 2 , - 4 , 12\right\rangle$

So now we need to find the angle between $A$ and $C$. We can use this formula $\cos \theta = \frac{u \cdot v}{| | u | | \setminus | | v | |}$.

$A \cdot C = \left\langle 8 , 1 , 4\right\rangle \cdot \left\langle 2 , - 4 , 12\right\rangle = 8 \cdot 2 + 1 \cdot - 4 + 4 \cdot 12 = 60$

$| | A | | = \sqrt{{8}^{2} + {1}^{2} + {4}^{2}} = \sqrt{81} = 9$
$| | C | | = \sqrt{{2}^{2} + {\left(- 4\right)}^{2} + {12}^{2}} = \sqrt{164} = 2 \sqrt{41}$

So putting these values together would get you $\cos \theta = \frac{60}{9 \cdot 2 \sqrt{41}} = \frac{60}{\sqrt{41}} = \setminus \frac{10}{3 \setminus \sqrt{41}}$

Now take the inverse of cosine function on both sides to obtain theta = cos^{-1}(frac{10}{3sqrt41})\approx58.6289°