# If A = <4 ,-9 ,5 >, B = <1 ,-3 ,-8 > and C=A-B, what is the angle between A and C?

C=<3,-6,13> -> cos theta=(<4,-9,5>*<3,-6,13>)/(sqrt(4^2+(-9)^2+5^2) sqrt(3^2+(-6)^2+13^2)
$\cos \theta = \frac{12 + 54 + 65}{\left(\sqrt{122}\right) \left(\sqrt{214}\right)} \to \theta = {\cos}^{-} 1 \left(\frac{131}{\sqrt{26108}}\right) = {35.83}^{\circ}$
$C = < 4 - 1 , - 9 - \left(- 3\right) , 5 - \left(- 8\right) > = < 3 , - 6 , 13 >$ Then find the dot product between vector A and C which is $A \cdot C = 4 \left(3\right) + \left(- 9\right) \left(- 6\right) + \left(5\right) \left(13\right)$. And divide that by the magnitude of both vector A and C and then take the cosine inverse to find the angle