# If A = <6 ,-7 ,8 >, B = <1 ,-5 ,-8 > and C=A-B, what is the angle between A and C?

Mar 14, 2016

$\theta = 33 , {4}^{o}$

#### Explanation:

$s t e p - 1 : \text{ find C=A-B}$
$C = \left({A}_{x} - {B}_{x} , {A}_{y} - {B}_{y} , {A}_{z} - {B}_{z}\right) = \left(6 - 1 , - 7 + 5 , 8 + 8\right)$
$C = \left(5 , - 2 , 16\right)$
$s t e p - 2 : \text{ find A.C (dot product)}$
$A \cdot C = {A}_{x} \cdot {C}_{x} + {A}_{y} \cdot {C}_{y} + {A}_{z} \cdot {C}_{z} = 30 + 14 + 128 = 172$
$s t e p - 3 : \text{ find magnitude of A}$
$| | A | | = \sqrt{{6}^{2} + {7}^{2} + {8}^{2}} = \sqrt{36 + 49 + 64} = \sqrt{149}$
$s t e p - 4 : \text{ find magnitude of C}$
$| | C | | = \sqrt{{5}^{2} + {2}^{2} + {16}^{2}} = \sqrt{25 + 4 + 256} = \sqrt{285}$
$s t e p - 5 : \text{ use dot product formula}$
$A \cdot C = | | A | | \cdot | | C | | \cdot \cos \theta$
$172 = \sqrt{149} \cdot \sqrt{285} \cdot \cos \theta$
$\cos \theta = \frac{172}{\sqrt{149 \cdot 285}}$
$\cos \theta = 0 , 8346678313$
$\theta = 33 , {4}^{o}$