# If A = <2 ,3 ,-4 >, B = <6 ,1 ,2 > and C=A-B, what is the angle between A and C?

Jan 3, 2016

${56.912}^{\circ}$

#### Explanation:

$C = A - B$
$= \left(2 , 3 , - 4\right) - \left(6 , 1 , 2\right)$
$= \left(- 4 , 2 , - 6\right)$.

Now to find the angle $\theta$ between any 2 vectors A and C, we may use either of the following 2 theorems involving algebraic products :

1. $A \times C = | | A | | \cdot | | C | | \cdot \sin \theta$
2. $A \cdot B = | | A | | \cdot | | B | | \cdot \cos \theta$

Since it is quicker to evaluate a Euclidean inner product as opposed to a vector cross product, I shall solve this problem using the latter theorem.

therefore cos theta = (A*C)/(|A||*||C||

$\therefore \theta = {\cos}^{- 1} \left(\frac{- 8 + 6 + 24}{\sqrt{{2}^{2} + {3}^{2} + {4}^{2}} \cdot \sqrt{{4}^{2} + {2}^{2} + {6}^{2}}}\right)$

$= {\cos}^{- 1} \left(\frac{22}{\sqrt{29} \cdot \sqrt{56}}\right)$

$= {56.912}^{\circ}$.