# If A = <7 ,9 ,4 >, B = <6 ,-9 ,-3 > and C=A-B, what is the angle between A and C?

Jan 27, 2018

$\theta = {\cos}^{-} 1 \left(\frac{197}{\sqrt{146} \sqrt{374}}\right) \approx 0.568$ radians or ${32.54}^{\circ}$

#### Explanation:

$\vec{c} = \left[7 - 6 , 9 - \left(- 9\right) , 4 - \left(- 3\right)\right] = \left[1 , 18 , 7\right]$

The angle between vectors is given by:

$\theta = {\cos}^{-} 1 \left(\frac{\vec{a} \cdot \vec{c}}{| | \vec{a} | | | | \vec{c} | |}\right)$

Calculating individual parts:
$\vec{a} \cdot \vec{c} = 7 \cdot 1 + 9 \cdot 18 + 4 \cdot 7 = 197$
$| | \vec{a} | | = \sqrt{{7}^{2} + {9}^{2} + {4}^{2}} = \sqrt{146}$
$| | \vec{c} | | = \sqrt{{1}^{2} + {18}^{2} + {7}^{2}} = \sqrt{374}$

So:

$\theta = {\cos}^{-} 1 \left(\frac{197}{\sqrt{146} \sqrt{374}}\right)$, which is about as good as it gets, but approximations are:

$\theta \approx 0.568$ radians or $\approx {32.54}^{\circ}$