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

Sep 18, 2016

${70.9}^{\circ}$

#### Explanation:

We have: $A = < 4 , 9 , 1 >$ and $B = < 5 , 8 , - 3 >$

First, let's determine $C$:

$\implies C = A - B$

$\implies C = < 4 , 9 , 1 > - < 5 , 8 , - 3 >$

$\implies C = < \left(4 - 5\right) , \left(9 - 8\right) , \left(1 - \left(- 3\right)\right) >$

$\implies C = < - 1 , 1 , 4 >$

Then, let's determine the angle between vectors $A$ and $C$.

The dot product between two vectors is given as $A \cdot B = \left\mid A \right\mid \left\mid B \right\mid \cos \left(\theta\right)$.

We can rearrange this to get:

$\implies \cos \left(\theta\right) = \frac{A \cdot B}{\left\mid A \right\mid \left\mid B \right\mid}$

$\implies \cos \left(\theta\right) = \frac{A \cdot C}{\left\mid A \right\mid \left\mid C \right\mid}$

$\implies \cos \left(\theta\right) = \frac{< 4 , 9 , 1 > \cdot < - 1 , 1 , 4 >}{\left\mid < 4 , 9 , 1 > \right\mid \left\mid < - 1 , 1 , 4 > \right\mid}$

$\implies \cos \left(\theta\right) = \frac{\left(4 \cdot \left(- 1\right)\right) + \left(9 \cdot 1\right) + \left(1 \cdot 4\right)}{\sqrt{{4}^{2} + {9}^{2} + {1}^{2}} \cdot \sqrt{{\left(- 1\right)}^{2} + {1}^{2} + {4}^{2}}}$

$\implies \cos \left(\theta\right) = \frac{- 4 + 9 + 4}{\sqrt{42} \cdot \sqrt{18}}$

$\implies \cos \left(\theta\right) = \frac{9}{\sqrt{756}}$

$\implies \cos \left(\theta\right) = \frac{9}{\sqrt{4 \cdot 3 \cdot 7 \cdot 9}}$

$\implies \cos \left(\theta\right) = \frac{9}{6 \sqrt{21}}$

$\implies \cos \left(\theta\right) = \frac{3}{2 \sqrt{21}}$

$\implies \theta = \arccos \left(\frac{3}{2 \sqrt{21}}\right)$

$\implies \theta \approx {70.9}^{\circ}$