# If A = <2 ,3 ,-4 >, B = <5 ,1 ,4 > and C=A-B, what is the angle between A and C?

Oct 20, 2016

$\theta \approx 0.8$ radians

#### Explanation:

To obtain the vector C, subtract the components of B from the respective components of a:

$C = < 2 - 5 , 3 - 1 , - 4 - 4 >$

$C = < - 3 , 2 , - 8 >$

There are two ways to compute $A \cdot C$, the first is to add the products of the respective components of vector A and C:

$A \cdot C = \left(2\right) \left(- 3\right) + \left(3\right) \left(2\right) + \left(- 4\right) \left(- 8\right)$

$A \cdot C = 32$

The second way is:

$A \cdot C = | A | | C | \cos \left(\theta\right)$

where $\theta$ is the angle between the two vectors.

We know the value of $A \cdot C$ but we need to compute |A| and |C|

$| A | = \sqrt{{2}^{2} + {3}^{2} + {\left(- 4\right)}^{2}}$

$| A | = \sqrt{29}$

$| C | = \sqrt{{\left(- 3\right)}^{2} + {2}^{2} + {\left(- 8\right)}^{2}}$

$| A | = \sqrt{77}$

$\theta = {\cos}^{-} 1 \left(\frac{A \cdot C}{| A | | C |}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{32}{\sqrt{29} \sqrt{77}}\right)$

$\theta \approx 0.8$ radians