# If A = <2 ,4 ,-1 >, B = <3 ,8 ,2 > and C=A-B, what is the angle between A and C?

Sep 30, 2016

$\theta \approx 2.437$ radians

#### Explanation:

$C = < 2 - 3 , 4 - 8 , - 1 - 1 >$

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

The rectangular definition of the dot-product is:

A•C = (A_x)(C_x) + (A_y)(C_y) + (A_z)(C_z)

A•C = (2)(-1) + (4)(-4) + (-1)(-2)

A•C = -2 + -16 + 2

A•C = -16

$| A | = \sqrt{{2}^{2} + {4}^{2} + {\left(- 1\right)}^{2}}$

$| A | = \sqrt{21}$

$| C | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 4\right)}^{2} + {\left(- 2\right)}^{2}}$

$| C | = \sqrt{1 + 16 + 4}$

$| C | = \sqrt{21}$

The polar definition of the dot-product is:

A•C = |A||C|cos(theta)

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

Substituting what we have into the above:

$- 16 = \sqrt{21} \sqrt{21} \cos \left(\theta\right)$

Solve for $\theta$:

$\theta = {\cos}^{-} 1 \left(- \frac{16}{21}\right)$

$\theta \approx 2.437$ radians