# What is the distance between  (–2, 1, 3)  and (3, –1, 1) ?

Feb 20, 2017

The distance between the two points is $\sqrt{33}$ or $5.745$ rounded to the nearest thousandth.

#### Explanation:

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2} + {\left(\textcolor{red}{{z}_{2}} - \textcolor{b l u e}{{z}_{1}}\right)}^{2}}$

Substituting the values from the points in the problem gives:

$d = \sqrt{{\left(\textcolor{red}{3} - \textcolor{b l u e}{- 2}\right)}^{2} + {\left(\textcolor{red}{- 1} - \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{3}\right)}^{2}}$

$d = \sqrt{{\left(\textcolor{red}{3} + \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{- 1} - \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{3}\right)}^{2}}$

$d = \sqrt{{5}^{2} + {\left(- 2\right)}^{2} + {\left(- 2\right)}^{2}}$

$d = \sqrt{25 + 4 + 4}$

$d = \sqrt{33} = 5.745$ rounded to the nearest thousandth.