# What is the distance between (3, –1, 1)  and (4, 1, –3) ?

Jan 4, 2016

$\sqrt{21}$

#### Explanation:

The 3-D version of the Pythagorean Theorem tells us that that distance between two points $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is
$\textcolor{w h i t e}{\text{XXXXX}} \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} + {\left(\Delta z\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

In this case with points $\left(3 , - 1 , 1\right)$ and $\left(4 , 1 , - 3\right)$
the distance is
$\textcolor{w h i t e}{\text{XXX}} \sqrt{{\left(4 - 3\right)}^{2} + {\left(1 - \left(- 1\right)\right)}^{2} + {\left(\left(- 3\right) - 1\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{1}^{2} + {2}^{2} + {\left(- 4\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \sqrt{21}$