# What is the distance between  (–2, 1, 3)  and (–6, 3, 1) ?

Sep 1, 2016

$2 \sqrt{6}$

#### Explanation:

Using the $\textcolor{b l u e}{\text{3-d version of the distance formula}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1} , {z}_{1}\right) \text{ and " (x_2,y_2,z_2)" are 2 coordinate points.}$

Here the 2 points are (-2 ,1 ,3) and (-6 ,3 ,1)

let $\left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(- 2 , 1 , 3\right) \text{ and } \left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(- 6 , 3 , 1\right)$

$d = \sqrt{{\left(- 6 + 2\right)}^{2} + {\left(3 - 1\right)}^{2} + {\left(1 - 3\right)}^{2}}$

$= \sqrt{16 + 4 + 4} = \sqrt{24} = = \sqrt{4 \times 6} = 2 \sqrt{6}$