# What is the distance between  (–6, 3, 1)  and (–1, 4, –2) ?

Apr 6, 2018

$\sqrt{35}$

#### Explanation:

The (Euclidean) distance between two points $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by the formula:

$\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

So for $\left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(- 6 , 3 , 1\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(- 1 , 4 , - 2\right)$ the distance is:

$\sqrt{{\left(\left(\textcolor{b l u e}{- 1}\right) - \left(\textcolor{b l u e}{- 6}\right)\right)}^{2} + {\left(\left(\textcolor{b l u e}{4}\right) - \left(\textcolor{b l u e}{3}\right)\right)}^{2} + {\left(\left(\textcolor{b l u e}{- 2}\right) - \left(\textcolor{b l u e}{1}\right)\right)}^{2}}$

$= \sqrt{{5}^{2} + {1}^{2} + {\left(- 3\right)}^{2}} = \sqrt{25 + 1 + 9} = \sqrt{35}$