# What is the distance between  (–2, 1, 3)  and (–2, 0, 1) ?

Apr 7, 2016

$\sqrt{5}$

#### Explanation:

By plotting this out in stages and calculating the projected images on the x,y,z planes you end up with a 3 variable equivalent of the Pythagoras theorem

Let the distance between points be $d$

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

$\implies d = \sqrt{{\left[- 2 - \left(- 2\right)\right]}^{2} + {\left[1 - 0\right]}^{2} + {\left[3 - 1\right]}^{2}}$

$\implies d = \sqrt{0 + 1 + 4}$

$\implies d = \pm \sqrt{5}$

But the negative side of $\sqrt{5}$ is not logical for this context so we are only interested in $+ \sqrt{5}$