Elianna Lawrence

2022-07-23

Distance between (-6,3,1) and (2,-3,1)?

grocbyntza

Expert

using the 3-d version of the distance formula
$\overline{\underline{\mid \frac{2}{2}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}\frac{2}{2}\mid }}$
where $\left({x}_{1},{y}_{1},{z}_{1}\right),\left({x}_{2},{y}_{2},{z}_{2}\right)\phantom{\rule{1ex}{0ex}}\text{are 2 coordinate points}$
$\text{the 2 points here are}\phantom{\rule{1ex}{0ex}}\left(-6,3,1\right)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\left(2,-3,1\right)$
$\text{let}\phantom{\rule{1ex}{0ex}}\left({x}_{1},{y}_{1},{z}_{1}\right)=\left(-6,3,1\right),\left({x}_{2},{y}_{2},{z}_{2}\right)=\left(2,-3,1\right)$
$d=\sqrt{{\left(2+6\right)}^{2}+{\left(-3-3\right)}^{2}+{\left(1-1\right)}^{2}}$
$d=\sqrt{64+36+0}$
$d=\sqrt{100}=10\phantom{\rule{1ex}{0ex}}\text{units}$

Do you have a similar question?