# Distance between (3,-1,1) and (1,-2,0)?

Distance between (3,-1,1) and (1,-2,0)?
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blackcat1314xb
Use the 3-d version of the distance formula
$\overline{\underline{|\frac{2}{2}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}\frac{2}{2}|}}$
where $\left({x}_{1},{y}_{1},{z}_{1}\right),\left({x}_{2},{y}_{2},{z}_{2}\right)\phantom{\rule{1ex}{0ex}}\text{are 2 coordinate points}$
$\text{the 2 points here are}\phantom{\rule{1ex}{0ex}}\left(3,-1,1\right)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\left(1,-2,0\right)$
$\text{let}\phantom{\rule{1ex}{0ex}}\left({x}_{1},{y}_{1},{z}_{1}\right)=\left(3,-1,1\right),\left({x}_{2},{y}_{2},{z}_{2}\right)=\left(1,-2,0\right)$
$d=\sqrt{{\left(1-3\right)}^{2}+{\left(-2+1\right)}^{2}+{\left(0-1\right)}^{2}}$
$d=\sqrt{6}\approx 2.45\phantom{\rule{1ex}{0ex}}\text{to 2 dec. places}$