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

Apr 18, 2017

$\sqrt{6} \approx 2.45 \text{ to 2 dec. places}$

Explanation:

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

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

$\text{the 2 points here are " (3,-1,1)" and } \left(1 , - 2 , 0\right)$

$\text{let } \left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(3 , - 1 , 1\right) , \left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(1 , - 2 , 0\right)$

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

$\textcolor{w h i t e}{d} = \sqrt{4 + 1 + 1}$

$\textcolor{w h i t e}{d} = \sqrt{6} \approx 2.45 \text{ to 2 dec. places}$