What is the distance between (3,15,-2) and (-4,15,4)?

Feb 16, 2016

$\sqrt{85} \approx 9.2195$

Explanation:

Use the three-dimensional distance formula, which can be derived by applying the Pythagorean Theorem twice. Given two points in 3-dimensional space written in rectangular coordinates $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$, the distance is:

$\mathrm{di} s t = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2} + {\left({z}_{1} - {z}_{2}\right)}^{2}}$

For the problem at hand, this becomes

$\mathrm{di} s t = \sqrt{{7}^{2} + {0}^{2} + {6}^{2}} = \sqrt{49 + 36} = \sqrt{85} \approx 9.2195$.