# Distance between (3,-1,1) and (4,1,-3)?

Distance between (3,-1,1) and (4,1,-3)?
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edytorialkp
The 3-D version of the Pythagorean Theorem tells us that that distance between two points $\left({x}_{1},{y}_{1},{z}_{1}\right)$ and $\left({x}_{2},{y}_{2},{z}_{2}\right)$ is
$\text{XXXXX}\sqrt{{\left(\Delta x\right)}^{2}+{\left(\Delta y\right)}^{2}+{\left(\Delta z\right)}^{2}}$
$\text{XXX}=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}$
In this case with points (3,-1,1) and (4,1,-3)
the distance is
$\text{XXX}\sqrt{{\left(4-3\right)}^{2}+{\left(1-\left(-1\right)\right)}^{2}+{\left(\left(-3\right)-1\right)}^{2}}$
$\text{XXX}=\sqrt{{1}^{2}+{2}^{2}+{\left(-4\right)}^{2}}$
$\text{XXX}=\sqrt{21}$