# What is the distance between (3, –1, 1)  and (–4, 0, 2) ?

Feb 13, 2016

We must calculate distance as the usual way, using generalized Pythagoras theorem.

#### Explanation:

For generalized Pythagoras theorem, we have:

${d}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}$

where $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ are both points.

Hence:
${d}^{2} = {\left(- 4 - 3\right)}^{2} + {\left(0 - \left(- 1\right)\right)}^{2} + {\left(2 - 1\right)}^{2} = 51$

And taking square roots:
$d = \sqrt{51}$