What is the distance between (–6, 3, 4)  and (–5, –1, 1) ?

Dec 19, 2017

$\sqrt{26}$

Explanation:

You may be familiar with the two-dimensional distance formula, which tells us that the distance between $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is:

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

There is a similar formula for three dimensions for the distance between $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$, namely:

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

So in our example, the distance between $\left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(- 6 , 3 , 4\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(- 5 , - 1 , 1\right)$ is:

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

$= \sqrt{{\left(\left(- 5\right) - \left(- 6\right)\right)}^{2} + {\left(\left(- 1\right) - 3\right)}^{2} + {\left(1 - 4\right)}^{2}}$

$= \sqrt{{1}^{2} + {\left(- 4\right)}^{2} + {\left(- 3\right)}^{2}}$

$= \sqrt{1 + 16 + 9}$

$= \sqrt{26}$