# What is the distance between (–2, 2, 6)  and (–1, 1, 3) ?

Jul 31, 2016

Distance between $\left(- 2 , 2 , 6\right)$ and $\left(- 1 , 1 , 3\right)$ is $\sqrt{11} = 3.317$

#### Explanation:

The distance between two points $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by

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

Hence distance between $\left(- 2 , 2 , 6\right)$ and $\left(- 1 , 1 , 3\right)$ is

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

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

= $\sqrt{{1}^{2} + 1 + 9}$

= $\sqrt{11}$

= $3.317$

Jul 31, 2016

The distance between to points $\left({x}_{1} , {x}_{2} , {x}_{3}\right)$ and $\left({y}_{1} , {y}_{2} , {y}_{3}\right)$ in the 3-dimensional space is sqrt((x_1-y_1)^2+ (x_2-y_2)^2+ (x_3-y_3)^2

#### Explanation:

Now simply replacing $\left({x}_{1} , {x}_{2} , {x}_{3}\right)$ and $\left({y}_{1} , {y}_{2} , {y}_{3}\right)$ by the values given we have:

$\sqrt{{\left(- 2 + 1\right)}^{2} + {\left(2 - 1\right)}^{2} + {\left(6 - 3\right)}^{2}} = \sqrt{{\left(- 1\right)}^{2} + {\left(1\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{11}$

which is the distance