# What is the distance between (2,15,7) and (2,4,-4)?

Jun 5, 2016

The distance between $\left(2 , 15 , 7\right)$ and $\left(2 , 4 , - 4\right)$ is $15.5562$

#### Explanation:

In a two dimensional plane, distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by

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

and in three dimensional space, 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, the distance between $\left(2 , 15 , 7\right)$ and $\left(2 , 4 , - 4\right)$ is

$\sqrt{{\left(2 - 2\right)}^{2} + {\left(4 - 15\right)}^{2} + {\left(- 4 - 7\right)}^{2}}$

= $\sqrt{{0}^{2} + {\left(- 11\right)}^{2} + {\left(- 11\right)}^{2}} = \sqrt{0 + 121 + 121} = 11 \sqrt{2} = 15.5562$