# What is the distance between (–6, 3, 1)  and (4, 4, 2) ?

Feb 16, 2017

The distance between $\left(- 6 , 3 , 1\right)$ and $\left(4 , 4 , 2\right)$ is $12.083$

#### Explanation:

Just like $\left(x , y\right)$ represents a point on a plane i.e. in $2$-dimension,

$\left(x , y , z\right)$ represents a point in $3$-dimension.

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

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

Therefore, distance between $\left(- 6 , 3 , 1\right)$ and $\left(4 , 4 , 2\right)$ is

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

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

= $\sqrt{{12}^{2} + {1}^{2} + {1}^{2}} = \sqrt{144 + 1 + 1} = \sqrt{146} = 12.083$