What is the distance between (8,1,-4) and (-3,6-2)?

Jun 9, 2016

The distance between $\left(8 , 1 , - 4\right)$ and $\left(- 3 , 6 , - 2\right)$ is $12.2475$

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 $\left(8 , 1 , - 4\right)$ and $\left(- 3 , 6 , - 2\right)$ is

$\sqrt{{\left(8 - \left(- 3\right)\right)}^{2} + {\left(1 - 6\right)}^{2} + {\left(- 4 - \left(- 2\right)\right)}^{2}}$

= $\sqrt{{11}^{2} + {\left(- 5\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{121 + 25 + 4} = \sqrt{150}$

= $\sqrt{2 \times 3 \times 5 \times 5} = 5 \sqrt{6} = 5 \times 2.4495 = 12.2475$