# What is the distance between  (–2, 1, 3)  and (2, 0, –1) ?

Jul 11, 2016

$s = \sqrt{33} \text{ unit}$

#### Explanation:

$A = \left(- 2 , 1 , 3\right)$
${A}_{x} = - 2 \text{ ; "A_y=1" ; } {A}_{z} = 3$

$B = \left(2 , 0 , - 1\right)$
${B}_{x} = 2 \text{ ; "B_y=0" ; } {B}_{z} = - 1$

$\text{the distance between A and B can be calculated using:}$

$s = \sqrt{{\left({B}_{x} - {A}_{x}\right)}^{2} + {\left({B}_{y} - {A}_{y}\right)}^{2} + {\left({B}_{z} - {A}_{z}\right)}^{2}}$

$s = \sqrt{{\left(2 + 2\right)}^{2} + {\left(0 - 1\right)}^{2} + {\left(- 1 - 3\right)}^{2}}$

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

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

$s = \sqrt{33} \text{ unit}$