# What is the distance between (–6, 3, 1)  and (–4, 0, 2) ?

Apr 22, 2016

$s = \sqrt{14}$

#### Explanation:

$A = \left(- 6 , 3 , 1\right)$
$\text{where : " A_x=-6" "A_y=3" } {A}_{z} = 1$

$B = \left(- 4 , 0 , 2\right)$
${B}_{x} = - 4 \text{ "B_y=0" } {B}_{z} = 2$

$\text{distance between (-6,3,1) and (-4,0,2) can be calculated using}$

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

$s = \sqrt{{\left(- 4 + 6\right)}^{2} + {\left(0 - 3\right)}^{2} + {\left(2 - 1\right)}^{2}}$

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

$s = \sqrt{4 + 9 + 1}$

$s = \sqrt{14}$