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

Jan 30, 2016

The distance between these points is given by $r = \sqrt{{\left(0 - \left(- 4\right)\right)}^{2} + {\left(4 - 0\right)}^{2} + {\left(\left(- 2\right) - 2\right)}^{2}}$ and is $4 \sqrt{3}$ or $6.93$ units.

#### Explanation:

The distance, $r$, between two points in 3 dimensions is given by:

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

Substituting in the coordinates for the two points given:

$r = \sqrt{{\left(0 - \left(- 4\right)\right)}^{2} + {\left(4 - 0\right)}^{2} + {\left(\left(- 2\right) - 2\right)}^{2}}$
= $\sqrt{{\left(- 4\right)}^{2} + {\left(4\right)}^{2} + {\left(- 4\right)}^{2}}$
= $\sqrt{16 + 16 + 16} = \sqrt{48} = 4 \sqrt{3} = 6.93$

Jan 30, 2016

6.928

#### Explanation:

suppose,
${x}_{1} = - 4$

${y}_{1} = 0$

${z}_{1} = 2$

${x}_{2} = 0$

${y}_{2} = 4$

${z}_{2} = - 2$

now, if we find out the position vector of the two points for main point O(0,0,0), we get,
$\vec{O A} = - 4 i + 2 k$

$\vec{O B} = 4 j - 2 k$

we know,

$\vec{A B} = \vec{O B} - \vec{O A}$

$= \left(4 j - 2 k\right) - \left(4 i + 2 k\right)$

$= - 4 i + 4 j - 2 k - 2 k$

$= - 4 i + 4 j - 4 k$

so, the distanse is,

$| \vec{A B} | = \sqrt{{\left(- 4\right)}^{2} + {4}^{2} + {\left(- 4\right)}^{2}}$

$= \sqrt{48}$

$= 6.928$