What is the distance between (-1,-1,-1) and (1,1,1)?

Mar 26, 2018

See a solution process below:

Explanation:

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2} + {\left(\textcolor{red}{{z}_{2}} - \textcolor{b l u e}{{z}_{1}}\right)}^{2}}$

Substituting the values from the points in the problem gives:

$d = \sqrt{{\left(\textcolor{red}{1} - \textcolor{b l u e}{- 1}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{- 1}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{- 1}\right)}^{2}}$

$d = \sqrt{{\left(\textcolor{red}{1} + \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{red}{1} + \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{red}{1} + \textcolor{b l u e}{1}\right)}^{2}}$

$d = \sqrt{{2}^{2} + {2}^{2} + {2}^{2}}$

$d = \sqrt{4 + 4 + 4}$

$d = \sqrt{12}$

$d = \sqrt{4 \cdot 3}$

$d = \sqrt{4} \sqrt{3}$

$d = 2 \sqrt{3}$

Or

$d \cong 3.464$