What is the distance between (-1,4,-4) and (13,15,-2)?

Jan 7, 2018

$d = \sqrt{321} \approx 17.92 \text{ to 2 dec. places}$

Explanation:

$\text{using the 3-d version of the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)

$\text{let } \left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(- 1 , 4 , - 4\right)$

$\text{and } \left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(13 , 15 , - 2\right)$

$d = \sqrt{{\left(13 + 1\right)}^{2} + {\left(15 - 4\right)}^{2} + {\left(- 2 + 4\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{196 + 121 + 4}$

$\textcolor{w h i t e}{d} = \sqrt{321} \approx 17.92 \text{ to 2 dec. places}$

Jan 7, 2018