# What is the distance between (-5,13,-14) and (-11,4,1)?

Apr 15, 2016

$\implies L = 3 \sqrt{38} \text{ " ~~" } 18.493$ to 3 decimal places

#### Explanation:

Treat the way you would a triangle using Pythagoras but with 3 values instead of two.

Let the length between the two points be L

Let point 1 $\to {P}_{1} \to \left({x}_{1} , {y}_{1} , {z}_{1}\right) \to \left(- 5 , 13 , - 14\right)$
Let point 2 $\to {P}_{2} \to \left({x}_{2} , {y}_{2} , {z}_{2}\right) \to \left(- 11 , 4 , 1\right)$

Then

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

So $L = \sqrt{{\left(- 11 - \left[- 5\right]\right)}^{2} + {\left(4 - 13\right)}^{2} + {\left(1 - \left[- 14\right]\right)}^{2}}$

$L = \sqrt{36 + 81 + 225} = \sqrt{342}$

But $342 = 2 \times {3}^{2} \times 19$ but both 19 and 2 are prime numbers

$\implies L = 3 \sqrt{38}$