Calculate distance in 3D space Imagine I want to determine the distance between points 0,0,0 and 1,2,3.
How is this calculated?
 A: By using the the Pythagorean theorem twice, you can show that $d((0,0,0),(1,2,3))=\sqrt{\left(\sqrt{1^2+2^2}\right)^2+3^2}=\sqrt{1^2+2^2+3^2}$.
In general, if you have two points $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ in $\mathbb{R}^n$, you can use the Pythagorean theorem $n-1$ times to show that the distance between them is
$$\sqrt{\displaystyle\sum_{i=1}^n (x_i -y_i)^2}$$
A: The distance between two points in three dimensions is given by:
        Given two points: point $a = (x_0, y_0, z_0)$; point $b = (x_1, y_1, z_1)$
        The distance is (in units):
         $$d = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2 + (z_1 - z_0)^2}$$
        For your given points: point $a = (0,0,0)$; point $b = (1,2,3)$
        Using substitution:
               $$d = \sqrt{(1-0)^2+(2-0)^2+(3-0)^2}$$
               $$d = \sqrt{(1 + 4 + 9)}$$
               $$d = \sqrt{(14)}$$
               $$d = 3.7$$
        Note: if one point, point a, is the origin $$(0,0,0)$$ then
              the equation reduces to d = $\sqrt{(x^2 + y^2 + z^2)}$
A: Here is an illustration:

You want to find $d$, where $d^2 = h^2 + z^2$, and $h^2 = x^2 + y^2$. So 
$d^2 = (x^2 + y^2) + z^2$, and therefore $d = \sqrt{x^2 + y^2 + z^2}$
A: It's Pythagorean theorem, just like with 2D space.
$||[0, 0, 0]-[1, 2, 3]|| = \sqrt{(0-1)^2+(0-2)^2+(0-3)^2} = \sqrt{1+4+9} = \sqrt{14}$
