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Imagine I want to determine the distance between points 0,0,0 and 1,2,3.

How is this calculated?

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    No idea. For some reason they don't learn us that at school…2011-06-01

4 Answers 4

49

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 $\sqrt{\displaystyle\sum_{i=1}^n (x_i -y_i)^2}$

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    A starting point might be [this article on Wikipedia](https://en.wikipedia.org/wiki/Euclidean_distance).2018-04-19
16

Here is an illustration:

3d Pythagorean theorem 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 $d = \sqrt{x^2 + y^2 + z^2}$

11

It's Pythagorean theorem, just like with 2D space.

$||[0, 0, 0]-[1, 2, 3]|| = \sqrt{(0-1)^2+(0-2)^2+(0-3)^2} = \sqrt{1+4+9} = \sqrt{14}$

5

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 = $\sqrt{(x^2 + y^2 + z^2)}$

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    Thanks, this is by far the clearest answer for us non-mathematicians. I needed that extra bit of hand holding.2018-11-23