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

How is this calculated?

  • 1
    How would you do it in two dimensions? – JavaMan Jun 1 '11 at 20:00
  • No idea. For some reason they don't learn us that at school… – Simon Verbeke Jun 1 '11 at 20:12
up vote 42 down vote accepted

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

  • How can we prove the last formula? I find it difficult to imagine a n-dimensional space geometrically when $n > 3$... – user3019105 Apr 19 at 10:30
  • 2
    @user3019105 We use induction, together with the fact that all the coordinate axes are orthogonal to each other. So the way the z-axis is orthogonal to the xy-plane (and hence to any line that has constant z-coordinate) generalizes. Or rather, that is how we define the geometry of higher dimensional space. – Aaron Apr 19 at 15:31
  • Thank you, where can I find some theory and examples about this topic? – user3019105 Apr 19 at 20:55
  • @user3019105 unfortunately, I don't know any references that talk about this in elementary terms, explaining why things are like they are. This is a basic example of a metric space (the Euclidean metric), and the metric is induced by a norm ($\ell^2$ norm), and that norm comes from an inner product, so this is an example in a lot of places, and I can give you references for generalizations, but I don't think they are what you are looking for. – Aaron Apr 19 at 21:30
  • Whatever you have, if you could share it, I will be thankful and take a look! – user3019105 Apr 19 at 21:33

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

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

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

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

  • For some basic information about writing math at this site see e.g. here, here, here and here. – Chantry Cargill Dec 15 '14 at 20:40
  • 1
    Does d equal exactly 3.7? I get 3.7416573867739413 doing the calculation in Python. – ThorSummoner Feb 9 '15 at 7:15

protected by Zev Chonoles Jan 7 '16 at 17:38

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.