Imagine I want to determine the distance between points 0,0,0 and 1,2,3.
How is this calculated?
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asked Jun 1, 2011 at 19:57
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1$\begingroup$ How would you do it in two dimensions? $\endgroup$– JavaManCommented Jun 1, 2011 at 20:00
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$\begingroup$ No idea. For some reason they don't learn us that at school… $\endgroup$– Simon VerbekeCommented Jun 1, 2011 at 20:12
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4 Answers
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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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$\begingroup$ How can we prove the last formula? I find it difficult to imagine a n-dimensional space geometrically when $n > 3$... $\endgroup$ Commented Apr 19, 2018 at 10:30
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2$\begingroup$ @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. $\endgroup$– AaronCommented Apr 19, 2018 at 15:31
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$\begingroup$ Thank you, where can I find some theory and examples about this topic? $\endgroup$ Commented Apr 19, 2018 at 20:55
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$\begingroup$ @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. $\endgroup$– AaronCommented Apr 19, 2018 at 21:30
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1$\begingroup$ A starting point might be this article on Wikipedia. $\endgroup$– AaronCommented Apr 19, 2018 at 21:37
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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 $d = \sqrt{x^2 + y^2 + z^2}$
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It's Pythagorean theorem, just like with 2D space.
$||[0, 0, 0]-[1, 2, 3]|| = \sqrt{(0-1)^2+(0-2)^2+(0-3)^2} = \sqrt{1+4+9} = \sqrt{14}$
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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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1$\begingroup$ Does d equal exactly
3.7? I get3.7416573867739413doing the calculation in Python. $\endgroup$ Commented Feb 9, 2015 at 7:15 -
$\begingroup$ Thanks, this is by far the clearest answer for us non-mathematicians. I needed that extra bit of hand holding. $\endgroup$ Commented Nov 23, 2018 at 15:01