Logo for NNM (Nemo Nisi Mors) Lake Runn, December 2008 (Photo: Anders Gustafson) Logo for NNM (Nemo Nisi Mors)
Revised

What is the relation between the length of the space diagonal \(a\) and the lengths of the edges \(x\), \(y\) and \(z\) in a three-dimensional box?

Figure 1: Space diagonal \(a\) in a box with sides \(x\), \(y\) and \(z\).

If we draw the face diagonal \(b\) for the rectangle defined by the edges \(x\) and \(y\), we can see that we have a right triangle (blue below) with the sides \(x\), \(y\) and the hypotenuse \(b\).

Figure 2: Space diagonal \(a\) and face diagonal \(b\) in a box, where \(b^2 = x^2 + y^2\).

Now we can see that the space diagonal \(a\) is the hypotenuse of a right triangle (red below), with the sides \(b\) and \(z\).

Figure 3: Space diagonal \(a\) and face diagonal \(b\) in a box, where \(a^2 = b^2 + z^2\).

Thanks to Pythagoras, we have these relations between the variables

\[b^2 = x^2 + y^2\]
\[a^2 = b^2 + z^2 = x^2 + y^2 + z^2\]
\[a = \sqrt{x^2 + y^2 + z^2}\]

So the length of the space diagonal in a box is expressed by a simple equation between the length of the three edges.

Diagonal in a cube

In the special case of a cube all edges are equal (\(x\) = \(y\) = \(z\))

\[a^2 = x^2 + x^2 + x^2 = 3x^2\]
\[a = \sqrt{3x^2} = x\sqrt{3}\]

Note: If you want to bring your 165 cm skis when traveling, but the maximum package dimension for a side is 100 cm, you can solve your problem by putting the skis in a 100x100x100 cm box.

Diagonal in a square

We leave the three-dimensional geometry and examine the two-dimension instead. For a square with edges of length \(x\), the face diagonal \(b\) is

\[b^2 = x^2 + x^2 = 2x^2\]
\[b = \sqrt{2x^2} = x\sqrt{2}\]

Diagonal in a line

Now to the one-dimension. For a line of length \(x\), the "diagonal" \(c\) can be expressed as

\[c^2 = x^2 = 1x^2\]
\[c = \sqrt{1x^2} = x\sqrt{1}\]

Note: The length of the "diagonal" of a line is of course equal to the length of the line itself, but we want to follow the pattern from the previous equations.

Diagonal in an n-dimensional hypercube

It seems to be so nice, that for the \(n\)-dimensional hypercube with edges of length \(x\), the diagonal \(d\) is calculated by

\[d = x\sqrt{n}\]
First published by Anders Gustafson 2018-03-03
Logo for NNM
NNM Logo
Friends in need
CCF Cheetah Conservation Fund
NNM Clock
Random quote
Indeed, astrology is not a science at all, but merely a collection of superstitions and hokum that tries to use some of the terminology of astronomy but which rejects the logical thinking that is at the heart of science.
Kaufmann and Freedman