Sunset Venice Beach (Florida, United States), December 2014 (Photo: Anders Gustafson)
Modified 2021-02-21T12:31:27Z

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
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