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The diagonals of a square intersect at a right angle.
Is that true in three dimensions? I.e. would the two diagonals of a cube, each running from one corner of the cube to its opposite corner and crossing in the center, also cross at right angles?

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  • $\begingroup$ Any two of them, but you already covered this! $\endgroup$
    – ThomasL
    Jul 23 '20 at 19:40
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    $\begingroup$ Depends on the point of view, any angle can vary between its planar normal value (obeserved normal to the plane it is inscribed in) and 180° (observed from coplanar point of view) $\endgroup$
    – qq jkztd
    Feb 1 at 15:57
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There are four diagonals of a cube, but none of them meet at right angles:

enter image description here image source

To see this, observe that

any two diagonals have four endpoints which can be split into two pairs, each pair of endpoints sharing one edge of the cube.

The two diagonals define a unique plane containing both of them. In that plane, the two cube diagonals are also the diagonals of a rectangle,

whose side lengths are respectively $1$ and $\sqrt{2}$ times the side-length of the cube. That's not a square, so the diagonals are not perpendicular.

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  • $\begingroup$ what about an n-dimensional cube? $\endgroup$ Jul 24 '20 at 6:26
  • $\begingroup$ @DmitryKamenetsky I guess the angle will get further and further from 90 degrees as the dimension increases. $\endgroup$ Jul 24 '20 at 6:35
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    $\begingroup$ @DmitryKamenetsky it will depend on the pair of diagonals you choose for even n, and is impossible for odd n. If you are familiar with vectors, each diagonal can be represented by an n-d vector whose values each 1 or -1. Then for n odd, the dot product of two diagonals is always the sum of an odd number of 1s and -1s and is hence odd and !=0, thus no two diagonals are perpendicular. For n even, then the diagonals corresponding to vectors (1,1,...,1) and (1,-1,1,-1,...,-1) have dot product 0 and are thus perpendicular. $\endgroup$
    – boboquack
    Jul 24 '20 at 7:10

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