Diagonals of a cube

I understand that using the dot product, the diagonals of a cube can be shown to intersect at an angle of

\cos^{-1}(\frac{1}{3})

but I don't get why it's not a right angle? If two lines intersect at anything other than a right angle, they form two angles (an acute one and an obtuse one). But since all vertices of a cube are equivalent, how can there be two different angles formed? There would be no way of choosing two different viewpoints of the cube to be able to distinguish two different angles from?

Reply 1

nuodai

Say all the diagonals intersected at right-angles. Then there would be four mutually perpendicular lines... if only that were possible in three dimensions! If you have any cubes handy it's fairly easy to convince yourself that the angles can't be right angles just by playing around with it.

Anyway yes, an acute and an obtuse angle are formed, but that still makes sense... I don't quite understand your thing about changing viewpoint. If you change your pair of diagonals, the same two angles are formed, thanks to the nature of three dimensions. If you tried the same thing with a tesseract, you'd find that if you fix one diagonal, then another diagonal can make an angle of either

\cos^{-1} \dfrac{1}{4}

or a right-angle with that diagonal.

(edited 13 years ago)

Reply 2

Plato's Trousers

So, using cartesian coords, let's say my first diagonal goes from the origin (0,0,0) to (1,1,1) Does it matter if i choose my second diagonal as (1,0,0) to (0,1,1) or as (0,1,0) to (1,0,1)?

there are only two types of diagonal in a cube, right? A face diagonal and a body diagonal?

Reply 3

nuodai

Original post by Plato's Trousers

Indeed.

So say you've got the diagonal from (0,0,0) to (1,1,1) fixed. Then there are three other diagonals to choose from:
(i) (1,0,0) to (0,1,1)
(ii) (0,1,0) to (1,0,1)
(iii) (0,0,1) to (1,1,0)

In each case, the pair of diagonals will pass through four vertices on the cube. These come in two pairs of vertices, which define two opposite edges of the cube. The remaining four edges have no lines through them; this is essentially where the lack of complete symmetry comes from here. Also, because no matter which diagonal you pick, the vertices will always lie on an edge adjacent to one of the vertices already "hit" by the diagonal from (0,0,0) to (1,1,1), it makes no difference which other diagonal you pick.

In four dimensions, say you fix the diagonal from (0,0,0,0) to (1,1,1,1). Then there are six other diagonals to choose from to get a pair from, three of which make an angle

\cos^{-1} \dfrac{1}{4}

and three of which form right-angles. This is because there is no longer a need for adjacency.

(edited 13 years ago)

Reply 4

Plato's Trousers

Original post by nuodai

\cos^{-1} \dfrac{1}{4}

and three of which form right-angles. This is because there is no longer a need for adjacency.

Eureka! Thank you. Your explanation does it.

Basically, the ends of the diagonals end up either on an edge (which in a unit cube would be 1 unit long) or on a face diagonal (with lenth sqrt 2), so there will be two different angles.

:biggrin: