Determining the Angle between the Apex Vectors in an n-d Pyramid

Upon lacking a better word, we are widening the use of the word ‘pyramid’ to name a regular triangle- or tetrahedron-type geometric object of any dimension n. By ‘regular’ we mean that all edges, sides, angles, surface areas, volumes, etc, whatever there might be in the object, they are all similar and of the same size and shape and texture, correspondingly. And by triangle- or tetrahedron-type we mean that there are no curved or other kind of body parts except straight ones, and further that the object is composed of only the least possible number of any body parts in order to make it an n-d body. That is to say, an n-dimensional pyramid has exactly n+1 apexes. If it has less apexes, then it would be a pyramid of smaller dimension instead. But it could not have more apexes, because all of the angles could not be equal any more, as any new apex should always point towards a new dimension.

So, our 2-d pyramid is a triangle with equal sides and equal angles, and our 3-d pyramid is the classic regular real tetrahedron. 4-d and up pyramids are more hard to imagine, but let us believe that they exist at least in our minds. By an apex vector we mean a vector drawn from the center point of the pyramid to any of the apexes. There are n+1 apex vectors in an n-d pyramid.

The purpose of this study is to determine the angle between any two of the apex vectors. Obviously, the regularity of the pyramid implies that all of the said angles are the same. The urge to this study rises from the fact that the H-O-H angle of the water molecule is known to be about 104.5°.

We are using a vector analysis technology. It is worth acknowledging that vector analysis is not regular mathematics, but rather a technological application solely in a three dimensional space. For instance, there is no content for the cross product of vectors, except in a 3-d space. However, being a scalar quantity, the dot product of two vectors is readily and without loss of conceptuality applicaple to any dimension. Namely, while the dot product applies the magnitudes of the vectors and the angles between, we may always be able to turn the object in such a way that the two vectors lie on our paper ahead, one by one.

The figure on the left shows a 2-d pyramid with three apex vectors A1, A2, A3. The figure on the right shows the same vectors when summed together, indicating that the sum is zero. This fact follows from the regularity of the pyramid (a separate study on the back burner is referred).

A1

A2

x A1

A2

A3

A3

More obvious is that the angles between the vectors are the same. The angle is named x, as it is the quantity we are looking for. While the figure shows a 2-d case, we will assume a general nd case without drawing a figure for each case separately. Accordingly, we have n+1 apex vectors A1, A2, A3, ... ,An+1 with mutual angles x = (A1,A2) = (A1,A3) = ... = (An,An+1).

The regularity also infers that the lengths of all apex vectors are the same; that would be =L.

We first state the sum of all of the apex vectors. As mentioned earlier, it is zero:

A1 + A2 + A3 + ... + An+1 = 0

When multiplying this equation by A1• we obtain the following:

A1•A1 + A1•A2 + A1•A3 + ... + A1•An+1 = A1•0

The dot product (scalar product) of any two vectors Ai and Ak(unless i = k) is defined as

Ai•Ak = Ai Ak cos(Ai,Ak) = L2 cos x

Ai’s are the lengths of the vectors Ai, respectively. For i = k = 1, the dot product = L2. Substituting the definition into the equation above yields:

L2 + n L2 cos x = 0

That will provide us with a solution:

cos x = -1/n

The following shows the angle for some dimensions:

n = 1: x = 180°

n = 2: x = 120°

n = 3: x = 109.4712°

n = 4: x = 104.4775°

n = 5: x = 101.537°

n : x 90°

Back burner. A1 + A2 + A3 + ... + An+1 = 0. One might wonder, wheather this is true. We’ll next show that it follows from the regularity of the pyramid, that is:

(A1,A2) = (A1,A3) = ... = (An,An+1). and Ai = Ak for all i, k.

Denoting A1 + A2 + A3 + ... + An+1 = Z we try to show that Z = 0. Multiplying it by A1• and A2•, and then subtracting, we have:

A1•A1 + A1•A2 + A1•A3 + ... + A1•An+1 = A1•Z

A2•A1 + A2•A2 + A2•A3 + ... + A2•An+1 = A2•Z

0 = (A1 - A2) • Z

Similarly, we do the same for all pairs of any two vectors, and have for all pairs of unequal i, k:

0 = (Ai - Ak) • Z

We must have all n(n+1)/2 equations simultaneously effective. Now, none of the difference vectors Ai - Ak is a null vector. If Z is not a null vector, then it must be perpendicular to all of the non-zero difference vectors simultaneously. It is only possible, if Z points out to a new dimension out of the sub-space, where the other vertors and their differences are located. However, Z is the sum of the vectors Ai, so it must lie in the same sub-space as vectors Ai. While Z can lie in the sub-space and point out of it at the same time only, if it is a null vector, we conclude the case shown.

The opposite is not valid, which is shown in 3-d case aside. Given the four equal-length vectors Ai, the sum of which is zero, it does not require their six mutual angles to be equal. They could be positioned in many ways. Here the vectors are located in a 2-d plane. Four angles are 90°, the remaining two are 180°. -ay