# Angles Distribution in 3D

1. Nov 11, 2008

### mabs239

Dear All,

My question goes as follows:

If we want equally distributed N-angles in a plane, then the angle between any two lines will be 360/N. e.g 5-equal angles will be calculated to 72-degs each.

Now if the same five lines (centred at one point) have to be equally spread in a sphere, then what should be the angle between any two lines?

OR

The angle between two lines of a tetrahedron is 109.47°. How is it calculated?

2. Nov 11, 2008

### HallsofIvy

Staff Emeritus
It s not correct that the angle between two lines of a tetrahedron is 109.47°. I would interpret the "lines" of a tetrahedron as being an edge and the angle between the edges is 60 degrees. Each of its faces is an equilateral triangle so each angle between two edges is 60 degrees or $\pi/3$ radians.

Do you mean lines from the center of a regular tetrahedron to its vertices, which is what you were saying at first?

I can't say this is simplest or best, but here is how I would do it off the top of my head: Set up coordinate system so that two vertices of the tetrahedron are at (0,0,0) and (1, 0, 0) and one face is in the xy-plane. Since the face is an equilateral triangle, it is easy to calculate that the third vertex of that face is at $(1/2, \sqrt{3}/2, 0)$. The fourth point of the tetrahedron is at the intersection of three spheres, centered at those points with radii 1. I won't go through the calculation here but it is easy to show that the point is $(1/2, \sqrt{3}/6, \sqrt{6}/3)$.

We can get the coordinates of the center (centroid) of a tetrahedron is just by averaging the coordinates of the vertices (that's not true of every figure but it is true of a triangle in 2 dimensions and a tetrahedron in 3). The centroid of this tetrahedron is at $(1/2, \sqrt{3}/6, \sqrt{6}/12)$.

A vector from the centroid to (0, 0, 0) is $<1/2, \sqrt{3}/6, \sqrt{6}/12>$ and a vector from the centroid to (1, 0, 0) is $<-1/2, \sqrt{3}/6, \sqrt{6}/12>$. Each of those has length $3\sqrt{6}/12$ and their dot product is -1/8. Since $\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}| cos(\theta)$, $cos(\theta)= (\vec{u}\cdot\vec{v}/(|\vec{u}||\vec{v}|)$.

Here, that is $cos(\theta)= (-1/8)(24/9)= -1/3$. The arccos of -1/3 is indeed 109.47.... degrees.

3. Nov 11, 2008

### lurflurf

working in degrees
we know among the vertices of a regular tetrahedron any 3 form a circle and the other is on a line perpendicular to the circle and passing through the center
suppose wlog that the 4 points are
P1=(0,0,1)
P2=(sin(x),0,cos(x))
P3=(sin(x)cos(120),sin(x)sin(120),cos(x))
P4=(sin(x)cos(120),-sin(x)sin(120),cos(x))
where we know (trig)
cos(120)=-1/2
sin(120)=sqrt(3)/2
now we choose x so that all the distances are equal
thus all dot products are equal in particular
P1.P2=c
P3.P4=1.5c^2-.5
where c=cos(x)
we solve
c=1.5c^2-.5
3c^2-2c-1=0
(3c+1)(c-1)
thus
c=-1/3
thus
x=Arccos(-1/3)=109.471220634 degrees

Last edited: Nov 11, 2008
4. Nov 12, 2008

### mabs239

Thak you so much for you help HallSofIvy and lurflurf,

Your explainations are great help. I am satisfied with what HallSof has said, in fact I understood it easily. You have done exactally what I was looking for. However my origional problem is to find five such angles (and may be the genral case for N angles).

I see that the symmetry of the hexagon is much help to solve it for four points. Perhaps more points would make the figure more complex to forsee in symmetry for help.

Quote: "We can get the coordinates of the center (centroid) of a tetrahedron is just by averaging the coordinates of the vertices"
I think same can be helpful for a star configuration also.

Dear lurflurf,
I have difficulty to figure out the coordinates of P1 through P4. Are these from 3-D parameteric equations of a sphere? If I put my problems:

How you come to 120deg angle?
What is c?

Anyway I appreciate your help and request for a little more elaboration.

5. Nov 17, 2008

### lurflurf

c was just cos(x) where x is the desired angle Arccos(-1/3)=109.471220634 degrees
S(x,y)=(sin(x)cos(y),sin(x)sin(y),cos(x))
is the parametric equation for a unit sphere
I let P1=S(0,0)=(0,0,1)
next we want P2,P3,and P4 to form an equilateral triangle
since each is also equidistant to P1 each will have the same 3rd coordinate (and first angle x)
Thus they lie on the circle
C(y)=S(x,y)
since they are equally spaced and 360/3=120 we can have
P2=S(x,y)
P3=S(x,y+120)
P4=S(x,y-120)
and since no generality is lost we can let y=0
P2=S(x,0)=(sin(x),0,cos(x))
P3=S(x,120)=(sin(x)cos(120),sin(x)sin(120),cos(x))
P4=S(x,-120)=(sin(x)cos(120),-sin(x)sin(120),cos(x))

now we need to solve for x it could be solved using distance, but dot products determine distance and are neater so
P1.P2=cos(x)
=c
P3.P4={(sin(x))^2}[(cos(120))^2-(sin(120)^2]+(cos(x))^2
={1-(cos(x))^2}[(cos(240)^2]+(cos(x))^2
={1-(cos(x))^2}[-.5]+(cos(x))^2
=1.5(cos(x))^2-.5
=1.5c^2-.5
we solve
c=1.5c^2-.5
3c^2-2c-1=0
(3c+1)(c-1)
thus
c=-1/3
thus
x=Arccos(-1/3)=109.471220634 degrees

Another neat thing
P1+P2+P3+P4=0
(since they are equidistant to 0)
so
0=(P1+P2+P3+P4).P1=P1.P1+P2.P2+P3.P3+P4.P1=1+3c
c=-1/3

This can be generalized but there are complications
One thing, the requirement that each point by equidistant from all others is impossible when n>4
we might however only require each point to be equidistant to some of the others
or require the inverse square forces to balance
then the n=5 case is an easy direct generalization
P1=(0,0,1)
P2=(sin(x),0,cos(x))
360/4=90
P3=(sin(x)cos(90),sin(x)sin(90),cos(x))=(0,sin(x),cos(x))
P4=(sin(x)cos(180),sin(x)sin(180),cos(x))=(-sin(x),0,cos(x))
P5=(sin(x)cos(270),sin(x)sin(270),cos(x))=(0,-sin(x),cos(x))
(algebra)
x=Arcos(-1/4)=104.477512186 Degrees

or

0=(P1+P2+P3+P4+P5).P1=P1.P1+P2.P2+P3.P3+P4.P1+P1.P5=1+4c
c=-1/4

The n=6 case is easy but different
P1=(0,0,1)
P2=(1,0,0)
P3=(0,1,0)
P4=(-1,0,0)
P5=(0,-1,0)
P6=(0,0,-1)

x=90