# Irregular tetrahedron (coordinate of vertices)

1. Aug 1, 2012

### Sebtimos

This is my first post and I wish to get help in finding an analytically way to get the coordinate of an irregular tetrahedron.

let ABCD be the 4 vertices of the tetrahedron in 3D, all vertices have different (x,y,z).

the coordinate of vertex D is known (Xd,Yd,Zd), and the 3 angle between faces at vertex D are also known angle adb , adc, cdb.

Coordinate of the other 3 vertices A, B, C are known on X, Y but not on Z. (ie Za,Zb,Zc are unknown).

As I can expect this system will give me 2 solutions +&-.
I've started by developing system of equation for the sides using cosines and as a function of Za,Zb,Zc, where (Zb-Za)^2= [Lda^2+Ldb^2-2Lda*Ldb*cos(adb)]-[(xb-xa)^2+(Yb-Ya)^2]

Any suggestions on how to approach this problem would be appreciated.

2. Aug 2, 2012

### haruspex

Lda = √((Xd-Xa)2+ etc.) etc.
So you can get three equations involving Za, Zb, Zc. For each pair of unknowns, Z1, Z2, you get an equation of the form
Squaring those gives you three quartics. If lucky, there may be some useful cancellation.
There may be a better way, but I doubt it.

3. Aug 2, 2012

### coolul007

The vertices should all be on the surface of a sphere. I don't know if that makes the problem easier, just a thought.

4. Aug 2, 2012

### haruspex

Why? We're told it is irregular (or, at, least, not necessarily regular).

5. Aug 2, 2012

### Sebtimos

Thank you for the contribution,
coolul007: unfortunately its not the case in irregular tetrahedron.

haruspex: It looks like I must go through these equations and see what i get, they look very ugly though.
do you think adding another equation using the volume of the tetrahedron, which is 1/6 of the volume of parallelepiped, with α, β, and γ are the internal angles between the edges, the volume is equations are given by ( http://en.wikipedia.org/wiki/Parallelepiped )
V=1/6 [Lad.Lbd.Lcd * √(1+2cos(α) cos(β) cos(γ)-cos(α)^2-cos(β)^2-cos(γ)^2)]

it is also equivalent to the absolute value of the determinant of a three dimensional matrix built using a, b and c:
V=1/6 [a.(b*c)]

thanks for the help

6. Aug 2, 2012

### haruspex

My experience of 3D trig is that quartics are par for the course. Working with volumes is likely to make matters worse, introducing 6th powers.

7. Aug 3, 2012

### coolul007

I hope I am not riding a dead horse here, but it seems that the bottm triangle is on a sphere and all that is left is the vertex to intersect with a sphere. I haven't tried finding it yet but it seems there has to be one.

8. Aug 3, 2012

### haruspex

Sorry, you're right. Just as every triangle defines a circle through its vertices, so there is a unique sphere through the vertices of a tetrahedron, regular or not.
However, I don't see that it helps.

9. Aug 3, 2012

### Sebtimos

Yes, coolul007 is correct, that would have helped alot if the center of the sphere can be defined.