tetrahedron-4th-vertex

Demonstration of finding the 4th vertex of a tetrahedron given its base and distances to vertex

Problem definition

Consider a Tetrahedron in the regular $0\vec{i}\vec{j}\vec{k}$ cartesian space defined by:

Points P, Q and R
Distances $||\vec{PE}||$ , $||\vec{QE}||$ and $||\vec{RE}||$ from the respectives points P, Q and R to the unknown 4th vertex E

The problem consists on finding the (x, y, z) coordinates of the point E.

Solution

Although there are other solutions using intersections of spheres or planes, this problem can be solved by regular trigonometric approach.

Note that θ is the angle between the unknow vector $\vec{PE}$ and the plane defined by the points P, Q and R. σ is the angle between the projection of $\vec{PE}$ on the same plane PQR and the vector $\vec{PR}$ .

As the image suggests, σ and θ can be obtained in a straightforward way from the tetrahedron height and elementary trigonometric properties, as shown below.

Finding $\vec{PE}$ angles σ and θ

Find the tetrahedron Volume using Calyer-Menger determinant:

$288 Volume^2 = \left|\begin{matrix}0 & 1 & 1 & 1 & 1\cr 1 & 0 & ||\vec{RE}||^{2} & ||\vec{PE}||^{2} & ||\vec{QE}||^{2}\cr 1 & ||\vec{RE}||^{2} & 0 & \tilde||\vec{QE}||^{2} & \tilde||\vec{PE}||^{2}\cr 1 & ||\vec{PE}||^{2} & \tilde||\vec{QE}||^{2} & 0 & \tilde||\vec{RE}||^{2}\cr 1 & ||\vec{QE}||^{2} & \tilde||\vec{PE}||^{2} & \tilde||\vec{RE}||^{2} & 0\end{matrix}\right|$

Find the Area of triangle P, Q, R using Heron's formula:

$Area using Heron's formula$

Find the tetrahedron height H using the relationship between Volume and Area:

$H=\frac{3\times Volume}{Area}$

Find θ:

$obtaining theta from height$

Once we have θ the next step is to find the length of the projections $\vec{PE'}$ and $\vec{RE'}$ onto the plane defined by P, Q and R:

$||\vec{PE'}||=\sqrt{||\vec{PE}||^2-H^2}$

$||\vec{RE'}||=\sqrt{||\vec{RE}||^2-H^2}$

Thus, using the Law of Cosines, σ is given by:

$\sigma = arccos\left (\frac{||\vec{PE'}||^2 - ||\vec{RE'}||^2 + ||\vec{PR}||^2}{2 ||\vec{PE'}|| \times ||\vec{PR}||}\right )$

Once we have P, $||\vec{PE}||$ , σ and θ we know everything we need to find E.

Finding E given σ, θ, P and $||\vec{PE}||$

There are several ways to obtain E(x, y, z), one of them is rotating $\vec{PR}$ by σ and then rotating again by θ, as demonstrated below.

Find the PQR triangle normal $\vec{n}$ :

$\vec{n} = \frac{\vec{PR}\times\vec{PQ}}{||\vec{PR}|| \times ||\vec{PQ}||}$

Rotate $\vec{PR}$ about $\vec{n}$ by -σ using Rodrigues' formula:

$\vec{PE'} = \vec{PR}cos(-\sigma) + (\vec{n} \times \vec{PR})\sin(-\sigma) + \vec{n}(\vec{n} \cdot \vec{PR}) (1 - cos(-\sigma))$

Find the normal $\vec{m}$ from $\vec{PE'}$ and $\vec{n}$ :

$\vec{m} = \frac{\vec{PE'}\times\vec{n}}{||\vec{PE'}|| \times ||\vec{n}||}$

Rotate $\vec{PE'}$ by -θ about $\vec{m}$ :

$\vec{PE_{dir}} = \vec{PE'}cos(-\theta) + (\vec{m} \times \vec{PE'})\sin(-\theta) + \vec{m}(\vec{m} \cdot \vec{PE'}) (1 - cos(-\theta))$

Get the unit vector from $\vec{PE_{dir}}$ and multiply it by $||\vec{PE}||$ in order to obtain $\vec{PE}$ :

$\vec{PE} = \frac{\vec{PE_{dir}}}{||\vec{PE_{dir}}||} \times ||\vec{PE}||$

Finally, E is given by

$E = \vec{PE} + P$

It is noteworthy that the symmetric solution $E_2$ can be find by rotating $\vec{PE'}$ about $\vec{m}$ by +θ (instead of -θ):

Different approaches

This same problem can be solved by finding the intersection of the 3 spheres centered on P, Q and R and respective radius $||\vec{PE}||$ , $||\vec{QE}||$ and $||\vec{RE}||$ .

Another approach is by finding the intersection of planes PER, PQE and REQ.

Future work

One of my future work is checking out if this approach is less computational intensive than others.

Acknowledgements

I would like to thanks to github markup team for not provide a simple way to write math expressions on markdown documents. Thus, in order to provide the formulas I ended up using this formula image generator: https://www.codecogs.com/latex/eqneditor.php. For more details see github/markup#897.

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