# Define Tetrahedron knowing one vertex, 3 vectors, opposite face.

by powederhound
Tags: define, knowing, tetrahedron, vectors, vertex
 Sci Advisor HW Helper PF Gold P: 3,289 Hopefully I've interpreted your problem correctly. If not, perhaps you can attach a picture showing the situation. If we take the origin (your point D) to be the zero vector, and ##a##, ##b##, and ##c## are the vectors from the origin to the three prisms, you know the unit vectors ##a/\|a\|##, ##b/\|b\|## ,and ##c/\|c\|##. And you also know the vectors between each pair of prisms, namely ##a-b##, ##b-c##, and ##a-c##. Your goal is to find ##\|a\|##, ##\|b\|##, and ##\|c\|##. Let's introduce some notation for the known unit vectors: ##u_a = a/\|a\|##, ##u_b = b/\|b\|##, and ##u_c = c/\|c\|##, and for the known difference vectors: ##v_1 = a - b##, ##v_2 = b - c##, and ##v_3 = a-c##. Then we have: \begin{align} u_a \|a\| - u_b \|b\| &= v_1 \\ u_b \|b\| - u_c \|c\| &= v_2 \\ u_a \|a\| - u_c \|c\| &= v_3 \\ \end{align} One possible way to solve this would be to take the inner (dot) product of each of these equations with each of ##u_a##, ##u_b##, and ##u_c## to obtain a system of 9 equations and 3 unknowns, of the form $$U\begin{bmatrix}\|a\|\\\|b\|\\\|c\|\end{bmatrix} = V$$ where ##U## is ##9 \times 3## and ##V## is ##9 \times 1##. If your measurements were perfect, six of the nine equations would be redundant, but in practice, they will be slightly inconsistent, so you should aim for a least-squares solution. One way to obtain such a solution is to multiply both sides by ##U^T##: $$U^TU\begin{bmatrix}\|a\|\\\|b\|\\\|c\|\end{bmatrix} = U^TV$$ Now ##U^T U## is ##3 \times 3## and positive semidefinite since for any ##x## we have ##x^T U^T U x = ||Ux||^2 \geq 0##. It will be positive definite provided that the null space of ##U## has dimension zero (##Ux = 0## implies ##x = 0##), in which case it can be inverted to obtain a solution: $$\begin{bmatrix}\|a\|\\\|b\|\\\|c\|\end{bmatrix} = (U^T U)^{-1} U^T V$$ An equivalent condition for the null space of ##U## to be zero is that ##U## must have full rank (3). In other words, at least three of the nine equations must be linearly independent. This will almost surely be the case based on the geometry have you described.
 P: 4 Thank you for your quick response! I think there may be one clarification necessary, although it may be my lack of understanding that is driving the confusion. When you set up your "difference vectors" you establish that v1 = a - b, v2 = b - c, and v3 = a - c. While that certainly can be done, we still don't know the magnitude of the vectors a, b, and c (we only know their direction as provided by the surveyor.) I do however know (to a high degree of precision) the magnitudes of the vectors v1, v2 and v3 from the CMM measurements that were taken on my prism housing (that houses all three prisms.) The way you set it up, v1, v2 and v3 are now vectors indicating the separation of the survey prisms which we know the magnitude of (from CMM), but not the direction (because the prism mounting assembly has been placed at an unknown location within the surveyor reference frame.) So when you set up your linear equation: $U \begin{pmatrix} ||a|| \\ ||b|| \\ ||c|| \end{pmatrix} = V$ You now have unknowns both in the matrix containing the magnitudes of a, b, and c, as well as in the Matrix V. The rest of the linalg is pretty straightforward but I'm not sure it will provide the result we are looking for. Thanks again for your help. *bit of a beginners learning curve on that mathtype :)
 Sci Advisor HW Helper PF Gold P: 3,289 Do you know the angles between the ##u## and ##v## vectors, by any chance? If so, then we can start with these equations: \begin{align} u_a \|a\| - u_b \|b\| &= v_1 \\ u_b \|b\| - u_c \|c\| &= v_2 \\ u_a \|a\| - u_c \|c\| &= v_3 \\ \end{align} and take the dot product of each equation with ##v_1##, ##v_2##, and ##v_3##, respectively: \begin{align} v_1 \cdot u_a \|a\| - v_1 \cdot u_b \|b\| &= \|v_1\|^2 \\ v_2 \cdot u_b \|b\| - v_2 \cdot u_c \|c\| &= \|v_2\|^2 \\ v_3 \cdot u_a \|a\| - v_3 \cdot u_c \|c\| &= \|v_3\|^2 \\ \end{align} Dividing the ##i##th equation by ##\|v_i\|## gives you \begin{align} \cos(\theta_{1a}) \|a\| -\cos(\theta_{1b}) \|b\| &= \|v_1\| \\ \cos(\theta_{2b}) \|b\| -\cos(\theta_{2c}) \|c\| &= \|v_2\| \\ \cos(\theta_{3a}) \|a\| -\cos(\theta_{3c}) \|c\| &= \|v_3\| \\ \end{align} where ##\theta_{nx}## denotes the angle between ##v_n## and ##u_x##. You can also obtain six other equations by taking the dot product of ##v_i## with equation ##j##, with ##i \neq j##. For example, taking the dot product of ##v_2## with equation 1 yields $$v_2 \cdot u_a \|a\| - v_2 \cdot u_b \|b\| = v_2 \cdot v_1 = \|v_1\| \|v_2\| \cos(\theta_{12})$$ Dividing by ##\|v_2\|## gives you $$\cos(\theta_{2a}) \|a\| - \cos(\theta_{2b}) \|b\| = \|v_1\|\cos(\theta_{12})$$ Hopefully my theta notation is clear. When there are two number subscripts such as ##\theta_{12}##, I mean the angle between ##v_1## and ##v_2##. When there is a number and a letter such as ##\theta_{1a}##, I mean the angle between ##v_1## and ##u_a##.
 Sci Advisor HW Helper PF Gold P: 3,289 Summarizing in case the above is not clear, you would have nine equations and three unknowns as before, but now the only information presumed known are the magnitudes of the ##v## vectors, the angles between the ##v## vectors, and the angles between the ##v## and ##u## vectors. The equation to solve would be $$\begin{bmatrix} \cos(\theta_{1a}) & -\cos(\theta_{1b}) & 0 \\ 0 & \cos(\theta_{2b}) & -\cos(\theta_{2c}) \\ \cos(\theta_{3a}) & 0 & -\cos(\theta_{3c}) \\ \cos(\theta_{2a}) & -\cos(\theta_{2b}) & 0 \\ \cdots & \cdots & \cdots \\ \end{bmatrix} \begin{bmatrix} \|a\| \\ \|b\| \\ \|c\| \end{bmatrix} = \begin{bmatrix} \|v_1\| \\ \|v_2\| \\ \|v_3\| \\ \|v_1\|\cos(\theta_{12})\\ \cdots \end{bmatrix}$$