Strange results after solving a system of equations

Question

I've got a plane in a 3D space (triangle, represented by 2 direction vectors a and b) and a vertex on the plane (v).

$\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{v}$ are vectors, $s$ and $t$ are scalars $$\boldsymbol{v} = s \, \boldsymbol{a} + t \, \boldsymbol{b}$$

$a = (a_x, a_y, a_z)$ etc.

$$v_x = s \, a_x + t \, b_x$$ $$ v_y = s \, a_y + t \, b_y$$ $$ v_z = s \, a_z + t \, b_z$$

I know a, b and v. And v is definitelly on the plane. I need to find the scalar-factors s and t. When I try to solve this I get something like

$$\begin{align}t &=& \frac{v_z - v_x \frac{ a_z }{ a_x} }{ b_z - b_x \frac{ a_z }{ a_x}}\\ s &=& \frac{v_x - t \, b_x }{ a_x} \end{align}$$

and other variations. And it seems to give me good results many times, but not always.

Problem:
Sometimes I get for $2$ different $v$ the very same values for $s$ and $t$. I expect each new vertex of the triangle to have a unique pair of $s$ and $t$.

I know I have too few unknown variables. What I don't understand just now: what does it mean for my results? What can I do to improve my procedure of solving this equation?

Answer 1

If you have the same values for $s$ and $t$ for different $v$, one of the $v$ cannot be on the plane.

In order to improve your procedure, I would recommend to introduce a third vector $c$ which is the cross product of $a$ and $b$, i.e. $c=a\times b$. Then try to solve $$ v_x = a_xs+b_xt+c_xu \\ v_y = a_ys+b_yt+c_yu \\ v_z = a_zs+b_zt+c_zu $$ If $a$ and $b$ actually span a plane (not only a line), then this system has always a unique solution. If $u=0$ then you can be sure that $v$ is on the plane and $s$ and $t$ are the desired results. If $u\neq 0$, then $v$ is not on the plane.

Use Cramer's rule and the Rule of Sarrus, if you want to have a closed-form expression for the solution.

Alternatively, multiply the first of your equations with $a_x$, the second with $a_y$ and the third with $a_z$ and take the sum of the results. Then do the same with $b_x$, $b_y$ and $b_z$. You will get two new equations: $$ (a^{2}_{x}+a^{2}_{y}+a^{2}_{z})s + (a_xb_x+a_yb_y+a_zb_z)t = (a_xv_x+a_yv_y+a_zv_z) \\ (a_xb_x+a_yb_y+a_zb_z)s + (b^{2}_{x}+b^{2}_{y}+b^{2}_{z})t = (b_xv_x+b_yv_y+b_zv_z) $$ If $a$ and $b$ span a plane, this system of equations is always solvable without distinction of cases.

Answer 2

If you have two linear independent vectors $a$, $b$ in $\mathbb{R}^3$ they indeed span a plane: $$ H = \langle a, b \rangle = \{ v = s a + t b \mid s, t \in \mathbb{R} \} $$ The scalar coefficients $s$ and $t$ can be interpreted as coordinates regarding the base vectors $a$ and $b$.

You have $a$, $b$ and want to recover $s$ and $t$ for a given $v \in H$. So we have $$ v = s a + t b $$ which in matrix form can be written as $$ \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \\ \end{pmatrix} \begin{pmatrix} s \\ t \end{pmatrix} = \begin{pmatrix} v_1 \\ v_2 \\ v_2 \end{pmatrix} \quad (*) $$ In general $(*)$ is an inhomogeneous linear system $Ax = v$.

While the matrix $A$ has three rows, only two of them can be linear independent, as the generated plane is a two-dimensional subspace of $\mathbb{R}$. It depends on the specific choice of $a$ and $b$ which two of the three rows are linear independent ones. Your given formula

$$\begin{align}t &= \frac{v_z - v_x \frac{ a_z }{ a_x} }{ b_z - b_x \frac{ a_z }{ a_x}}\\ s &= \frac{v_x - t \, b_x }{ a_x} \end{align}$$

will only work for the special case where $a_x \ne 0$. It will not handle the case $a = (0,1,0)^T, b=(0,0,1)^T$. That is why one has to use a more flexible method to solve the system.

We can use the following short hand notation for the system $(*)$: $$ [A\mid v] = \left[ \begin{array}{rr|r} a_1 & b_1 & v_1 \\ a_2 & b_2 & v_2 \\ a_3 & b_3 & v_3 \end{array} \right] $$ On this augmented matrix we perform a Gauss-Jordan elimination, using a couple of elementary row operations. This will result in a augmented matrix in reduced row echelon form: $$ \left[ \begin{array}{rr|r} 1 & 0 & s \\ 0 & 1 & t \\ 0 & 0 & 0 \end{array} \right] $$ It allows to read the solution from this matrix. In practice the finite representation of real numbers might introduce errors. This is the subject of numerical mathematics.

Sometimes I get for 2 different v the very same values for s and t. I expect each new vertex of the triangle to have a unique pair of s and t.

This indicates an error in your solver. The coordinates $s$ and $t$ are unique for a basis $a$, $b$ of the plane.

Answer 3

If your vectors $a$, $b$ and $v$ are coplanar, than $v$ is a unique linear representation of vectors $a$ and $b$ if and only if your vectors $a$ and $b$ are not collinear.

So if your vectors $a$ and $b$ are not parallel you should always have a unique solution $(s,t)$, while if they are parallel that would make the system either have no solutions (if your vector $v$ is not parallel to $a$ and $b$) or infintelly many (if vector $v$ is parallel to vectors $a$ and $b$).

Basically what you need to ensure, is that your vectors $a$ and $b$ are not multiples of each other (so that would mean, they are not parallel to each other, or they actually span a plain not just a line).

Edit: The other problem may be, that if you got two different $v_1$ and $v_2$ for the same $(s,t)$ you should just check in the end, which one of those two actually equals $sa+tb$. The other one probably just does not lie on the plain.