Recovering three dimensional vectors after projection and cross product Suppose $e_i \in \mathbb{R}^3$, $1\leq i \leq 3$ with $\Vert e_i \Vert=1$.
Suppose $u,v \in \mathbb{R}^3$, $u^T v=0$, $e_i^T u \neq 0$, $\Vert u \Vert =1$. Suppose $k\in \mathbb{R}$.
Define the projection on the plane orthogonal to $e_i$
$P_i= I-e_i e_i^T$
where $I$ is the $\mathbb{R}^{3\times 3}$ identity matrix.
Suppose $e_i$ and
$\displaystyle q_i = k \frac{e_i \times u}{e_i^T u} + P_i v $
are known for $1\leq i \leq 3$.
Is it possible to recover $k$, $u$ and $v$?
(and, even before that, is the function from $(k,u,v)$ to $(q_1,q_2,q_3)$ injective? It is clearly not surjective in $\mathbb{R}^9$)
 A: We can eliminate $v$ from the equations. Let $Q$ and $E$ be the augmented matrices $[q_1|q_2|q_3]$ and $[e_1|e_2|e_3]$ respectively. I assume that $E$ is nonsingular, otherwise the system of equations in question is clearly not uniquely solvable. From the equations $u^Tv=0$ and
$$
q_i=k\frac{e_i\times u}{e_i^Tu}+P_iv,\tag{1}
$$
we obtain $u^Tq_i=-(e_i^Tu)(e_i^Tv)$ for each $i$. Therefore
$$
Q^Tu=-\operatorname{diag}(e_1^Tu,\,e_2^Tu,\,e_3^Tu)E^Tv.
$$
Since $E$ is invertible and $e_i^Tu\ne0$, we can solve $v$ in terems of $u$:
$$
v=-(E^T)^{-1}\operatorname{diag}\left(\frac{1}{e_1^Tu},\,\frac{1}{e_2^Tu},\,\frac{1}{e_3^Tu}\right)Q^Tu.\tag{2}
$$
The system of equations and inequations $e_i^Tu\ne0,\,u^Tv=0$ and $(1)$ is now equivalent to
\begin{cases}
e_i^Tu\ne0,\\
u^T(E^T)^{-1}\operatorname{diag}\left(\frac{1}{e_1^Tu},\,\frac{1}{e_2^Tu},\,\frac{1}{e_3^Tu}\right)Q^Tu=0,\\
q_i=k\frac{e_i\times u}{e_i^Tu}-P_i(E^T)^{-1}\operatorname{diag}\left(\frac{1}{e_1^Tu},\,\frac{1}{e_2^Tu},\,\frac{1}{e_3^Tu}\right)Q^Tu.
\end{cases}
(The condition $\|u\|=1$ is useless, as $(1)$ is homogeneous in $u$. We can always solve for $u$ first and normalise it later.) If we put $x=E^Tu,\,S= (E^TE)^{-1},\,R=Q^T(E^T)^{-1},\,C_i=[e_i]_\times(E^T)^{-1}$ and $L_i=P_i(E^T)^{-1}$, the above system can be rewritten as
$$
\begin{cases}
x_i\ne0,\\
x^TS\operatorname{diag}\left(\frac{1}{x_1},\,\frac{1}{x_2},\,\frac{1}{x_3}\right)Rx=0,\\
q_i=\frac{k}{x_i}C_ix-L_i\operatorname{diag}\left(\frac{1}{x_1},\,\frac{1}{x_2},\,\frac{1}{x_3}\right)Rx.
\end{cases}
$$
Note that if $(k,x)$ is a solution, so is $(k,tx)$ for all nonzero $t$. Since $x_i\ne0$ for each $i$, there always exists a $t$ such that $(tx_1)(tx_2)(tx_3)=1$. Therefore, we can replace the inequality constraint $x_i\ne0$ above by $x_1x_2x_3=1$. If we also clear the denominators, the problem will reduce to a system of one degree-$3$ polynomial equation and ten degree-$4$ equations in four unknowns $x_1,x_2,x_3$ and $k$:
$$
\begin{align}
&x_1x_2x_3=1,\tag{3}\\
&x^TS\operatorname{diag}\left(x_2x_3,\,x_1x_3,\,x_1x_2\right)Rx=0,\tag{4}\\
&q_i=k\left(\prod_{j\ne i}x_j\right)C_ix-L_i\operatorname{diag}\left(x_2x_3,\,x_1x_3,\,x_1x_2\right)Rx.\tag{5}
\end{align}
$$
A: I'm sure this needs checking.
The strategy is to find three equations entirely in $u$ variables $u_1,u_2,u_3$ and solve them.
$$\displaystyle q_i = \frac{e_i \times u}{e_i^T u} + P_i v \tag{1}$$
$$u \cdot v = 0 \tag{2}$$
If $P_i$ is invertible then:
$$\displaystyle v = -{P_i}^{-1}\frac{e_i \times u}{e_i^T u} + {P_i}^{-1} q_i \tag{3}$$
$v$ is expressed in terms of $u$.
Substituting $v$ into $(2)$ gives one equation entirely in $u$.
If $P$ is not invertible then row reductions can be performed to find a row echelon form that will have one or more zero rows.
$P \rightarrow \begin{bmatrix} a & b & c\\ 0 & d & e\\ 0 & 0 & 0\end{bmatrix}$ or
$\begin{bmatrix} a & b & c\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$ or
$\begin{bmatrix}  0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$ or other forms.
Each zero row produces an equation in $u$ variables only (no $v$ variables).
$$\displaystyle q_{ik} = \frac{(e_i \times u)_k}{e_i^T u} \: \: with\: row\: reductions\tag{4}$$
$$e_i \times u = \begin{bmatrix} e_{i2}u_3 - e_{i3}u_2  \\ e_{i3}u_1 - e_{i1}u_3 \\ e_{i1}u_2 - e_{i2}u_1\end{bmatrix} \tag{5}$$
Some or many of the $e_{ik}$ values could be zero so select non zero rows of $e_i \times u$.
In the case where $P_i$ is invertible $(3)$ substituted into $(2)$ has a common scalar denominator ${e_i^T u} $ that can be multiplied into the numerator:
$$u_1 \cdot v_1 +  u_2 \cdot v_2 + u_3 \cdot v_3 = $$
$$u_1 [{P_i}^{-1}{(e_i \times u)} - {e_i^T u}{P_i}^{-1} q_i]_1 + u_2 [{P_i}^{-1}{(e_i \times u)} - {e_i^T u}{P_i}^{-1} q_i]_2 + u_3 [{P_i}^{-1}{(e_i \times u)} - {e_i^T u}{P_i}^{-1} q_i]_3 = 0 \tag{6}$$
The $u$ order of equation $(6)$ is $2$ i.e. it has terms of the form ${u_1}^2$, $u_1u_2$ etc... Its a quadratic.
This produces equations of the form:
$$ c_{11}{u_1}^2 + c_{22}{u_2}^2 + ... + c_{12}u_1u_2 ... = 0  \tag{7}$$
If all $P_i$ are invertible there will be three quadratic equations of the form $(7)$.
From $(4)$ if some $P_i$ where not invertible there will be equations of the form:
$$c_1 u_1 + c_2 u_2 + c_3 u_3 = 0 \tag{8}$$
If the conditions are not degenerate (bad $e_i$) then these equations should be solvable.
