Recovering three dimensional vectors after projection and cross product

Question

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 $\mathbb{R}^9$)

Answer 1

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 $k$: $$ \begin{align} &x_1x_2x_3=1,\tag{3}\\ &x^TS\operatorname{diag}\left(x_2x_3,\,x_1x_3,\,x_1x_2\right)Rx=0,\tag{4}\\ &q_i=k\left(\prod_{j\ne i}x_j\right)C_ix-L_i\operatorname{diag}\left(x_2x_3,\,x_1x_3,\,x_1x_2\right)Rx.\tag{5} \end{align} $$

Answer 2

Maxima could not solve it but here are the $10$ equations in $6$ unkowns all equal to $0$.

The $q_i$ equations are multiplied by the denominator scalar $\displaystyle e_i^T u$.

Gröbner basis might be feasable for this size problem.

Maxima:

load("vect");
cross(u, v) := matrix(u[2] * v[3] - v[2] * u[3], v[1] * u[3] - u[1] * v[3],u[1] * v[2] - v[1] * u[2]);
dot(u, v) := u[1] * v[1] + u[2] * v[2] + u[3] * v[3];
u : matrix ([u[1]] , [u[2]], [u[3]]);
v : matrix ([v[1]] , [v[2]], [v[3]]);
q1 : matrix ([q1[1]] , [q1[2]], [q1[3]]);
e1 : matrix ([e1[1]] , [e1[2]], [e1[3]]);
P1 : matrix([1,0,0],[0,1,0],[0,0,1]) -  e1 . transpose(e1);
E1 : cross(e1,u) + transpose(e1) . u * ( P1 . v   - q1) ;
q2 : matrix ([q2[1]] , [q2[2]], [q2[3]]);
e2 : matrix ([e2[1]] , [e2[2]], [e2[3]]);
P2 : matrix([1,0,0],[0,1,0],[0,0,1]) -  e2 . transpose(e2);
E2 : cross(e2,u) + transpose(e2) . u * ( P2 . v   - q2) ;
q3 : matrix ([q3[1]] , [q3[2]], [q3[3]]);
e3 : matrix ([e3[1]] , [e3[2]], [e3[3]]);
P3 : matrix([1,0,0],[0,1,0],[0,0,1]) -  e3 . transpose(e3);
E3 : cross(e3,u) + transpose(e3) . u * ( P3 . v   - q3) ;
solve([E1[1][1],E1[2][1],E1[3][1],E2[1][1],E2[2][1],E2[3][1],E3[1][1],E3[2][1],E3[3][1], u . v],[u[1],u[2],u[3],v[1],v[2],v[3]]);
E1[1][1];
tex(%);
E1[2][1];
tex(%);
E1[3][1];
tex(%);
E2[1][1];
tex(%);
E2[2][1];
tex(%);
E2[3][1];
tex(%);
E3[1][1];
tex(%);
E3[2][1];
tex(%);
E3[3][1];
tex(%);
u . v;
tex(%);

$$\left({\it e_1}_{3}\,u_{3}+{\it e_1}_{2}\,u_{2}+{\it e_1}_{1}\,u_{1 }\right)\,\left(-{\it e_1}_{1}\,{\it e_1}_{3}\,v_{3}-{\it e_1}_{1}\, {\it e_1}_{2}\,v_{2}+\left(1-{\it e_1}_{1}^2\right)\,v_{1}-{\it q_1} _{1}\right)+{\it e_1}_{2}\,u_{3}-u_{2}\,{\it e_1}_{3}$$

$$\left({\it e_1}_{3}\,u_{3}+{\it e_1}_{2}\,u_{2}+{\it e_1}_{1}\,u_{1 }\right)\,\left(-{\it e_1}_{2}\,{\it e_1}_{3}\,v_{3}+\left(1- {\it e_1}_{2}^2\right)\,v_{2}-{\it q_1}_{2}-{\it e_1}_{1}\,v_{1}\, {\it e_1}_{2}\right)-{\it e_1}_{1}\,u_{3}+u_{1}\,{\it e_1}_{3}$$

$$\left({\it e_1}_{3}\,u_{3}+{\it e_1}_{2}\,u_{2}+{\it e_1}_{1}\,u_{1 }\right)\,\left(\left(1-{\it e_1}_{3}^2\right)\,v_{3}-{\it q_1}_{3}- {\it e_1}_{2}\,v_{2}\,{\it e_1}_{3}-{\it e_1}_{1}\,v_{1}\,{\it e_1} _{3}\right)+{\it e_1}_{1}\,u_{2}-u_{1}\,{\it e_1}_{2}$$

$$\left({\it e_2}_{3}\,u_{3}+{\it e_2}_{2}\,u_{2}+{\it e_2}_{1}\,u_{1 }\right)\,\left(-{\it e_2}_{1}\,{\it e_2}_{3}\,v_{3}-{\it e_2}_{1}\, {\it e_2}_{2}\,v_{2}+\left(1-{\it e_2}_{1}^2\right)\,v_{1}-{\it q_2} _{1}\right)+{\it e_2}_{2}\,u_{3}-u_{2}\,{\it e_2}_{3}$$

$$\left({\it e_2}_{3}\,u_{3}+{\it e_2}_{2}\,u_{2}+{\it e_2}_{1}\,u_{1 }\right)\,\left(-{\it e_2}_{2}\,{\it e_2}_{3}\,v_{3}+\left(1- {\it e_2}_{2}^2\right)\,v_{2}-{\it q_2}_{2}-{\it e_2}_{1}\,v_{1}\, {\it e_2}_{2}\right)-{\it e_2}_{1}\,u_{3}+u_{1}\,{\it e_2}_{3}$$

$$\left({\it e_2}_{3}\,u_{3}+{\it e_2}_{2}\,u_{2}+{\it e_2}_{1}\,u_{1 }\right)\,\left(\left(1-{\it e_2}_{3}^2\right)\,v_{3}-{\it q_2}_{3}- {\it e_2}_{2}\,v_{2}\,{\it e_2}_{3}-{\it e_2}_{1}\,v_{1}\,{\it e_2} _{3}\right)+{\it e_2}_{1}\,u_{2}-u_{1}\,{\it e_2}_{2}$$

$$\left({\it e_3}_{3}\,u_{3}+{\it e_3}_{2}\,u_{2}+{\it e_3}_{1}\,u_{1 }\right)\,\left(-{\it e_3}_{1}\,{\it e_3}_{3}\,v_{3}-{\it e_3}_{1}\, {\it e_3}_{2}\,v_{2}+\left(1-{\it e_3}_{1}^2\right)\,v_{1}-{\it q_3} _{1}\right)+{\it e_3}_{2}\,u_{3}-u_{2}\,{\it e_3}_{3}$$

$$\left({\it e_3}_{3}\,u_{3}+{\it e_3}_{2}\,u_{2}+{\it e_3}_{1}\,u_{1 }\right)\,\left(-{\it e_3}_{2}\,{\it e_3}_{3}\,v_{3}+\left(1- {\it e_3}_{2}^2\right)\,v_{2}-{\it q_3}_{2}-{\it e_3}_{1}\,v_{1}\, {\it e_3}_{2}\right)-{\it e_3}_{1}\,u_{3}+u_{1}\,{\it e_3}_{3}$$

$$\left({\it e_3}_{3}\,u_{3}+{\it e_3}_{2}\,u_{2}+{\it e_3}_{1}\,u_{1 }\right)\,\left(\left(1-{\it e_3}_{3}^2\right)\,v_{3}-{\it q_3}_{3}- {\it e_3}_{2}\,v_{2}\,{\it e_3}_{3}-{\it e_3}_{1}\,v_{1}\,{\it e_3} _{3}\right)+{\it e_3}_{1}\,u_{2}-u_{1}\,{\it e_3}_{2}$$

$$u_{3}\,v_{3}+u_{2}\,v_{2}+u_{1}\,v_{1}$$

Answer 3

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.

Answer 4

$$ \displaystyle q_i = \frac{e_i \times u}{e_i^T u} + P_i v \tag{1}$$

$(1)$ Expanded produces:

$$-{\it e_{11}}\,{\it e_{13}}\,{\it v_3}-{\it e_{11}}\,{\it e_{12}}\, {\it v_2}+\left(1-{\it e_{11}}^2\right)\,{\it v_1}+{{{\it e_{12}}\, {\it u_3}-{\it e_{13}}\,{\it u_2}}\over{{\it e_{13}}\,{\it u_3}+ {\it e_{12}}\,{\it u_2}+{\it e_{11}}\,{\it u_1}}}-{\it q_{11}}$$

$$-{\it e_{12}}\,{\it e_{13}}\,{\it v_3}+\left(1-{\it e_{12}}^2 \right)\,{\it v_2}-{\it e_{11}}\,{\it e_{12}}\,{\it v_1}+{{ {\it e_{13}}\,{\it u_1}-{\it e_{11}}\,{\it u_3}}\over{{\it e_{13}}\, {\it u_3}+{\it e_{12}}\,{\it u_2}+{\it e_{11}}\,{\it u_1}}}- {\it q_{12}}$$

$$\left(1-{\it e_{13}}^2\right)\,{\it v_3}-{\it e_{12}}\,{\it e_{13}} \,{\it v_2}-{\it e_{11}}\,{\it e_{13}}\,{\it v_1}+{{{\it e_{11}}\, {\it u_2}-{\it e_{12}}\,{\it u_1}}\over{{\it e_{13}}\,{\it u_3}+ {\it e_{12}}\,{\it u_2}+{\it e_{11}}\,{\it u_1}}}-{\it q_{13}}$$

$$-{\it e_{21}}\,{\it e_{23}}\,{\it v_3}-{\it e_{21}}\,{\it e_{22}}\, {\it v_2}+\left(1-{\it e_{21}}^2\right)\,{\it v_1}+{{{\it e_{22}}\, {\it u_3}-{\it e_{23}}\,{\it u_2}}\over{{\it e_{23}}\,{\it u_3}+ {\it e_{22}}\,{\it u_2}+{\it e_{21}}\,{\it u_1}}}-{\it q_{21}}$$

$$-{\it e_{22}}\,{\it e_{23}}\,{\it v_3}+\left(1-{\it e_{22}}^2 \right)\,{\it v_2}-{\it e_{21}}\,{\it e_{22}}\,{\it v_1}+{{ {\it e_{23}}\,{\it u_1}-{\it e_{21}}\,{\it u_3}}\over{{\it e_{23}}\, {\it u_3}+{\it e_{22}}\,{\it u_2}+{\it e_{21}}\,{\it u_1}}}- {\it q_{22}}$$

$$\left(1-{\it e_{23}}^2\right)\,{\it v_3}-{\it e_{22}}\,{\it e_{23}} \,{\it v_2}-{\it e_{21}}\,{\it e_{23}}\,{\it v_1}+{{{\it e_{21}}\, {\it u_2}-{\it e_{22}}\,{\it u_1}}\over{{\it e_{23}}\,{\it u_3}+ {\it e_{22}}\,{\it u_2}+{\it e_{21}}\,{\it u_1}}}-{\it q_{23}}$$

$$-{\it e_{31}}\,{\it e_{33}}\,{\it v_3}-{\it e_{31}}\,{\it e_{32}}\, {\it v_2}+\left(1-{\it e_{31}}^2\right)\,{\it v_1}+{{{\it e_{32}}\, {\it u_3}-{\it e_{33}}\,{\it u_2}}\over{{\it e_{33}}\,{\it u_3}+ {\it e_{32}}\,{\it u_2}+{\it e_{31}}\,{\it u_1}}}-{\it q_{31}}$$

$$-{\it e_{32}}\,{\it e_{33}}\,{\it v_3}+\left(1-{\it e_{32}}^2 \right)\,{\it v_2}-{\it e_{31}}\,{\it e_{32}}\,{\it v_1}+{{ {\it e_{33}}\,{\it u_1}-{\it e_{31}}\,{\it u_3}}\over{{\it e_{33}}\, {\it u_3}+{\it e_{32}}\,{\it u_2}+{\it e_{31}}\,{\it u_1}}}- {\it q_{32}}$$

$$\left(1-{\it e_{33}}^2\right)\,{\it v_3}-{\it e_{32}}\,{\it e_{33}} \,{\it v_2}-{\it e_{31}}\,{\it e_{33}}\,{\it v_1}+{{{\it e_{31}}\, {\it u_2}-{\it e_{32}}\,{\it u_1}}\over{{\it e_{33}}\,{\it u_3}+ {\it e_{32}}\,{\it u_2}+{\it e_{31}}\,{\it u_1}}}-{\it q_{33}}$$

Note that these equations are linear in $v = [v_1,v_2,v_3]^T$

Degenerate equations will occur.

The solvability of $v_k$ depends on the rank of the matrix of the coefficients of $v_k$ variables.

Reducing all $v_k$ out of the equations will leave equations in $u$.

$u \cdot v = 0$ can now be added to the equations.

Since there are three divisors $e_i \cdot u$ multiplying the numerators with these divisors will result in cubic equations in $u_k$ with mixed terms e.g. $u_i u_j u_k$.

The problem reduces to solving a system of cubic equations in three variables $u_1,u_2,u_3$.

There does not seem to be a way to find these equations without knowledge of $e_i$.

Answer 5

Example: $e_1=(0,-\frac1{2},\frac{\sqrt{3}}{2}), \: e_2=(\frac{\sqrt{3}}{4},\frac1{4},\frac{\sqrt{3}}{2}), \: e_3=(-\frac{\sqrt{3}}{4}, \frac1{4}, \frac{\sqrt{3}}{2})$

Maxima:

load("vect");
cross(u, v) := matrix(u[2] * v[3] - v[2] * u[3], v[1] * u[3] - u[1] * v[3],u[1] * v[2] - v[1] * u[2]);
dot(u, v) := u[1] * v[1] + u[2] * v[2] + u[3] * v[3];
u : matrix ([u1] , [u2], [u3]);
v : matrix ([v1] , [v2], [v3]);
q1 : matrix ([q11] , [q12], [q13]);
e1 : matrix ([0] , [-1/2], [sqrt(3)/2]);
P1 : matrix([1,0,0],[0,1,0],[0,0,1]) -  e1 . transpose(e1);
E1 : cross(e1,u)/(transpose(e1) . u ) +  P1 . v   - q1 ;
q2 : matrix ([q21] , [q22], [q23]);
e2 : matrix ([sqrt(3)/4] , [1/4], [sqrt(3)/2]);
P2 : matrix([1,0,0],[0,1,0],[0,0,1]) -  e2 . transpose(e2);
E2 : cross(e2,u)/(transpose(e2) . u) + P2 . v   - q2 ;
q3 : matrix ([q31] , [q32], [q33]);
e3 : matrix ([-sqrt(3)/4] , [1/4], [sqrt(3)/2]);
P3 : matrix([1,0,0],[0,1,0],[0,0,1]) -  e3 . transpose(e3);
E3 : cross(e3,u)/(transpose(e3) . u ) +  P3 . v   - q3 ;
A1  : augcoefmatrix(E1[1],[v1,v2,v3]);
A2  : augcoefmatrix(E1[2],[v1,v2,v3]);
A3  : augcoefmatrix(E1[3],[v1,v2,v3]);
A4  : augcoefmatrix(E2[1],[v1,v2,v3]);
A5  : augcoefmatrix(E2[2],[v1,v2,v3]);
A6  : augcoefmatrix(E2[3],[v1,v2,v3]);
A7  : augcoefmatrix(E3[1],[v1,v2,v3]);
A8  : augcoefmatrix(E3[2],[v1,v2,v3]);
A9  : augcoefmatrix(E3[3],[v1,v2,v3]);
A : matrix(A1[1],A2[1],A3[1],A4[1],A5[1],A6[1],A7[1],A8[1],A9[1]);
AA : A;
for k:1 thru 9  do AA[k][4] : ratsimp(AA[k][4]);
BB : copy(AA);
S1 : copy(BB[1][4]);
S2 : copy(BB[2][4]);
S3 : copy(BB[3][4]);
S4 : copy(BB[4][4]);
S5 : copy(BB[5][4]);
S6 : copy(BB[6][4]);
S7 : copy(BB[7][4]);
S8 : copy(BB[8][4]);
S9 : copy(BB[9][4]);
BB[1][4] : R1;
BB[2][4] : R2;
BB[3][4] : R3;
BB[4][4] : R4;
BB[5][4] : R5;
BB[6][4] : R6;
BB[7][4] : R7;
BB[8][4] : R8;
BB[9][4] : R9;
for k:2 thru 9  do AA : rowop(AA,k,1,AA[k][1]);
for k:2 thru 9  do AA[k] : AA[k]/AA[k][2];
for k:3 thru 9  do AA : rowop(AA,k,2,1);
for k:4 thru 9  do AA[k] : AA[k]/AA[k][3];
for k:5 thru 9  do AA : rowop(AA,k,4,1);
AA : rowop(AA,2,4,AA[2][3]);
for k:1 thru 9  do AA[k][4] : ratsimp(AA[k][4]);
AA : rowswap(AA,3,4);
for k:2 thru 9  do BB : rowop(BB,k,1,BB[k][1]);
for k:2 thru 9  do BB[k] : BB[k]/BB[k][2];
for k:3 thru 9  do BB : rowop(BB,k,2,1);
for k:4 thru 9  do BB[k] : BB[k]/BB[k][3];
for k:5 thru 9  do BB : rowop(BB,k,4,1);
BB : rowop(BB,2,4,BB[2][3]);
for k:1 thru 9  do BB[k][4] : ratsimp(BB[k][4]);
BB : rowswap(BB,3,4);

$$\displaystyle q_i = \frac{e_i \times u}{e_i^T u} + P_i v \tag{1} $$

Matrix form of $(1)$

$$AA = \pmatrix{1&0&0&-{{\left(\sqrt{3}\,{\it q_{11}}+1\right)\,{\it u_3}+ \left(\sqrt{3}-{\it q_{11}}\right)\,{\it u_2}}\over{\sqrt{3}\, {\it u_3}-{\it u_2}}}\cr 0&{{3}\over{4}}&{{\sqrt{3}}\over{4}}&-{{ \sqrt{3}\,{\it q_{12}}\,{\it u_3}-{\it q_{12}}\,{\it u_2}-\sqrt{3}\, {\it u_1}}\over{\sqrt{3}\,{\it u_3}-{\it u_2}}}\cr 0&{{\sqrt{3} }\over{4}}&{{1}\over{4}}&-{{\sqrt{3}\,{\it q_{13}}\,{\it u_3}- {\it q_{13}}\,{\it u_2}-{\it u_1}}\over{\sqrt{3}\,{\it u_3}- {\it u_2}}}\cr {{13}\over{16}}&-{{\sqrt{3}}\over{16}}&-{{3}\over{8}} &-{{\left(2\,\sqrt{3}\,{\it q_{21}}-1\right)\,{\it u_3}+\left( {\it q_{21}}+2\,\sqrt{3}\right)\,{\it u_2}+\sqrt{3}\,{\it q_{21}}\, {\it u_1}}\over{2\,\sqrt{3}\,{\it u_3}+{\it u_2}+\sqrt{3}\,{\it u_1} }}\cr -{{\sqrt{3}}\over{16}}&{{15}\over{16}}&-{{\sqrt{3}}\over{8}}&- {{\left(2\,\sqrt{3}\,{\it q_{22}}+\sqrt{3}\right)\,{\it u_3}+ {\it q_{22}}\,{\it u_2}+\left(\sqrt{3}\,{\it q_{22}}-2\,\sqrt{3} \right)\,{\it u_1}}\over{2\,\sqrt{3}\,{\it u_3}+{\it u_2}+\sqrt{3}\, {\it u_1}}}\cr -{{3}\over{8}}&-{{\sqrt{3}}\over{8}}&{{1}\over{4}}&- {{2\,\sqrt{3}\,{\it q_{23}}\,{\it u_3}+\left({\it q_{23}}-\sqrt{3} \right)\,{\it u_2}+\left(\sqrt{3}\,{\it q_{23}}+1\right)\,{\it u_1} }\over{2\,\sqrt{3}\,{\it u_3}+{\it u_2}+\sqrt{3}\,{\it u_1}}}\cr {{ 13}\over{16}}&{{\sqrt{3}}\over{16}}&{{3}\over{8}}&-{{\left(2\,\sqrt{ 3}\,{\it q_{31}}-1\right)\,{\it u_3}+\left({\it q_{31}}+2\,\sqrt{3} \right)\,{\it u_2}-\sqrt{3}\,{\it q_{31}}\,{\it u_1}}\over{2\,\sqrt{ 3}\,{\it u_3}+{\it u_2}-\sqrt{3}\,{\it u_1}}}\cr {{\sqrt{3}}\over{16 }}&{{15}\over{16}}&-{{\sqrt{3}}\over{8}}&-{{\left(2\,\sqrt{3}\, {\it q_{32}}-\sqrt{3}\right)\,{\it u_3}+{\it q_{32}}\,{\it u_2}+ \left(-\sqrt{3}\,{\it q_{32}}-2\,\sqrt{3}\right)\,{\it u_1}}\over{2 \,\sqrt{3}\,{\it u_3}+{\it u_2}-\sqrt{3}\,{\it u_1}}}\cr {{3}\over{8 }}&-{{\sqrt{3}}\over{8}}&{{1}\over{4}}&-{{2\,\sqrt{3}\,{\it q_{33}} \,{\it u_3}+\left({\it q_{33}}+\sqrt{3}\right)\,{\it u_2}+\left(1- \sqrt{3}\,{\it q_{33}}\right)\,{\it u_1}}\over{2\,\sqrt{3}\, {\it u_3}+{\it u_2}-\sqrt{3}\,{\it u_1}}}\cr } \tag{2}$$

More readable:

$$BB = \pmatrix{1&0&0&{\it R_1}\cr 0&{{3}\over{4}}&{{\sqrt{3}}\over{4}}& {\it R_2}\cr 0&{{\sqrt{3}}\over{4}}&{{1}\over{4}}&{\it R_3}\cr {{13 }\over{16}}&-{{\sqrt{3}}\over{16}}&-{{3}\over{8}}&{\it R_4}\cr -{{ \sqrt{3}}\over{16}}&{{15}\over{16}}&-{{\sqrt{3}}\over{8}}&{\it R_5} \cr -{{3}\over{8}}&-{{\sqrt{3}}\over{8}}&{{1}\over{4}}&{\it R_6}\cr {{13}\over{16}}&{{\sqrt{3}}\over{16}}&{{3}\over{8}}&{\it R_7}\cr {{ \sqrt{3}}\over{16}}&{{15}\over{16}}&-{{\sqrt{3}}\over{8}}&{\it R_8} \cr {{3}\over{8}}&-{{\sqrt{3}}\over{8}}&{{1}\over{4}}&{\it R_9}\cr } \tag{3}$$

Reduced row echelon form:

$$BB = \pmatrix{1&0&0&{\it R_1}\cr 0&1&0&{{16\,{\it R_4}+8\,\sqrt{3}\, {\it R_2}-13\,{\it R_1}}\over{5\,\sqrt{3}}}\cr 0&0&1&-{{16\,\sqrt{3} \,{\it R_4}+4\,{\it R_2}-13\,\sqrt{3}\,{\it R_1}}\over{5\,\sqrt{3}}} \cr 0&0&0&{{12\,{\it R_3}-4\,\sqrt{3}\,{\it R_2}}\over{3^{{{3}\over{ 2}}}}}\cr 0&0&0&-{{80\,{\it R_5}-112\,\sqrt{3}\,{\it R_4}-128\, {\it R_2}+32\,3^{{{3}\over{2}}}\,{\it R_1}}\over{35\,\sqrt{3}}}\cr 0 &0&0&{{40\,\sqrt{3}\,{\it R_6}+16\,3^{{{3}\over{2}}}\,{\it R_4}+32\, {\it R_2}-8\,3^{{{3}\over{2}}}\,{\it R_1}}\over{5\,3^{{{3}\over{2}}} }}\cr 0&0&0&{{16\,{\it R_7}+16\,{\it R_4}-26\,{\it R_1}}\over{5}} \cr 0&0&0&-{{80\,{\it R_8}-112\,\sqrt{3}\,{\it R_4}-128\,{\it R_2}+ 86\,\sqrt{3}\,{\it R_1}}\over{35\,\sqrt{3}}}\cr 0&0&0&{{40\,\sqrt{3} \,{\it R_9}+16\,3^{{{3}\over{2}}}\,{\it R_4}+32\,{\it R_2}-2\,3^{{{7 }\over{2}}}\,{\it R_1}}\over{5\,3^{{{3}\over{2}}}}}\cr } \tag{4}$$

The first three rows solve for $v_1,v_2,v_3$.

These can be substituted into $u \cdot v = 0$

Rows four to nine are equations entirely in $u$.

Since there are only three denominator expressions in $R$ then multiplying the equations by them will result in cubic equations in $u$.