Extract vector from vector-matrix equation

Question

We have an equation of the following form:

$$\frac{(u \times w)^T}{(u \times w) \cdot v} \cdot \frac{v \times w}{(v \times w) \cdot u} = \omega$$

where $u,v$ known unit vectors and $\omega$ are known scalar.

Is it possible to solve this equation with respect to an unknown vector $w$?

Answer 1

First I assume $u,v$ are not a multiple of each other. Thus the triplet $\left(u,v,z=\frac{u\times v}{|u\times v|}\right)$ form a complete right-hand frame (coordinate system) in 3D where $|u\times v|$ is the length of the $u\times v$. Now we have $w=au+bv+cz$ where $(a,b,c)$ must be obtained from the given equation. Now from

$$\frac{(u \times w)^T}{(u \times w) \cdot v} \cdot \frac{v \times w}{(v \times w) \cdot u} = \omega$$

we have

$$\frac{(u \times (au+bv+cz))^T}{(u \times (au+bv+cz)) \cdot v} \cdot \frac{v \times (au+bv+cz)}{(v \times (au+bv+cz)) \cdot u} = \omega$$

and with the help from $u\times(u\times v) = (u\cdot v)u-(u\cdot u)v$ and $v\times(u\times v) = (v\cdot v)u-(v\cdot u)v$ we get

$$\frac{\left( b|u \times v|z+c(u\cdot v)u -c\Vert u\Vert^2v \right)}{c\left((u\cdot v)^2-\Vert u\Vert^2\Vert v\Vert^2 \right)/|u\times v|} \cdot \frac{\left(-a|u \times v|z +c\Vert v\Vert^2u -c(v\cdot u)v\right )}{-c\left((u\cdot v)^2-\Vert u\Vert^2\Vert v\Vert^2 \right)/|u\times v|} = \omega$$

Therefore

$$\frac{-ab|u\times v|^2+c^2(u\cdot v)\Vert u \Vert^2-c^2(u \cdot v)^3-2c^2\Vert u\Vert^2\Vert v\Vert^2(u\cdot v)}{\frac{-c^2\left((u\cdot v)^2-\Vert u\Vert^2\Vert v\Vert^2 \right)^2}{|u\times v|^2}} = \omega$$

You have to choose $a,b,c$ such that this equation be satisfied. There are many solutions to this, since you have three unknowns $a,b,c$ and only one scalar equation. The non-linear system is under-determined.

If we assume the linear independence does not hold, i.e. $u=\beta v$ then $(u\times w)\cdot v = \beta(v\times w)\cdot v = 0$ and the problem is ill-posed, no solution exists.

Answer 2

Just a suggestion, too long for a comment.

After some straigtforward computations, we have to solve $$\langle u,v\rangle\|w\|^2-\langle u,w\rangle\langle v,w\rangle=-\omega\langle w,u\times v\rangle^2$$ for $w$. Assume that $w$ is a unit vector. If $u$ and $v$ are linearly independent, get from the orthogonal basis $$\left\{u,v-\frac{\langle u,v\rangle}{\|u\|}\frac{u}{\|u\|},u\times v\right\}$$ an orthonormal one. Express $w$ through that basis and solve for the coefficients.

Answer 3

To avoid confusing $\omega$ with $w$, I will write $c$ for $\omega$.

The denominators in the equations are, up to a sign, $(u\times v)\cdot w$. Therefore you need $\{u,v\}$ to be linearly independent in order that the problem is well posed. Suppose this is indeed the case. Let $p=u\times v$. Then the equation can be rewritten as $$ (u\cdot v)\|w\|^2 - (u\cdot w)(v\cdot w) + c(p\cdot w)^2 = 0,\tag{1} $$ subject to the constraint that $p\cdot w\ne0$. Then $(1)$ can be rewritten as $$ w^T\,\underbrace{\left[(u\cdot v)I_3-\frac12(uv^T+vu^T)+cpp^T\right]}_A\,w=0.\tag{2} $$ Since $A$ is symmetric, it can be orthogonally diagonalised as $A=QDQ^T$. Let $x=Q^Tw$ and $q=Q^Tp$. The constraint $p\cdot w\ne0$ thus becomes $q^Tx\ne0$. Hence the problem boils down to solving \begin{align} d_1x_1^2+d_2x_2^2+d_3x_3^2&=0,\tag{3}\\ q_1x_1+q_2x_2+q_3x_3&\ne0.\tag{4} \end{align} Clearly, $(3)$ is solvable only if $d_i=0$ for some $i$ or $d_id_j<0$ for some $i,j$, i.e. only if $A$ is singular or indefinite. And when $(3)$ is solvable, it is not hard to see if it is still solvable with the constraint $(4)$. Once a solution $x$ is found, we may set $w=Qx$.