Solve vectorial linear system with cross product

Question

Question: Does the system bellow have any solution for $\vec{F}_a$ and $\vec{F}_b$? If yes, how can I find them all?

$$ \begin{cases}\vec{F}_a + \vec{F}_b = \vec{F}_r \\ \vec{p}_a \times \vec{F}_a + \vec{p}_b \times \vec{F}_b = \vec{M}_r \end{cases} $$

I tried to transform it into a block linear system, but it seems that $\det K =0$:

$$ \underbrace{\begin{bmatrix} \left[I_3\right] & \left[I_3\right] \\ \left[\vec{p}_a\right]_{\times} & \left[\vec{p}_b\right]_{\times} \\ \end{bmatrix}}_{K} \begin{bmatrix} \vec{F}_a \\ \vec{F}_b \\ \end{bmatrix}= \begin{bmatrix} \vec{F}_r \\ \vec{M}_r \\ \end{bmatrix} $$

$[I_{3}]$ is the identity matrix and $[u]_{\times}$ is the cross product matrix of $u$:

$$ u \times a = \left[u\right]_{\times} \cdot a $$ $$ u = \left(u_x, \ u_y, \ u_z\right) \Longrightarrow \left[u\right]_{\times} = \begin{bmatrix}0 & -u_z & u_y \\ u_z & 0 & -u_x \\ -u_y & u_x & 0\end{bmatrix} $$

Answer 1

By substituting $\vec{F}_b = \vec{F}_r - \vec{F}_a$ into the second equation, you only need to solve $$ \underbrace{\left(\vec{p}_a-\vec{p}_b\right)}_u\times\underbrace{\vec{F}_a}_x =\underbrace{\vec{M}_r-\vec{p}_b\times \vec{F}_r}_v $$ which is in the form of $$ u\times x=v\tag{1} $$ where $u$ and $v$ are given and $x$ is the unknown vector. Since the linear map $x\mapsto u\times x$ is singular, equation $(1)$ has either no solution or infinitely many solutions. More specifically,

• if $v=0=u$, every vector $x$ is a solution;
• if $v=0\ne u$, the general solution is given by $x=cu$ where $c$ is an arbitrary scalar;
• if $v\ne0$, the equation is solvable if and only if $0\ne u\perp v$. When this condition is satisfied, the general solution is given by $x=\frac{\|v\|}{\|v\times u\|}v\times u+cu$ where $c$ is arbitrary.

Answer 2

Indeed the determinant of $K$ is $0$ as mentioned by @Theo. That means this system has no unique solution.

We change the first equation to

$$ F_a = F_r - F_b $$

$$ \Rightarrow p_a \times (F_r - F_b) + p_b \times F_b = M_r $$

$$ \Rightarrow (p_b - p_a) \times F_b = M_r - p_a \times F_r $$

As shown in this question, there's no unique solution to this equation.

If $(F_{a0}, \ F_{b0})$ is a solution, then $\forall \omega \in \mathbb{R}$

$$F_a = F_{a0} - \omega \cdot (p_b - p_a) \\ F_b = F_{b0}+\omega \cdot (p_b-p_a)$$

is also a solution.

Extra: For the specific case that $M_r = p_r \times F_r$, the equations would be

$$ F_a + F_b = F_r $$ $$ p_a \times F_a + p_b \times F_b = p_r \times F_r $$

We can rewrite the equations to get

$$ (p_b - p_a) \times F_b = (p_r - p_a) \times F_r $$ $$ (p_b - p_a) \times F_a = (p_b - p_r) \times F_r $$

It still doesn't have a specific solution. But we expect (for physical reasons) that

• If $p_r = p_a$, then $F_{a0} = F_r$ and $F_{b0} = 0$
• If $p_r = p_b$, then $F_{a0} = 0 $ and $F_{b0} = F_r$
• If $p_r = \frac{1}{2}(p_a+p_b)$, then $F_{a0}=F_{b0}=\frac{1}{2}F_r$

As all the solutions lies on the line $p_b-p_a$, then we can create a parameter $t \in \mathbb{R}$ which tells the relative distance between $p_r$, $p_a$ and $p_b$.

$$ t = \dfrac{\langle p_r - p_a , \ p_b - p_a \rangle}{\langle p_b - p_a, \ p_b - p_a \rangle} \Rightarrow \begin{cases}p_r = p_a \Rightarrow t = 0 \\ p_r = p_b \Rightarrow t = 1\end{cases} $$

Then we set the solution to:

\begin{align*} F_{a0} & = (1-t) \cdot F_r \\ F_{b0} & = t \cdot F_r \end{align*}