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My question is based on "What is the geometric meaning of singular matrix" posted here some years ago.

To make this a bit more intuitive I would like to add an example.

A three-dimensional force vector $F$ applied at a point $P$ with coordinates $(x, y, z)$ creates a moment $M$ at, say, point $(0, 0, 0)$. This moment $M$ is again a vector with components $M_x$, $M_y$ and $M_z$ for which, the following hold:

$$ \begin{bmatrix} 0 & F_z & -F_y \\ -F_z & 0 & F_x \\ F_y & -F_x & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} M_x \\ M_y \\ M_z \\ \end{bmatrix} $$

As it can be seen, the above matrix is singular. Math tells me that as a result it cannot be inverted and consequently, given the forces and moments one cannot solve for the coordinates of $P$, aka given the freedom to apply a given force wherever, not all moments vectors can be produced.

Two questions come to mind here:

  • How does one derive what (sub)part of $\mathbb R^3$ this given matrix embeds to? What points are reachable?
  • How about the moments that can be produced? How would one solve for those?

One can easily construct a solution-triplet by setting $P$ to $(1, 1, 1)$ from which, $M=\begin{bmatrix} F_z - F_y \\ F_x - F_z \\ F_y - F_x \\ \end{bmatrix}$

but assuming that instead of the $P$ we knew this $M$ how could $P$ be located?

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The moment at $O = (0, 0, 0)$ is $\vec{M} = \vec{P} \times \vec{F}$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $\vec{F}$ and the vector $\vec{P}$ (right-hand rule), and length equal to the area of the parallelogram with sides $OF$ and $OP$. Thus, if the force is fixed, but you can vary $P$, you can achieve any point on the plane through 0, perpendicular to $\vec{F}$.

On the other hand, infinitely many points $P$ can give the same moment. We know $P$ must be on the plane through 0, perpendicular to $\vec{M}$. Further, we want the parallelogram with sides $OP$ and $OF$ to have area equal to $||\vec{M}||$. So $P$ must be on a line parallel to $\vec{F}$, at a distance $\frac{||\vec{M}||}{||\vec{F}||}$ from $O$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.

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  • $\begingroup$ Did you mean to write $\vec{M} = \vec{P} \times \vec{F}$? $\endgroup$ – Ev. Kounis Jan 14 at 12:20
  • $\begingroup$ @Ev.Kounis Yes, indeed $\endgroup$ – Todor Markov Jan 14 at 12:36

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