I would like to find a reference point (or Neutral Point) in 3D space where the net torque is zero.
Starting from: $$ \tau = \mathbf r \times \mathbf F $$ $$ \sum_{i=1}^n ( \mathbf r_i - \mathbf p ) \times \mathbf F_i = 0 $$
where I want to know $ \mathbf p $.
I was able to rearrange it like so:
$$ \implies \sum_{i=1}^n ( \mathbf r_i \times \mathbf F_i - \mathbf p \times \mathbf F_i ) = 0 $$ $$ \implies \sum_{i=1}^n \mathbf r_i \times \mathbf F_i - \sum_{i=1}^n \mathbf p \times \mathbf F_i = 0 $$ $$ \implies \sum_{i=1}^n \mathbf r_i \times \mathbf F_i = \sum_{i=1}^n \mathbf p \times \mathbf F_i $$
and because $ \sum_{i=1}^n \mathbf p \times \mathbf F_i = \mathbf p \times \mathbf F_1 + \mathbf p \times \mathbf F_2 + \cdots + \mathbf p \times \mathbf F_n = \mathbf p \times \sum_{i=1}^n \mathbf F_i $
$$ \implies \sum_{i=1}^n \mathbf r_i \times \mathbf F_i = \mathbf p \times \sum_{i=1}^n \mathbf F_i \tag 1 $$
Here is where I get stuck. I do not know how to isolate the vector $ \mathbf p $.
I tried converting to matrix multiplication: $$ \sum_{i=1}^n \mathbf r_i \times \mathbf F_i = \left[F\right]_x^T \mathbf p $$ But since the matrix is a 3x3 skew-symmetric matrix, the determinate is zero so there is no inverse and I still cannot move it to the other side.
Extra context: This is not homework. It is a personal project that I am playing with at home.
I want to display the total force and total torque exerted on a rigid body submerged in a fluid, but I do not know where to draw the Force "arrow". The magnitude and direction comes from summing the forces, but the application point is determined by the net torque. If the net torque is zero, then the application point is through the center of mass. Otherwise, the application point is offset from the center of mass by some amount, which I represented above as the vector $ \mathbf p $.
The body is represented by a 3D mesh, so I can calculate the forces on each triangle of the mesh, which gives me the set of forces and their application points.
Use the following:
$$ \boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F}\|^2 } $$
The point is return is the point closest to the origin on the line of action of the force.
Proof
$$ \require{cancel} \boldsymbol{F} \times \boldsymbol{\tau} = \boldsymbol{F} \times (\boldsymbol{r} \times \boldsymbol{F} ) = \boldsymbol{r} ( \boldsymbol{F} \cdot \boldsymbol{F} ) - \boldsymbol{F} (\cancelto{0}{ \boldsymbol{r} \cdot \boldsymbol{F}} ) = \boldsymbol{r} \| \boldsymbol{F} \|^2$$
So this ingores all components of $\boldsymbol{r}$ that are parallel to $\boldsymbol{F}$, since $\boldsymbol{r} \cdot \boldsymbol{F} =0 $.
Bonus Content
You can do something similar to get the location of instant rotation axis
$$ \boldsymbol{r} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v} }{ \| \boldsymbol{\omega}\|^2 } $$