I was experimenting with the triple scalar product and forces in equilibrium when I came to this result:
Consider 4 forces $ \pmb{F_i}$ for $i=1,2,3,4$.
$\pmb{F_i}=F_i\hat{e_i}$ where $\hat{e_i}$ is the unit vector in the direction of the corresponding force and $F_i$ is the magnitude.
$$F_1\hat{e_1}+F_2\hat{e_2}+F_3\hat{e_3}+F_4\hat{e_4}=0$$
Take the scalar product of the system with $e_3 \times e_4$ (the vector product of $e_3,e_4$).
$$F_1\hat{e_1}\cdot (e_3 \times e_4)+F_2\hat{e_2}\cdot (e_3 \times e_4)+F_3\hat{e_3}\cdot (e_3 \times e_4)+F_4\hat{e_4}\cdot (e_3 \times e_4)=0$$
$$\Rightarrow F_1\hat{e_1}\cdot (e_3 \times e_4)+F_2\hat{e_2}\cdot (e_3 \times e_4)+0+0=0$$
$$\Rightarrow \frac{F_1}{\hat{e_2}\cdot (e_3 \times e_4)}=\frac{F_2}{-\hat{e_1}\cdot (e_3 \times e_4)}$$
$$\Rightarrow \frac{F_1}{V_1}=\frac{F_2}{V_2}$$
Where $V_1$ is the volume of the parallelepiped with edges $\hat{e_{2}},\hat{e_{3}},\hat{e_{4}}$.
This can be extended to any other pair of forces.
If one expresses Lami's theorem as
If one has 3 forces in equilibrium acting at a single point in 2D then the magnitude of each force is proportional to the area of the parallelogram made from the unit vectors of the other 2 forces.
Then the similarities between my result and this are pretty clear.
This also leads me to conjecture for that for two forces in 1 dimension, the forces are proportional to the magnitude of the direction of the other force.
Firstly, is my result correct and valid? Does this theorem have a name, or are there any resources on it I could read?
Can this formulation and result be extended into other (n?) dimensions?
Best Answer
The statement generalizes to all dimensions.
Given a vector space $\mathbb{R}^n$ with the usual Euclidean metric, we represent a $k$-dimensional subplane spanned by the vectors $v_1,\dots,v_k$ with $v_1 \wedge v_2 \wedge \dots \wedge v_k$, where the wedge $\wedge$ is the antisymmetric linear product of the exterior algebra. Also, we have that the volume of an $n$-dimensional "parallelogramm" spanned by $v_1,\dots,v_n$ is just the $\lvert V \rvert$ in
$$ v_1 \wedge v_2 \wedge \dots \wedge v_n = V e_1 \wedge \dots \wedge e_n $$
where $e_i$ is the standard basis. Additionally, we have the Hodge star $\star$, which, in particular, sends a $n$-plane $v_1 \wedge \dots \wedge v_n$ to its volume $\mathrm{vol}(v_1 \wedge \dots \wedge v_n)$.The $n$-dimensional version of Lami's theorem is now:
where the $j_{\circ}$ are a permutation of the indices except $i$, and likewise the $l_\circ$ a permutation of the indices except $k$.
Proof: Choose any two indices $i,j$, let them w.l.o.g. be $1,2$, and then apply $\wedge v_3 \wedge v_4 \wedge \dots \wedge v_{n+1}$ to the equation $\sum_i v_i = 0$. Antisymmetry of the wedge implies that a summand becomes zero as soon as one of the $v_i$ appears twice. Only the summands $v_1$ and $v_2$ survive this process, and we get
$$ v_1 \wedge v_3 \wedge \dots \wedge v_n = - v_2 \wedge v_3 \wedge \dots \wedge v_n $$
Now, write $v_i = \lvert v_i \rvert \hat{v}_i$ where $\hat{v_i}$ is a unit vector. Linearity of the wedge implies $v_1 \wedge v_3 \wedge \dots \wedge v_n = \lvert v_1 v_3 \dots v_n \rvert \hat{v}_1 \wedge \hat{v}_3 \wedge \dots \wedge \hat{v}_n$, and thus we arrive at
$$ \lvert v_1 \rvert \hat{v}_1 \wedge \hat{v}_3 \wedge \dots \wedge \hat{v}_n = \lvert v_2 \rvert \hat{v}_2 \wedge \hat{v}_3 \wedge \dots \wedge \hat{v}_n$$
and applying the Hodge star to both sides yields the desired result.