Samuel Humeau

An expression of the vector cross product in non-orthogonal basis systems

I came accross this problem in an old french textbook: Traité de mathématiques spéciales, Georges Cagnac, Joanny Commeau, Edmond Ramis, 1967, Masson.

I got stuck on it, finding one reference to it online in the form of a question at https://math.stackexchange.com. I ended up finding a solution that I will now detail.

We are in $\mathbb{R}^3$. Suppose you are given a basis $(\vec{u},\vec{v},\vec{t})$. Let us note $\alpha=(\vec{v},\vec{t});~~~\beta=(\vec{t},\vec{u});~~~\gamma=(\vec{u},\vec{v})$ the corresponding angles.

In order to simplify let me suppose $\vec{u}$, $\vec{v}$, and $\vec{t}$ are unit vectors. The solution will be complicated enough. It is always possible to swap the vectors so as to get a right-handed basis. Thus, without lost of generality (wlog) the considered basis is supposed right-handed.

Then let $\vec{t'}$ be a unit vector of same direction and sense than $\vec{u}\times\vec{v}$, and let $\vec{v'}$ be a unit vector in the same plane than $\vec{u}$ and $\vec{v}$ such that $\vec{u}$, $\vec{v'}$, $\vec{t'}$ is orthonormal. Note that as $\vec{t'}$ is colinear to $\vec{u}\times\vec{v}$ it is orthogonal to the plane containing $\vec{u}$ and $\vec{v}$ so that $\vec{v'}=\vec{t'}\times\vec{u}$ exists. Finally let $\delta$ be the angle $(\vec{t},\vec{t'})$.

Now that notations are set let me express $\vec{t}$ in the basis $\vec{u}$, $\vec{v'}$, $\vec{t'}$. As it is orthonormal we can use the scalar product: $\vec{t}\cdot\vec{u}=\cos\beta;~~~\vec{t}\cdot\vec{t'}=\cos\delta$ Finding $\vec{t}\cdot\vec{v'}$ is a bit more difficult: we need to express $v'$. As it is in the same plane than $\vec{u}$ and $\vec{v}$, as it is orthogonal to $\vec{u}$, and as $\alpha=(\vec{u},\vec{v})$, we get $v=\cos\gamma\vec{u}+\sin\gamma\vec{v'}$ But $\sin\gamma\neq 0$ as $\vec{u}$ and $\vec{v}$ are not colinear. Thus, $\vec{v'}=\frac{1}{\sin\gamma}\vec{v}-\frac{\cos\gamma}{\sin\gamma}\vec{u}$ giving $\vec{t}\cdot\vec{v'}=-\frac{\cos\gamma\cos\beta}{\sin\gamma}+\frac{\cos\alpha}{\sin\gamma}$ and $\vec{t}=\cos\beta\vec{i}+\left(\frac{\cos\alpha-\cos\gamma\cos\beta}{\sin\gamma}\right)\vec{v'}+\cos\delta\vec{t'}~~~~(1)$ This equality between vectors means equality of their squared norm: $1=\cos^2\beta+\left(\frac{\cos\alpha-\cos\gamma\cos\beta}{\sin\gamma}\right)^2+\cos^2\delta$ I will spare the details but it gives $\cos\delta=\frac{\sqrt{1-2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma}}{\sin\gamma}$ Note that the choice of orientation of the two basis ensure that $\cos\delta\ge 0$ making sure we chose the right squared root.

By definition of $\vec{t'}$ we have $\vec{u}\times\vec{v}=\sin\gamma\vec{t'}$ Using $(1)$ and after some more involved calculus we get $\vec{u}\times\vec{v}=\frac{(\cos\alpha\cos\gamma-\cos\beta)\vec{u}+(\cos\gamma\cos\beta-\cos\alpha)\vec{v}+\sin^2\gamma\vec{t}}{\sqrt{1-2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma}}$ Using symmetry arguments this easily gives $\vec{v}\times\vec{t}$ and $\vec{t}\times\vec{u}$. Note also that when the original basis is orthonormal, all cosines are equal to $0$ and the sine is equal to $1$: we find the usual $\vec{u}\times\vec{v}=\vec{t}$.