Division using cross vector

Question

Given $4$ vectors $\vec a$ and $\vec b$, $\vec c$, $\vec d$, if

$$\vec u=\vec a \times \vec b= \hat n \ ab \sin θ \tag 1$$ where $\hat n$ is a normal unit vector of both $\vec a$ and $\vec b$, and $θ$ is the angle formed by the two vectors. Same for $\vec u$. I have $$\vec v=\vec c \times \vec d= (-\hat n) \ cd \sin θ' \tag 2$$

If I consider

$$\frac{\vec u}{\vec v}=\frac{\hat n \ ab \sin θ}{-\hat n \ cd \sin θ'}=\frac{\hat n \ ab \sin θ}{-\hat n \ ab \sin θ'}$$

Is it correct this appoach?

$$\frac{\vec u}{\vec v}=\frac{\color{red}{\not{\hat n}} \ ab \sin θ}{-\color{red}{\not{\hat n}}\ cd \sin θ'}=k \frac{\sin θ}{\sin θ'}, \quad k=-ab/cd \tag 3$$

the value of the $(1)$ is different for different values of $θ$ (and $θ'$) and therefore the division $(3)$ not uniquely determined? Is it correct?

Answer 1

As J.G. mentioned in the comments, your definition of division isn't very general because the only way the $\vec n$'s are the same (and thus cancellable) is if all of the vectors are in the same plane. And I suspect it's not well-defined even that context either.

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But if you're interested in division of vectors, you should look into Geometric Algebra. There is a new vector product $\vec u\vec v$ called the geometric product which has the property $$\vec u\vec v = \vec u\cdot \vec v + \vec u \wedge \vec v$$ That $\vec u\wedge \vec v$ is probably unfamiliar to you as well, but roughly it's a different (and more general) way to think about the cross product. It's not exactly the same though. But anyway, because $\vec v\wedge\vec v = 0$ for all $\vec v$, we can define the inverse of a vector $\vec v$ as $$\vec v^{\ -1} = \frac{\vec v}{\|\vec v\|^2}$$

This works because $$\begin{align} \vec v\vec v^{\ -1} &= \vec v\frac{\vec v}{\|\vec v\|^2} \\ &= \frac{\vec v\vec v}{\|\vec v\|^2} \\ &= \frac{\vec v\cdot \vec v + \vec v\wedge \vec v}{\|\vec v\|^2} \\ &= \frac{\|\vec v\|^2 + 0}{\|\vec v\|^2} \\ &= 1 \end{align}$$

Answer 2

There is no division of vectors in terms of the inverse of the cross product.

Essentially, when using the cross product you get a vector, but contrary to real numbers where its division produces another number, such an operation does not seems to exist with vectors. For example, when using real numbers the division of a number by itself is one, but the closest thing we have with vectors is a normalization (where what it becomes unity is the length of the vector, but do not cancel the direction). In terms of products $a\: .\: 1=a$ for $a\in \mathbb{R}$, but there is not such a unit vector satisfying that property for the cross product, ergo the division it is not defined in the same sense.