Strategy for breaking up a vector product

Question

Given is the equality

$$\vec{A}\times\vec{B}=\vec{C}$$ What is the strategy for breaking up the vector product and expressing $\vec{B}$? Given this equality, what does $\vec{B}$ alone equal to?

I know one strategy, say $\vec{A}$ is a constant vector and $\vec{B}$ is the position vector $\vec{x}$:$$\vec{A}\times\vec{x}=\vec{C}$$ Take curl of both sides: $$\nabla\times(\vec{A}\times\vec{x})=\nabla \times \vec{C},$$ then $$2\vec{A}=\nabla \times \vec{C}, $$ or $$\vec{A}=\frac{1}{2}\left( \nabla \times \vec{C} \right)$$

Do you have a strategy for expressing either A or B when both $\vec{A}$ and $\vec{B}$ are not constant? How do we eliminate $\vec{A}$ form the LHS to express $\vec{B}$.

Answer 1

Attempted answer:

If $$\vec{A}\times\vec{B}=\vec{C},$$ then $$\vec{B}=-\frac{\vec{A}\times\vec{C}}{|A|^2}+\lambda\vec{A},$$ where $\lambda$ is an arbitrary constant.

Check if true:

$$\vec{A}\times\vec{B}=$$ $$=-\frac{1}{|A|^2}\vec{A}\times(\vec{A}\times\vec{C})+\lambda(\vec{A}\times\vec{A})$$ $$=-\frac{1}{|A|^2}\vec{A}\times(\vec{A}\times\vec{C})$$ $$=-\frac{1}{|A|^2}\left[(\vec{A}\cdot\vec{C})\vec{A}-(\vec{A}\cdot\vec{A})\vec{C} \right]$$

since $\vec{A}\times\vec{B}=\vec{C},$ the vectors A and C are orthogonal, $(\vec{A}\cdot\vec{C})=0$ and we are getting $$=\vec{C}$$

Answer 2

Short answer to

How do we eliminate $A$ form the LHS to express $B$?

is that you can't.

Geometrically, the cross product $A \times B$ is a vector perpendicular to both $A$ and $B$ whose length is the area of the parallelogram determined by $A$ and $B$. Then it's clear that there are many vectors $X$ for which $A \times B = A \times X$.