If $\mathbf{A} \times \mathbf{B} = \mathbf{C}$, $\mathbf{A}$ and $\mathbf{B}$ are unique?

Question

As in the title, I know that a vector $\mathbf{C}$ is obtained by two vectors $\mathbf{A}$ and $\mathbf{B}$: by hypothesis, they are both entirely lying in a plane orthogonal to $\mathbf{C}$ and they are mutually orthogonal, that is $\mathbf{A} \perp \mathbf{B}$.

With such conditions, are $\mathbf{A}$ and $\mathbf{B}$ unique?

(I say: no, because can exist another couple of vectors $\mathbf{A}',\mathbf{B}'$ mutually orthogonal and such that $|\mathbf{A}| = |\mathbf{A}'|, |\mathbf{B}| = |\mathbf{B}'|$ which can produce $\mathbf{C}$ as well. Isn't it?)

Answer 1

They aren't unique for sure. You can scale $A$ up by $\alpha$ and scale $B$ down by $\alpha$ and their cross product will be the same, for instance. Alternatively, the cross product of $\hat{i}$ and $\hat{j}$ is $\hat{k}$ but so is the cross product of $\frac{1}{\sqrt 2}(\hat{i}-\hat{j})$ and $\frac{1}{\sqrt 2}(\hat{i}+\hat{j})$.

Answer 2

As the comments and the answers clearly explain, the answer is no, they are not unique.

If you understand the geometric meaning of $\mathbf{C}=\mathbf{A} \times \mathbf{B}$, then given $\mathbf{C}$ you can solve geometrically for all possible pairs $\mathbf{A},\mathbf{B}$.

Namely: $\mathbf{C}$ is the vector perpendicular to the plane spanned by $\mathbf{A}$ and $\mathbf{B}$, pointing in the direction that obeys the right hand rule, whose magnitude equals the area of the parallelogram spanned by $\mathbf{A}$ and $\mathbf{B}$.

So, start with $\mathbf{C}$, take a perpendicular plane, draw any parallelogram in that plane of area equal to the magnitude of $\mathbf{C}$, and take $\mathbf{A}$, $\mathbf{B}$ to be the vectors represented by two sides of that parallelogram, with their directions specified so that $\mathbf{A}, \mathbf{B}, \mathbf{C}$ obey the right hand rule.

Working this out specifically for $\mathbf{C}=\langle 0,0,1 \rangle$ is fun.