$\vec{y}=\vec{a}\times \vec{x}$. is it possible to define $\vec{x}$ as function of $\vec{y}$ $let\quad\vec{a} \in \mathbb{R^3}$ is it possible to define $\vec{x}$ as function of $\vec{y}$? $$\vec{y}=\vec{a}\times \vec{x}$$
So according to the solution the answer is not and I would like to focus on the geometric explanation.
the solution says that  if we use the magnittude definition we can get the relation $||\vec{x}||sin(\theta)=\frac{||\vec{y}||}{||\vec{a}||}$ so for a fixed $\vec{y}$ and $\vec{a}$ there is endless values for $\vec{x}$ that will satisfy the equation. And I really not understand how? I have fixed value on the right side wich means I also know $\theta$, so how is it possible?
 A: $\mathbf y = \mathbf a\times \mathbf x$ allows us to define $y$ as a function of $x.$
But this function has a non-trivial kernel.  That is, if $\mathbf x$ is a scalar multiple of $\mathbf a$ then $y = \mathbf a\times \mathbf x = \mathbf 0$
This means that the function $\mathbf y=\mathbf a\times \mathbf x$ is not 1-1 and not invertible.
Considering the geometric interpretation, the direction of $y$ is perpendicular to the plane containing $a,x$ and the magnitude of $y$ is the area of the paralelogram formed by the vectors $a$ and $x.$  For a given plane, containing $a,x$ there a multiple paralelograms that are the same area.
A: For fixed $\mathbf{a}$, the function $\mathbf{F}_{\mathbf{a}}:\mathbb{R}^3\to\mathbb{R}^3$ which maps $\mathbf{x}$ to $\mathbf{y}:=\mathbf{a}\times\mathbf{x}$ is obviously not one to one. For example, $\mathbf{F}_\mathbf{a}$ maps $\mathbf{0}$ to $\mathbf{0}$, but it also maps $\mathbf{a}$ to $\mathbf{0}$.
Thus $\mathbf{F}_\mathbf{a}$ is not invertible.
A: The relation $$\|\vec{x}\|\sin \theta =\frac{\|\vec{y}\|}{\|\vec{a}\|}$$ implies that given $\vec{y}$ and $\vec{a}$, you can find the product $\|\vec{x}\|\sin \theta$. In particular, you can take many values for $\theta$ and so also for $\|\vec{x}\|$.
For example, if $$\frac{\|\vec{y}\|}{\|\vec{a}\|}=1, $$then you can take say $\theta=\pi/2$ and $\|\vec{x}\|=1$ or $\theta=\pi/4$ and $\|\vec{x}\|=\sqrt2$.
