A not that obvious proof of vector property I want to show that if $\vec a  \cdot \vec b=0$, then there must be some vector $\vec x$ to let $\vec x\times\vec b=\vec a$ which $\vec a  ,\vec b,$ and $\vec x$ are not $\vec 0$ in $\Bbb R^3$.
The reverse is obvious. If $\vec x\times\vec b=\vec a$ then $\vec b\cdot(\vec x\times\vec b)=\vec b\cdot\vec a$
Where $\vec b\cdot(\vec x\times\vec b)=\vec x\cdot(\vec b\times\vec b)=0=\vec b\cdot\vec a$
But I don't find anything that is helpful to use $\vec a  \cdot \vec b=0$
I know this is the truth. Please don't give me an intuitive way, thanks.

Sketch: 


*

*The case where $\vec{a}$ is $\vec{0}$ is trivial, so this case can be dealt with easily.  

*The case where $\vec{b}$ is $\vec{0}$ is a counterexample (unless $\vec{a}$ also equals $\vec{0}$), so we must assume that $\vec{b}\not=\vec{0}$.

*Observe that $\vec{a}\cdot\vec{b}=0$ means that $\vec{a}$ and $\vec{b}$ are perpendicular.  In particular, they point in different directions.  

*Let $\vec{y}=\vec{a}\times\vec{b}$. Then, $\vec{y}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.  Since $\vec{a}$ and $\vec{b}$ do not point in the same direction, $\vec{y}$ is not $\vec{0}$.

*Consider $\vec{y}\times\vec{b}$.  This is a vector perpendicular to both $\vec{y}$ and $\vec{b}$.  

*Note that the set of vectors perpendicular to both $\vec{y}$ and $\vec{b}$ are on a line.  This line includes $\vec{a}$.  Now, scale $\vec{y}$ appropriately to get $\vec{x}$.
