How to find $\vec b$ from $\vec a \times \vec b$, $\vec a$ and $\alpha=\angle(\vec a;\vec b)$ purely algebraically?
I was able to do so, but I needed to use geometric considerations. Let us for now choose coordinate system such that all vectors originate from (0;0;0) point, $\vec a$ coincides with $\hat x$ and $\vec c$ coincides with $\hat y$. $\vec a \times \vec b$ coincides with $\hat y$. Try to prove this.
Clearly we can calculate $$a1)~~|b|=\frac{|c|}{|a|sin \alpha}$$, using crossproduct properties.
Clearly $\vec b$ is in the plane formed by $\vec a$ and $\vec a \times \vec c$ (It easy see from drawing), and we know $\angle(\vec a; \vec b)=\alpha$, so $$a2)~~proj_x \vec b=|b|cos \alpha \frac{\vec a}{|a|}$$ and $$a3)~~proj_y \vec b= |b|sin \alpha \frac{\vec a \times \vec c}{|\vec a \times \vec c|}$$. $$a4)~~\vec b=proj_x \vec b+proj_y \vec b$$ From a1-a4 we infer $$\vec b=\frac{|c|}{|a| sin \alpha}(cos \alpha \frac{\vec a}{|a|}+sin \alpha \frac{\vec a \times \vec c}{|\vec a \times \vec c|})$$
Although all steps clear to me, I want to do it formally without using any geometric intuition.I attempted to do so. So let’s try to solve this again. $$b1)~~\alpha=\angle(\vec a;\vec b)=\frac{\vec a \cdot \vec b}{|a||b|}$$ $$cos\alpha=\frac{a_x b_x+a_y b_y+a_z b_z}{|a||b|}$$ $$b2)~~\vec c \cdot \vec b=0$$, because $\vec c=\vec a\times \vec b$, so $$c_x b_x+c_y b_y+c_z b_z=0$$ $$b3)~~ |b|^2= b_x^2+b_y^2+b_z^2$$
So we need to solve system of equations b1, b2, b3, as result we should get three very big quadratic equations with $b_x$, $b_y$ and $b_z$ variables, all of them can be solved. Is there easier way? I am grasping that it is possible to solve this system of equations without decomposing vectors to its components. $$ \left\{ \begin{array}{c} \vec c=\vec a \times \vec b \\ \angle(\vec a;\vec b)=\alpha \end{array} \right. $$ where $\vec a$, $\vec c$ and $\alpha$ are known.
