Equation on cross product

Question

I have to solve the equation $\vec x\times\vec a=\vec b-\vec x$ , where $\vec x, \vec a,\vec b$ are vectors and the last two are known.

I have proven that $\vec a\cdot \vec x=\vec a\cdot\vec b$, but that is all.

Thank you in advance !

<< Symbol : $\vec a \vec b = \vec a \cdot \vec b$ >>

Answer 1

Given:

$$\vec{x} = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}, \vec{a} = \begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix} ~\text{and}~ \vec{b} = \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix},$$ the equation $$\vec x\times\vec a=\vec b-\vec x$$

corresponds to:

$$\begin{cases} a_3 x_2 - a_2 x_3 +x_1 - b_1 = 0\\ a_1 x_3 - a_3 x_1 +x_2 - b_2 = 0\\ a_2 x_1 - a_1 x_2 +x_3 - b_3 = 0 \end{cases}.$$

The solution is:

$$x_1 = \frac{a_1(\vec{a} \cdot \vec{b}) + b_1 + a_2 b_3 - a_3 b_2}{|\vec{a}|^2 + 1},$$

$$x_2 = \frac{a_2(\vec{a} \cdot \vec{b}) + b_2 + a_3 b_1 - a_1 b_3}{|\vec{a}|^2 + 1},$$

$$x_2 = \frac{a_3(\vec{a} \cdot \vec{b}) + b_3 + a_1 b_2 - a_2 b_1}{|\vec{a}|^2 + 1}.$$

Note that $\vec{a} \cdot \vec{b}$ is the scalar product, while $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$ is the squared norm of the vector $\vec{a}$.

Answer 2

If $\vec{x}=\vec{y},\,\vec{x}=\vec{y}+\vec{z}$ both work, $\vec{z}\times\vec{a}=-\vec{z}$, so $\vec{z}$ is self-orthogonal and vanishes, and the solution $\vec{x}$ is unique. Now try the Ansatz $\vec{x}=A\vec{a}+B\vec{b}+C\vec{a}\times\vec{b}$. As I noted in a comment on @thecandyman's answer, you should find$$\vec{x}=\frac{(\vec{a}\cdot\vec{b})\vec{a}+\vec{b}+\vec{a}\times\vec{b}}{a^2+1}.$$