Can the vector vec x be uniquely determined from vec x×vec…

Can the vector $\vec x$ be uniquely determined from $\vec x\times\vec a=\vec b$, where $\vec a,\vec b$ are given?

Step 1: Existence

$\vec x\times\vec a$ is always perpendicular to $\vec a$. Hence a solution exists only if

\[ \vec a\cdot\vec b=0\qquad(\vec a\neq\vec 0). \]

(If $\vec a=\vec 0$, a solution exists only for $\vec b=\vec 0$, and then every $\vec x$ works.)

Step 2: A particular solution

Multiply $\vec x\times\vec a=\vec b$ by $\vec a$ from the left and use $\vec a\times(\vec x\times\vec a)=\vec x\,(\vec a\cdot\vec a)-\vec a\,(\vec a\cdot\vec x)$: \[\vec a\times\vec b=|\vec a|^2\vec x-(\vec a\cdot\vec x)\,\vec a\ \Longrightarrow\ \vec x=\frac{\vec a\times\vec b}{|\vec a|^2}+\mu\,\vec a,\quad \mu=\frac{\vec a\cdot\vec x}{|\vec a|^2}.\]

\[\vec x=\frac{\vec a\times\vec b}{|\vec a|^2}+\mu\,\vec a,\qquad \mu\in\mathbb R\qquad(\vec a\cdot\vec b=0)\]

Step 3: Check

For $\vec x_0=\dfrac{\vec a\times\vec b}{|\vec a|^2}$, with $(\vec u\times\vec v)\times\vec w=\vec v\,(\vec u\cdot\vec w)-\vec u\,(\vec v\cdot\vec w)$: \[\vec x_0\times\vec a=\frac{(\vec a\times\vec b)\times\vec a}{|\vec a|^2}=\frac{\vec b\,(\vec a\cdot\vec a)-\vec a\,(\vec b\cdot\vec a)}{|\vec a|^2}=\vec b.\ \checkmark\]

Adding $\mu\vec a$ changes nothing, because $\vec a\times\vec a=\vec 0$.

Step 4: Answer

No, $\vec x$ is not unique. The solutions form a line parallel to $\vec a$ through $\vec x_0$. The equation fixes only the component of $\vec x$ perpendicular to $\vec a$. One extra scalar condition, such as the value of $\vec x\cdot\vec a$ or $\vec x\cdot\vec c$ for a vector $\vec c$ with $\vec c\cdot\vec a\neq0$, makes it unique (see Problem 106).