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If vectors $A$ and $B$ are fixed, then by definition of the cross product, I know that if $B$ is not perpendicular to $A$ then there will be no solutions.

So assume $B$ is perpendicular to $A$, then surely $x$ will have a fixed length and can take any direction (except $A$) in the plane perpendicular to $B$ containing vector $A$.

So does this mean that the set of possible $X$ vectors is a circle of fixed radius?

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Let's first solve the homogenous equation $A \times x = 0$. It's easy to see that if $x$ isn't parallel to $A$, then $A \times x \neq 0$, so the only solutions of equation $A\times x=0$ are vectors $x = \lambda A$, $\lambda\in\mathbb R$.

Let's now assume we have two solutions $x_1$ and $x_2$ of the nonhomogenous equation, $A\times x_1 = A\times x_2 = B$. We can notice that $A \times (x_2-x_1) = 0$, that is $x_2-x_1$ is a solution of homogenous equation that is $x_2-x_1=\lambda A$ for some $\lambda\in\mathbb R$. That means that if you have one solution $x_1$ of the nonhomogenous equation, all other solutions will have form $x = x_1 + \lambda A $, so they will form a line parallel to $A$.

Let us find that line, assuming that $B$ is perpendicular to $A$ (otherwise there's no solutions). We can use the fact that $A\times (A\times B) = (A\cdot B)A- (A\cdot A) B = -|A|^2 B$ to find one solution of the equation $A\times x=B$: $x_1 = -\frac{1}{|A|^2}A\times B$. By the previous arguments, all soultions of this equations have form: $$ x = -\frac{1}{|A|^2}A\times B + \lambda A, \qquad \lambda\in\mathbb R$$ and form a line.

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$A\times x_1=A\times x_2\iff A\times (x_1-x_2)=0\iff (A=0\lor x_1-x_2\in\Bbb RA)$.

Therefore the solution set of the equation $A\times x=B$ is:

  • the empty set if either $\langle B,A\rangle\ne 0$ or $A=0\land B\ne 0$.

  • $\Bbb R^3$ if $A=0$ and $B=0$

  • the $1$-dimensional affine subspace $\overline x+\Bbb RA$ if $A\ne 0\land \langle A,B\rangle=0$, where $\overline x$ is the vector in $\operatorname{Span}(A,B)^\perp$ such that $\lVert\overline x\rVert=\frac{\lVert B\rVert}{\lVert A\rVert}$ and such that either $\overline x=0$ or $A, \overline x, B$ is a positive basis of $\Bbb R^3$. It turns out that such vector may be written as $$\overline x=\frac1{\langle A,A\rangle}B\times A$$

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Firstly, by your geometric argument all such $x$ must lie in a fixed plane perpendicular to B. Since the cross product magnitude is (twice) the area of the triangle formed by the two vectors, we know that all such x form a triangle with a of area b/2. Therefore there are two such classes of solutions; the difference of any pair in the same class is a multiple of a. Geometrically, these two classes are two lines both parallel to a.

Edit: as pointed out in the comments, the magnitude is really the signed area of the triangle. Thus only one of the two lines is actually correct, and the other one isn’t (use the right hand rule to work out which) (the cross would be -A)

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  • $\begingroup$ I know x must lie in a plane perpendicular to B, but why also A? $\endgroup$ – Shree May 21 at 8:30
  • $\begingroup$ Only one of those two lines will be correct, I think. On the other line, $A\times x=-B$. $\endgroup$ – TonyK May 21 at 8:34
  • $\begingroup$ @Shree: I think auscrypt meant perpendicular to $B$. $\endgroup$ – TonyK May 21 at 8:36
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$\vec x$ lies in the plane perpendicular to $B$, so does $\vec A$.

$|\vec A|\cdot |\vec x| \sin \theta =|\vec B|$.

Hence for $\vec x$ not parallel or antiparallel to $\vec A$, i.e. $\theta \in (0,π)$:

$|\vec x|\sin \theta =\dfrac{ |\vec B|}{|\vec A|}$.

Consider a $x,y$ coordinate system In the plane perpendicular to $\vec B$. Let the positive $x-$axis be along $\vec A$, choose positive $y-$axis along $\vec B ×\vec A$.

Polar coordinates:

$r\sin \theta =C$, where $C=|\vec B|/|\vec A|>0$.

This is the line $y = C$.

Equation of this line in 3D:

Direction vector: $\vec A.$

Point $(0,C)$: $\dfrac{C}{|\vec A × \vec B|}\vec B × \vec A $

$\vec x= \dfrac{1}{|A|^2} \vec B × \vec A +t \vec A$, $t$ real.

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