= Cross product equation with a fixed unit vector
{title2=$r\times g=f\iff r=g\times f+tg,\quad f\cdot g=0$}
For a fixed <unit vector> $g\in\mathbb R^3$, the <linear map> $r\mapsto r\times g$ has <kernel> $\mathbb Rg$ and <image> the perpendicular <plane> $g^\perp$. Thus $r\times g=f$ is solvable exactly when $f\cdot g=0$, and then its full solution set is
$$
r=g\times f+tg,\qquad t\in\mathbb R.
$$
For the proof, a <cross product> with $g$ must be perpendicular to $g$. Conversely the <vector triple product identity> gives $g\times(r\times g)=r-(r\cdot g)g$, which determines the perpendicular <orthogonal projection> of $r$. The remaining parallel component is arbitrary. Finally $(g\times f)\times g=f$ when $f\cdot g=0$, verifying sufficiency.
Geometrically, the solution set is a <straight line> with direction $g$. Its nearest point to the origin is $g\times f$ and its distance from the origin is $|f|$. For a general nonzero fixed <vector> $a$, the corresponding formula is $r=(a\times f)/|a|^2+ta$, again subject to $f\cdot a=0$.
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