= Solvability of an affine rank-one vector equation
{title2=$x=a+(b\cdot x)c$}
Let $a,b,c\in\mathbb R^n$, with $c\ne0$. The equation $x=a+(b\cdot x)c$ is the <linear system> $(I-cb^T)x=a$, a <rank-one matrix update> of the <identity matrix>. Every solution has $x=a+\lambda c$. Substitution reduces the entire problem to
$$
\lambda(1-b\cdot c)=b\cdot a.
$$
If $b\cdot c\ne1$, there is one solution, $x=a+c(b\cdot a)/(1-b\cdot c)$. If $b\cdot c=1$ and $b\cdot a=0$, every $\lambda\in\mathbb R$ works and the solution set is the <straight line> $a+\mathbb Rc$. If $b\cdot c=1$ and $b\cdot a\ne0$, there is no solution.
The first reduction follows directly from the original equation in any dimension. In three dimensions it can also be obtained by taking the <cross product> with $c$, giving $(x-a)\times c=0$. The singular denominator therefore represents two different possibilities, inconsistency or a family of solutions, and must not simply be discarded as an undefined formula.
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