 Solvability of an affine rank-one vector equation

Let $a,b,c\in {\mathbb{R}}^n$, with $c\ne 0$. The equation $x=a+(b\cdot x)c$ is the linear system $(I-c{b}^T)x=a$, a rank-one matrix update of the identity matrix. Every solution has $x=a+\lambda c$. Substitution reduces the entire problem toIf $b\cdot c\ne 1$, 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 $a+\mathbb{R}c$. If $b\cdot c=1$ and $b\cdot a\ne 0$, 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.