= Inverse of a scalar plus a cross-product operator
{title2=$\alpha\mathbf x+\beta\mathbf n\times\mathbf x=\mathbf b$}
For a <unit vector> $\mathbf n$, real $\alpha\ne0$, and real $\beta$, the displayed <linear equation> has the unique <vector> solution
$$
\mathbf x=\frac{\alpha\mathbf b-\beta\mathbf n\times\mathbf b+(\beta^2/\alpha)\mathbf n(\mathbf n\cdot\mathbf b)}{\alpha^2+\beta^2}.
$$
To prove it, split <vectors> into components parallel and <orthogonal> to $\mathbf n$. On the parallel line, the <linear map> is multiplication by $\alpha$. On the perpendicular plane, the operator $N\mathbf v=\mathbf n\times\mathbf v$ obeys $N^2=-I$, by the <vector triple product>. Therefore $(\alpha I-\beta N)(\alpha I+\beta N)=(\alpha^2+\beta^2)I$ on that plane. Combining its inverse with the parallel solution gives the formula. This avoids component calculations in equations with an axial <cross product> term.
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