 Cross-product operator with a longitudinal rank-one term

For any real three-dimensional vectors $a,b$, the displayed linear equation has the unique solutionTaking a dot product with $b$ first yields $(1+|b{|}^2)(b\cdot x)=a\cdot b$. Taking a cross product with $b$ and using the vector triple product identity givesSubstitution into the original equation cancels the longitudinal terms and proves the formula. Conversely direct substitution verifies it for all $b$, including $b=0$, since the denominator is positive. If $C_{b}x=b\times x$, the underlying identity for linear maps is $(I-C_b)(I+C_b+b{b}^T)=(1+|b{|}^2)I$, using $C_b^2=b{b}^T-|b{|}^{2}I$.