= Cross-product operator with a longitudinal rank-one term
{title2=$x+(b\cdot x)b+b\times x=a$}
For any real three-dimensional <vectors> $a,b$, the displayed <linear equation> has the unique solution
$$
x=\frac{a-b\times a}{1+|b|^2}.
$$
Taking 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> gives
$$
b\times x=b\times a-b(b\cdot x)+|b|^2x.
$$
Substitution 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 $C_bx=b\times x$, the underlying identity for <linear maps> is $(I-C_b)(I+C_b+bb^T)=(1+|b|^2)I$, using $C_b^2=bb^T-|b|^2I$.
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