= Solution
Put $d=b\cdot x$ and $B=|b|^2$. Taking the <dot product> of the original equation with $b$ eliminates $b\times x$ and gives $d=a\cdot b-Bd$. Thus
$$
\boxed{b\cdot x=\frac{a\cdot b}{1+|b|^2}.}
$$
Next take its <cross product> with $b$. The term $d b$ disappears, and the <vector triple product identity> gives
$$
b\times x=b\times a-b\times(b\times x)
=b\times a-bd+Bx.
$$
Insert this expression back into $x=a-db-b\times x$. The two longitudinal terms cancel, leaving $(1+B)x=a-b\times a$. Consequently
$$
\boxed{x=\frac{a-b\times a}{1+|b|^2}=\frac{a+a\times b}{1+|b|^2}.}
$$
The denominator is strictly positive for every real $b$, including $b=0$. Substitution using the same triple-product identity verifies the solution and establishes uniqueness. This is the <cross-product operator with a longitudinal rank-one term>.
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