Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 15 / a

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 $d=a\cdot b-Bd$. ThusNext take its cross product with $b$. The term $db$ disappears, and the vector triple product identity givesInsert this expression back into $x=a-db-b\times x$. The two longitudinal terms cancel, leaving $(1+B)x=a-b\times a$. ConsequentlyThe 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.