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

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.

Let $d=b\cdot x$. Taking a dot product with $b$ reduces the vector equation to $(1-b\cdot c)d=b\cdot a$. If $b\cdot c\ne 1$, this determines a unique scalar and hence a unique vector:Direct substitution verifies it, so necessity has also produced sufficiency. The formula includes $c=0$, when $x=a$.

If $b\cdot c=1$, the reduced scalar equation becomes $0=b\cdot a$. When $b\cdot a\ne 0$, there is no solution. When $b\cdot a=0$, all solutions are $x=a+dc$ with arbitrary $d\in \mathbb{R}$: their dot product with $b$ is indeed $d$. Since $b\cdot c=1$ forces $c\ne 0$, this is an entire affine line, not a single point. These alternatives are the solvability of an affine rank-one vector equation.