 Solution

Take the dot product of the equation with $a$. Since $a\cdot (a\times r)=0$, we obtain $a\cdot r=(a\cdot c)/\lambda$. Next take its cross product with $a$ and apply the vector triple product identity:Substitute $a\times r=c-\lambda r$ and the known dot product, then rearrange:Here the vectors and $\lambda \ne 0$ are real, so the denominator is positive. The formula also covers $a=0$, giving $r=c/\lambda$. To check existence as well as uniqueness, write its numerator as $N$. The vector triple product identity gives $a\times N=\lambda a\times c-a(a\cdot c)+|a{|}^{2}c$, so $a\times N+\lambda N=({\lambda }^2+|a{|}^2)c$. Substitution verifies the original equation. This is an instance of the inverse of a scalar plus a cross-product operator.