Past exam of the elasticity and plasticity course of the University of Oxford / 2010 / 2 / a / Solution

Assume $\mathbf{k}\ne 0$. Dotting the decomposition with $\mathbf{k}$ givesCrossing it on the left by $\mathbf{k}$ and using $\mathbf{B}\cdot \mathbf{k}=0$ gives $\mathbf{k}\times \mathbf{a}=|\mathbf{k}{|}^2\mathbf{B}$. ThusIndeed $(\mathbf{k}\times \mathbf{a})\times \mathbf{k}/|\mathbf{k}{|}^2=\mathbf{a}-A\mathbf{k}$, verifying the decomposition and the imposed transverse condition.