# Cross product of the vector r and second order tensor

Position vector r is given as $$\vec r=x_i\hat e_i$$ and the second order tensor T is given as: $$\overline{\overline{T}}=\frac{\delta_{ij}\hat e_i\hat e_j}{r}+\frac{x_ix_j \hat e_i\hat e_j}{r^3}$$.

How to determine (cross product) $$\vec r\times\overline{\overline{T}}$$ by using index notation rules? I would appreciate any suggestions as I don't know whether I can dot the vector into the two parts of the tensor separately as $$\vec r\times\overline{\overline{T}}=\vec r\times\overline{\overline{T_1}}+\vec r\times\overline{\overline{T_2}}$$ or not.

Actually I have get this answer, but I don't know how I can simplify it more. $$\vec r\times\overline{\overline{T}}=(\delta_{ij}+\frac{x_ix_j}{r^2})\frac{x_m}{r}\epsilon_{mik}\hat e_k\hat e_j$$.

The main variables are \eqalign{ \rho &= \|r\| \cr n &= \rho^{-1}r \,\,\,\,\,\,\,\,\,\,{\rm [\,unit\,vector\,]} \cr r &= \rho n \cr T &= \rho^{-1}(I + nn^T) \cr } and we'd like to evaluate the matrix $$\,\,N = r\times T$$
Since the cross product distributes over addition, we can use the fact that $$n\times n=0$$ to simplify the calculation to $$N = n\times(I + nn^T) = n\times I$$ Switching to index notation $$N_{jp} = -n_{i}\varepsilon_{ijk}\delta_{kp} = -n_{i}\varepsilon_{ijp}$$ or in matrix notation $$N =\begin{bmatrix}0&-n_3&n_2\\n_3&0&-n_1\\-n_2&n_1&0\end{bmatrix} =\frac{1}{\rho}\begin{bmatrix}0&-r_3&r_2\\r_3&0&-r_1\\-r_2&r_1&0\end{bmatrix}$$