Curl of cross product $(\underline d \times \underline r)$

Question

When finding the curl of a vector cross product such as $$\underline\nabla\times(\underline d\times \underline r)$$, I can use the identity $$\underline\nabla\times(\underline d\times \underline r)=(\underline\nabla\cdot \underline r)\underline d+(\underline r\cdot \underline\nabla)\underline d-(\underline\nabla\cdot \underline d)\underline r-(\underline d\cdot \underline\nabla)\underline r$$If $\underline r = x_1\underline e_1 +x_2\underline e_2+x_3\underline e_3$ (the position vector) and $\underline d$ is a constant vector then the identity becomes:$$\underline\nabla\times(\underline d\times \underline r)=(3)\underline d+(0)\underline d-(0)\underline r-(0)\underline r $$as $\underline\nabla\cdot \underline r = 3 $ and $\underline r\cdot \underline \nabla = 0$ and divergence of $\underline d$ and $\underline d\cdot \underline\nabla$ are zero.

Hence the final result I obtain becomes $\underline\nabla\times(\underline d\times \underline r)=3\underline d$.

However, the answer I was supposed to obtain is $2\underline d$.

Where have I gone wrong?

$Edit: why\:is\: (\underline d\cdot \underline\nabla)\underline r=\underline d\:and\:not\:0?$

Answer 1

$\nabla \times (\overrightarrow{d} \times \overrightarrow{r}) = \nabla \times ((d_y z - d_z y)\overrightarrow{i} + (d_z x - d_x z)\overrightarrow{j} + (d_x y - d_y x)\overrightarrow{k}) = (d_x + d_x)\overrightarrow{i} + (d_y + d_y)\overrightarrow{j} + (d_z + d_z)\overrightarrow{k} = 2\overrightarrow{d}.$

A new comment from February 16, 2020: $$\nabla \times (\overrightarrow{d} \times \overrightarrow{r}) \equiv \nabla \times \begin{array}{|ccc|} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ d_x & d_y & d_z\\ x & y & z\\ \end{array}=\\\nabla \times ((d_y z - d_z y)\overrightarrow{i} + (d_z x - d_x z)\overrightarrow{j} + (d_x y - d_y x)\overrightarrow{k}) \equiv \begin{array}{|ccc|} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ d_y z - d_z y &d_z x - d_x z & d_x y - d_y x\\ \end{array} = \\ (d_x + d_x)\overrightarrow{i} + (d_y + d_y)\overrightarrow{j} + (d_z + d_z)\overrightarrow{k} = 2\overrightarrow{d}.$$