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So i need to find the curl of a vector field $v=(a\cdot r)a\times r$ where $a$ is some constant vector and $r=(x,y,z)$ is the position vector.

So i know the curl is given by the cross product $$\nabla \times v =\nabla \times ((a\cdot r)a\times r)$$

I know I can't use regular properties of dot and cross products since the $\nabla$ operator is non-commutative and the furthest i can get is$$\nabla \times v =\nabla \times(a\cdot r) (a\times r)$$

Since $a \cdot r$ is a scalar. So what should i do?

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    $\begingroup$ You can use the standard definition of the curl of a vector field. $\endgroup$ – Dog_69 Nov 9 '18 at 14:47
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    $\begingroup$ Learn the identities for curl $(\Omega \mathbf F)$ and curl $(\mathbf F \times \mathbf G)$, etc. See e.g. Bourne & Kendall's "Vector Analysis & Cartesian Tensors" $\endgroup$ – Richard Martin Nov 9 '18 at 14:49
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The $\nabla\times$ trick for computing the curl is only meant for the simplest situations. This mnemonic device cannot be integrated into more complicated vector algebra constructions, as are present in the example at hand.

In order to simplify the computations we introduce in ${\mathbb R}^3$ a new righthanded orthonormal frame $({\bf e}_1,{\bf e}_2,{\bf e}_3)$ with ${\bf e}_1$ in the direction of ${\bf a}$, so that ${\bf a}=\alpha{\bf e}_1$, $\>\alpha>0$. Then $${\bf v}({\bf x})=\alpha^2({\bf e}_1\cdot{\bf x})\bigl({\bf e}_1\times{\bf x}\bigr)=\alpha^2 x_1(0,-x_3,x_2)=\alpha^2(0,-x_1x_3,x_1x_2)\ .$$ Using the standard partial derivatives formula for computing the curl one obtains $${\rm curl}\,{\bf v}({\bf x})=\alpha^2(2x_1,-x_2,-x_3)=3\alpha^2({\bf e}_1\cdot{\bf x}){\bf e}_1-\alpha^2{\bf x}\ .$$ This can be rewritten in the coordinate free form $${\rm curl}\,{\bf v}({\bf x})=3({\bf a}\cdot{\bf x}){\bf a}-|{\bf a}|^2{\bf x}\ .\tag{1}$$ If you don't believe this compute ${\rm curl}\,{\bf v}$ with respect to the given coordinates $(x_1,x_2,x_3)$. In this setup ${\bf a}=(a_1,a_2,a_3)$, and you first have to compute the components $v_i$ of$${\bf v}({\bf x})=\bigl(v_1(x_1,x_2,x_3),v_1(x_1,x_2,x_3),v_1(x_1,x_2,x_3)\bigr)\ .$$ Then start differentiating. At the end you will see that the result can be simplified to $(1)$:

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  • $\begingroup$ Hi thanks for the answer, but what about the other possible components of $\textbf{a}$? $\endgroup$ – Questlove Nov 9 '18 at 21:38
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    $\begingroup$ See my edit. There are no other components of ${\bf a}$. $\endgroup$ – Christian Blatter Nov 10 '18 at 7:38
  • $\begingroup$ Thanks again and sorry for the trouble. $\endgroup$ – Questlove Nov 11 '18 at 23:38

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