If $\mathbf a$ is a constant vector in the 3-dimensional space and $\mathbf s=x\mathbf e_x+y\mathbf e_y +z\mathbf e_z$, I want to show that $$\nabla \land \left(\mathbf a \land \mathbf s\right) = 2\mathbf a. $$

I have done as follows: $$\nabla \land \left(\mathbf a \land \mathbf s\right)=(\nabla \cdot \mathbf s)\mathbf a\ -\ (\nabla \cdot \mathbf a)\mathbf s=3\mathbf a\ -\ (\nabla \cdot \mathbf a)\mathbf s $$

But I am confused as to how the last part is computed. Could you explicitly show how $(\nabla \cdot \mathbf a)\mathbf s$ equals $\mathbf a$ or point out any other mistake?

  • $\begingroup$ Your formula for $\nabla \wedge ({\bf a} \wedge {\bf s})$ is incorrect in view of $\nabla$ being a differential operator. It is not a simple triple cross product, the product rule will generate extra terms. $\endgroup$
    – Ivo Terek
    Commented Jan 3, 2018 at 17:00

1 Answer 1


We have the general formula $$\nabla \wedge ({\bf a}\wedge {\bf s}) = {\bf a}(\nabla \cdot {\bf s}) - {\bf s}(\nabla \cdot {\bf a}) + ({\bf s}\cdot \nabla){\bf a} - ({\bf a}\cdot \nabla){\bf s}.$$Now let's see each piece.

  • Since ${\bf s} = (x,y,z)$, clearly $\nabla \cdot {\bf s} = 3$.
  • Since ${\bf a}$ is constant, we have $\nabla \cdot {\bf a} = 0$.
  • Since ${\bf a}$ is constant and ${\bf s}\cdot \nabla$ is a differential operator, $({\bf s}\cdot \nabla){\bf a} = {\bf 0}$.
  • This last part we compute directly, calling ${\bf a} = (a_1,a_2,a_3)$, acting componentwise as follows: $$({\bf a}\cdot \nabla){\bf s} = \left(a_1\partial_x+a_2\partial_y+a_3\partial_z\right)(x,y,z) = (a_1 \cdot 1, a_2 \cdot 1, a_3 \cdot 1) = {\bf a}.$$

So everything boils down to $$\nabla \wedge ({\bf a}\wedge {\bf s}) =3{\bf a} - {\bf 0} + {\bf 0} - {\bf a} = 2{\bf a},$$as wanted.


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