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Calculate $\nabla\cdot F$ and $\nabla\wedge$$F$ for the vector function $F(x,y,z)=(a\wedge r)\wedge r$ where $r=xi+yj+zk$ and $a$ is a constant vector.

So little stuck here, I have used the fact that $F(x,y,z)=(a\wedge r)\wedge r$ $= (a\cdot r)r-(r\cdot r)a$ then I broke all the terms down using $a_{1}, a_{2}, a_{3}$ etc however this gets very tedious and there most be a better way, I must be missing some identity,

any help would be much appreciated, many thanks.

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  • $\begingroup$ Is $\wedge$ a common notation in Multivariable Calculus? If not, please define it. I think it may not be well-known. (I say this solely because I've not a single clue where it came from.) $\endgroup$
    – 000
    Commented Mar 29, 2012 at 20:21
  • $\begingroup$ Also known as the cross product $\endgroup$
    – user24930
    Commented Mar 29, 2012 at 22:00

1 Answer 1

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\begin{align*} \boldsymbol \nabla \boldsymbol \cdot \mathbf F &= \boldsymbol \nabla \boldsymbol \cdot \Big((\mathbf a \boldsymbol \times \mathbf r)\boldsymbol \times \mathbf r)\Big) \\ &= \boldsymbol \nabla \boldsymbol \cdot \Big((\mathbf a \boldsymbol \cdot \mathbf r)\mathbf r \boldsymbol - (\mathbf r \boldsymbol \cdot \mathbf r)\mathbf a\Big) \\ &= (\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \cdot \mathbf r + \boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \cdot \mathbf r - (\mathbf r \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \cdot \mathbf a - \boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) \boldsymbol \cdot \mathbf a \tag{$1$}\\ &= 3 \mathbf a \boldsymbol \cdot \mathbf r + \mathbf a \boldsymbol \cdot \mathbf r - 0 - 2 \mathbf r \boldsymbol \cdot \mathbf a \tag{$2$} \\ &= 2 \mathbf a \boldsymbol \cdot \mathbf r\\\\ \boldsymbol \nabla \boldsymbol \times \mathbf F &= \boldsymbol \nabla \boldsymbol \times\Big((\mathbf a \boldsymbol \times \mathbf r)\boldsymbol \times \mathbf r)\Big) \\ &= \boldsymbol \nabla \boldsymbol \times \Big((\mathbf a \boldsymbol \cdot \mathbf r)\mathbf r \boldsymbol - (\mathbf r \boldsymbol \cdot \mathbf r)\mathbf a\Big) \\ &= (\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \times \mathbf r \boldsymbol + \boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r) \boldsymbol \times \mathbf r \boldsymbol- (\mathbf r \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \times \mathbf a \boldsymbol - \boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) \boldsymbol \times \mathbf a \tag{$3$}\\ &= \mathbf 0 \boldsymbol+ \mathbf a \boldsymbol \times\mathbf r \boldsymbol- \mathbf 0 \boldsymbol- 2\mathbf r \boldsymbol \times \mathbf a \tag{$4$}\\ &= 3\mathbf a \boldsymbol \times \mathbf r, \end{align*} where

  • $(1)$ follows since $\boldsymbol \nabla \boldsymbol \cdot (\mathbf F + \mathbf G) = \boldsymbol \nabla \boldsymbol \cdot \mathbf F + \boldsymbol \nabla \boldsymbol \cdot \mathbf G$ and $\boldsymbol \nabla \boldsymbol \cdot (\phi \mathbf F) = \phi \boldsymbol \nabla \boldsymbol \cdot \mathbf F + \boldsymbol \nabla \phi \boldsymbol \cdot \mathbf F$;
  • $(2)$ follows since $\begin{aligned}[t]&\boldsymbol \nabla \boldsymbol \cdot \mathbf r = \boldsymbol \nabla \boldsymbol \cdot (x,y,z)=3, \\ &\boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r) = \boldsymbol \nabla(a_1x + a_2y+a_3y)=(a_1,a_2,a_3)=\mathbf a, \text{ and}\\ &\boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) = \boldsymbol \nabla\left(x^2+y^2+z^2\right) = (2x,2y,2z)=2\mathbf r;\end{aligned}$
  • $(3)$ follows since $\boldsymbol \nabla \boldsymbol \times (\mathbf F + \mathbf G) = \boldsymbol \nabla \boldsymbol \times \mathbf F + \boldsymbol \nabla \boldsymbol \times \mathbf G$ and $\boldsymbol \nabla \boldsymbol \times (\phi \mathbf F) = \phi \boldsymbol \nabla \boldsymbol \times \mathbf F + \boldsymbol \nabla \phi \boldsymbol \times \mathbf F$; and
  • $(4)$ follows since $\boldsymbol \nabla \boldsymbol \times \mathbf r = \boldsymbol \nabla \boldsymbol \times(x,y,z)=\mathbf 0$.
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