What is a content vector?

Question

If $\vec{a}$ is a content vector, prove that $\nabla \times (\vec{a} \times \vec{r}) = 2\vec{a}$

I don't understand what is a content vector?

Answer 1

We know that $$\vec{∇}\times(\vec u \times \vec v)=\vec u(\vec∇\cdot\vec v)-\vec v(\vec∇\cdot\vec u)+(\vec v\cdot \vec∇)\vec u-(\vec u\cdot \vec∇)\vec v$$

Here if we take $~\vec a=a_1\hat i+a_2\hat j+a_3\hat k~$(say) is constant vector, then $$\vec{∇}\times(\vec a \times \vec r)=\vec a(\vec∇\cdot\vec r)-\vec r(\vec∇\cdot\vec a)+(\vec r\cdot \vec∇)\vec a-(\vec a\cdot \vec∇)\vec r$$ $$=3~\vec a-0+0-\vec a~~~~~~~~~~~~~~~~~~~~~~$$ $$=2~\vec a~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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$$\vec∇\cdot\vec r=\left(\frac{\partial}{\partial x}\hat i+\frac{\partial}{\partial y}\hat j+\frac{\partial}{\partial z}\hat k\right)\cdot\left(x\hat i+y\hat j+z\hat k\right)=3$$

$$\vec∇\cdot\vec a=\left(\frac{\partial}{\partial x}\hat i+\frac{\partial}{\partial y}\hat j+\frac{\partial}{\partial z}\hat k\right)\cdot\left(a_1\hat i+a_2\hat j+a_3\hat k\right)=0$$

By similar manner $$(\vec r\cdot \vec∇)\vec a=0$$

$$(\vec a\cdot \vec∇)\vec r=\vec a$$