What is the value of $\nabla\times (\vec a\times \vec r)$?

Question

Given that $\vec r=xi+yj+zk$ & $\vec a$ is a constant vector then what is the value of $\nabla\times (\vec a\times \vec r)$ ?

a) $\ -2\vec a\quad $ b) $\ 2\vec a\quad $ c) $\ 3 a\vec r\quad $ d) $\ -3 a\vec r\quad $

My try:

$\nabla\times (\vec a\times \vec r)$

$=\vec a(\nabla\cdot \vec r)-\vec r(\nabla\cdot \vec a)$

$=\vec a(3)-\vec r(0)$ (because divergence of a constant vector $\vec a$ is zero.)

$=3\vec a$

But my answer doesn't match any options given above. My teacher says option (b) is correct answer but i don't know how.

Please correct me if i am wrong & help me solve this question. Thanks

Answer 1

You missed two terms: $$ \nabla\times (\vec a\times \vec r) = \vec a(\nabla\cdot \vec r)-\vec r(\nabla\cdot \vec a) +(\vec r\cdot\nabla)\vec a-(\vec a\cdot\nabla)\vec r \\ = \vec a(\nabla\cdot \vec r)-(\vec a\cdot\nabla)\vec r=3\vec a-\vec a=2\vec a. $$

Note that for the last term we use: $$ (\vec a\cdot\nabla)\vec r=(a_1\partial_x+a_2\partial_y+a_3\partial_z)r =(a_1,a_2,a_3)=\vec a $$

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[Added later by request.] Since $\vec a$ is a constant vector, one has $$ \nabla\cdot \vec a = 0,\quad (\vec r \cdot\nabla)a=0, $$ which explains why the second and third terms vanish.