Finding the result of $\nabla$ $×$ ($k$ $×$ $r$)

Question

I am working through the Boas mathematical methods book (Chapter 8, section 7, question 18). I am working on the vector calculus chapter.

The question is:

For $r = xi + y j + zk$,

Find: $\nabla$ $×$ ($k$ $×$ $r$)

My attempt:

Begin with expanding the triple cross product:

$(\nabla \cdot r)k$ - $(\nabla \cdot k)r$

However, now I am confused. If I take $\nabla$ = $\frac{\partial}{\partial x}i$ + $\frac{\partial}{\partial y}j$ + $\frac{\partial}{\partial z}k$, then how can I find $(\nabla \cdot k)$? The vectors are of different dimensions.

Any guidance is much appreciated.

Answer 1

You can just calculate it explicitly, step by step. $$ k \times r = \begin{vmatrix} i & j & k \\ 0 & 0 & 1 \\ x & y & z \end{vmatrix} = -yi + xj$$ then we can just directly compute the curl of the resulting vector-valued function. $$\nabla \times (k \times r) = \begin{vmatrix}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & 0\end{vmatrix} = \boxed{2k}$$

Answer 2

Let's first understand how the curl of a cross product of "normal" vectors works: $$[\nabla\times(A\times B)]_i=\epsilon_{ijk}\partial_j(A\times B)_k=\epsilon_{ijk}\epsilon_{klm}\partial_jA_lB_m=\left(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\right)\partial_j(A_lB_m)\\=\partial_j(A_iB_j)-\partial_j(A_jB_i)=(\nabla\cdot B)A_i+(B\cdot\nabla)A_i-(A\cdot\nabla)B_i-(\nabla\cdot A)B_i,$$where we sum over repeated indices. (By "normal", I mean let's not have any differential operators in $A$ or $B$.) Thus$$\nabla\times(A\times B)=(\nabla\cdot B)A+(B\cdot\nabla)A-(A\cdot\nabla)B-(\nabla\cdot A)B.$$In particular,$$\nabla\times(k\times r)=(\nabla\cdot r)k+(r\cdot\nabla)k-(k\cdot\nabla)r-(\nabla\cdot k)r=3k+0-\partial_zr-0=3k-k=2k.$$