How find the divergence and Curl of the following:

$(\vec{a} \cdot \vec{r}) \vec{b}$,

where $\vec{a}$ and $\vec{b}$ are the constant vectors and $\vec{r}$ is the radius vector.

I have tried solving this by supposing $\vec{r} = (x,y,z)$ and got answer as

div($(\vec{a} \cdot \vec{r}) \vec{b}$) = $\vec{a} \cdot \vec{b}$

but I was wondering if anybody could help me to solve it by using the formulas involving Nabla Operator.



If we use the well-known formula

$\nabla \cdot (f \vec X) = \nabla f \cdot \vec X + f \nabla \cdot \vec X, \tag{1}$

where $f$ is scalar function and $\vec X$ is a vector field (see http://en.m.wikipedia.org/wiki/Vector_calculus_identities), we find, since $\vec b$ is constant,

$\nabla \cdot ((\vec a \cdot \vec r) \vec b) = \nabla (\vec a \cdot \vec r) \cdot \vec b; \tag{2}$

at this point we pretty much have to use

$\vec a \cdot \vec r = a_x x + a_y y+ a_z z, \tag{3}$


$\vec a = (a_x, a_y, a_z)^T \tag{4}$

to obtain

$\nabla (\vec a \cdot r) = \vec a, \tag{5}$


$\nabla \cdot ((\vec a \cdot \vec r) \vec b = \vec a \cdot \vec b. \tag{6}$

We may find $\nabla \times ((\vec a \cdot \vec r) \vec b$ in an analogous manner, from (see the same linked citing)

$\nabla \times (f \vec X) = \nabla f \times \vec X + f\nabla \times X; \tag{7}$

again since $\vec b$ is constant, and by (5)

$\nabla \times ((\vec a \cdot \vec r) \vec b) = \nabla (\vec a \cdot \vec r) \times \vec b = \vec a \times \vec b. \tag{8}$

We can use the vector identities to simplify the calculations somewhat, but at some point the actual coordinate expression for $\vec a \cdot \vec r$ must be invoked to finalize the work.



$ \nabla \cdot (\phi \vec b)=\vec b \cdot \nabla \phi+\phi \nabla \cdot \vec b \qquad$ and $ \qquad \nabla \times (\phi \vec b)=\phi(\nabla \times\vec b) + (\nabla \phi) \times \vec b $

Where $\phi$ is the scalar function $(\vec a\cdot \vec r)$ (so you can easely find $\nabla \phi=\vec a$) and $\vec b$ is a constant.


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