Context
I am studying radiation fields using [1]. In Eq. 20.98 in [1] I came across an identity (see Eq 1). I have attempted to use the the chain rule, and product rule and other identities found [2], [3], and [4] to demonstrate this identity and have not succeeded.
Question
Given two vector fields $\mathbf{r}$ and $\mathbf{r}^\prime$, where $\mathbf{r} \in \mathbb{R}^n$ and $\mathbf{r}^\prime \in \mathbb{R}^n$, the unit vector $\hat{\mathbf{r}}$, and that the differential operator $\boldsymbol{\nabla}$ operates on $\mathbf{r}$, can you demonstrate that
$$\boldsymbol{\nabla} \left( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime \right)
= \hat{\mathbf{r}} \times\left(
\hat{\mathbf{r}} \times \frac{\mathbf{r}^\prime}{r}
\right)
?\tag{1}$$
My Attempt
I know from [2] that
$$
\nabla( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime
) =\ ( \hat{\mathbf{r}} \cdot \nabla)\mathbf{r}^\prime \,+\, (\mathbf{r}^\prime \cdot \nabla) \hat{\mathbf{r}} \,+\, \hat{\mathbf{r}} {\times} (\nabla {\times}\mathbf{r}^\prime) \,+\, \mathbf{r}^\prime {\times} (\nabla {\times} \hat{\mathbf{r}} ) .
$$
The third term on the right is zero, because the del operator acts on $\mathbf{r}$ and does not act on $\mathbf{r}^\prime$. The fourth term is zero according to the definition of the curl in spherical coordinates.
Now, I know also from [2] that in spherical coordinates
\begin{align}
(\mathbf{A} \cdot \boldsymbol{\nabla})\mathbf{B}
=
\left(
A_r \frac{\partial B_r}{\partial r}
+ \frac{A_\theta}{r} \frac{\partial B_r}{\partial \theta}
+ \frac{A_\varphi}{r\sin\theta} \frac{\partial B_r}{\partial \varphi}
– \frac{A_\theta B_\theta + A_\varphi B_\varphi}{r}
\right) &\hat{\mathbf r} \\
+ \left(
A_r \frac{\partial B_\theta}{\partial r}
+ \frac{A_\theta}{r} \frac{\partial B_\theta}{\partial \theta}
+ \frac{A_\varphi}{r\sin\theta} \frac{\partial B_\theta}{\partial \varphi}
+ \frac{A_\theta B_r}{r} – \frac{A_\varphi B_\varphi\cot\theta}{r}
\right) &\hat{\boldsymbol \theta} \\
+ \left(
A_r \frac{\partial B_\varphi}{\partial r}
+ \frac{A_\theta}{r} \frac{\partial B_\varphi}{\partial \theta}
+ \frac{A_\varphi}{r\sin\theta} \frac{\partial B_\varphi}{\partial \varphi}
+ \frac{A_\varphi B_r}{r}
+ \frac{A_\varphi B_\theta \cot\theta}{r}
\right) &\hat{\boldsymbol \varphi}.
\end{align}
Thus,
\begin{align}
(\hat{\mathbf{r}} \cdot \boldsymbol{\nabla})\mathbf{r}^\prime
= \mathbf{0}
\end{align}
I arrive at
$$
\boxed{
\boldsymbol{\nabla}( \hat{\mathbf{r}} \cdot \mathbf{r}^\prime
) = (\mathbf{r}^\prime \cdot \boldsymbol{\nabla}) \hat{\mathbf{r}} .}
$$
This is not that which I was supposed to prove.
Bibliography
[1] Zangwill, Modern Electrodynamics, 2013, p. 733.
[2] https://en.wikipedia.org/wiki/Vector_calculus_identities
[3] https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
Best Answer
Let me write $\mathbf{a} = \mathbf{r}'$ to better distinguish $\mathbf{r}$ and $\mathbf{r}'$. Then by BAC-CAB rule,
$$ \hat{\mathbf{r}} \times \left( \frac{\hat{\mathbf{r}}}{r} \times \mathbf{a} \right) = \frac{\hat{\mathbf{r}} (\hat{\mathbf{r}} \cdot \mathbf{a}) - \mathbf{a} (\hat{\mathbf{r}} \cdot \hat{\mathbf{r}})}{r}. \tag{1} $$
On the other hand, writing $\mathbf{r} = (x_1, x_2, x_3)$ and noting that $r = \sqrt{x_1^2 + x_2^2 + x_3^2}$,
\begin{align*} \nabla (\hat{\mathbf{r}} \cdot \mathbf{a}) &= \sum_{i} \mathbf{e}_i \frac{\partial}{\partial x_i} \left( \frac{\mathbf{r}}{r} \cdot \mathbf{a} \right) \\ &= \sum_{i} \mathbf{e}_i \left[ \left( \frac{\mathbf{e}_i}{r} - \frac{\mathbf{r}x_i}{r^3} \right) \cdot \mathbf{a} \right] \\ &= \frac{\sum_{i} \mathbf{e}_i (\mathbf{e}_i \cdot \mathbf{a})}{r} - \frac{\left( \sum_{i} \mathbf{e}_i x_i \right)(\mathbf{r} \cdot \mathbf{a})}{r^3} \\ &= \frac{\mathbf{a}}{r} - \frac{\mathbf{r} (\mathbf{r}\cdot\mathbf{a})}{r^3} \tag{2} \end{align*}
From this, it actually turns out that
$$ \nabla (\hat{\mathbf{r}} \cdot \mathbf{a}) = -\hat{\mathbf{r}} \times \left( \frac{\hat{\mathbf{r}}}{r} \times \mathbf{a} \right). $$
Mathematica 13 also verifies this: