If $\bf a$ is a constant vector field, and $\bf r$ is the position vector, prove that: $\nabla (\bf a\cdot \bf r)=\bf a$ If $\bf a$ is a constant vectorial field (constant magnitude and direction), and $\bf r$ is the position vector, prove that:
$$\nabla (\mathbf a \cdot \bf r)=\mathbf a $$

I assume that by $ar$ the OP means $\mathbf a \cdot \mathbf r$, where $\mathbf r$ is the position vector.  Then with $\mathbf a = (a_x, a_y, a_z)^T$ and $\mathbf r = (x, y, z)^T$, we have
$\mathbf a \cdot \mathbf r = a_x x + a_y y + a_z z; \tag{1}$
and since
$\nabla (\mathbf a \cdot \mathbf r) = (\dfrac{\partial \mathbf a \cdot \mathbf r}{\partial x}, \dfrac{\partial \mathbf a \cdot \mathbf r}{\partial y}, \dfrac{\partial \mathbf a \cdot \mathbf r}{\partial z})^T, \tag{2}$
we merely need observe that
$\dfrac{\partial \mathbf a \cdot \mathbf r}{\partial x} = \dfrac{\partial (a_x x + a_y y + a_z z)}{\partial x} = a_x, \tag{3}$
with the analogous results holding for partial differentiation with respect to $y$ and $z$.  Thus
$\nabla (\mathbf a \cdot \mathbf r) = (a_x, a_y, a_z)^T, \tag{4}$
as was to be shown.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
