# Calculating the gradient of the dot product of two vectors

I'm trying to calculate $$\vec\nabla(\vec k.\vec r)$$ where $$\vec k =k_x \hat{i}+k_y\hat{j}+k_z\hat{k}$$ is a constant vector and $$\vec r=x\hat{i}+y\hat{j}+z\hat{k}$$ is the position vector.

I tried doing this in the following two ways:

First, I used the known formula for Gradient of the dot product between two vectors:$$\vec \nabla(\vec k.\vec r)=\vec k \times(\vec \nabla\times\vec r)\space+\vec r \times(\vec \nabla\times\vec k)\space+(\vec\nabla.\vec k)\vec r+\space(\vec\nabla.\vec r)\vec k \space$$ The first term of this expression is $$\vec0$$ since the curl of the position vector ($$\vec \nabla\times\vec r$$) is $$\vec 0$$.The second and third terms are also both $$\vec 0$$ since they are respectively the Divergence and the Curl of a constant vector. This leaves me with the fourth term which evaluates to $$3\vec k$$ since the divergece of the position vector ($$\vec\nabla\times\vec r$$) is $$3$$. So using the formula the answer comes out to be $$3\vec k$$.

Then, I did it "manually":

$$\vec\nabla(\vec k.\vec r) = \frac {\partial (\vec k.\vec r)}{\partial x} \hat {i} +\frac {\partial (\vec k.\vec r)}{\partial y} \hat {j}+\frac {\partial (\vec k.\vec r)}{\partial z} \hat {k}$$

and $$(\vec k.\vec r)=k_xx+k_yy+k_zz$$

So, $$\vec\nabla(\vec k.\vec r)=\frac {\partial (k_xx+k_yy+k_zz)}{\partial x} \hat {i} +\frac {\partial (k_xx+k_yy+k_zz)}{\partial y} \hat {j}+\frac {\partial (k_xx+k_yy+k_zz)}{\partial z} \hat {k}=k_x \hat{i}+k_y\hat{j}+k_z\hat{k}=\vec k$$

To summarize, the first method gives me $$3\vec k$$ and the second gives me $$\vec k$$. Please tell me which one is right and which one is wrong.

Your first inequality is wrong, the second to last term should be $$(\vec{k}.\vec{\nabla)}\vec{r}=k_x\partial_x\vec r+k_y\partial_y\vec r+k_z\partial_z\vec r=\vec k$$ by the same reason the last term is zero, not $$3\vec k$$
• @Wolphramjonny Thank you. I thought $(\vec k . \vec \nabla)\vec r$ was the same as $(\vec \nabla . \vec k)\vec r$ Aug 28, 2020 at 17:21