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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.

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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 $3\vec k$

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  • $\begingroup$ The last two terms are wrong, in fact. (It's just that the last one is zero even in its correct form) :) $\endgroup$
    – Philip
    Aug 28, 2020 at 16:36
  • $\begingroup$ @Philip I showed that the third it is not zero,or I do not get what you mean $\endgroup$
    – user65081
    Aug 28, 2020 at 16:37
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    $\begingroup$ Your answer is perfectly correct! I just meant that it's not only the second to last term that's wrong, but also the last term. However I see now that you've actually mentioned that in your answer as well, don't know how I missed it. Apologies! :) $\endgroup$
    – Philip
    Aug 28, 2020 at 16:43
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    $\begingroup$ @Wolphramjonny Thank you. I thought $(\vec k . \vec \nabla)\vec r$ was the same as $(\vec \nabla . \vec k)\vec r$ $\endgroup$
    – ElonTusk
    Aug 28, 2020 at 17:21

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