How can we compute $$(\vec{B} \cdot \nabla)\vec{A} $$ if A and B are defined as following: $$\vec{A}=2yz\ \vec{i} -x^2y\ \vec{j} + xz^2\ \vec{k}$$ $$\vec{B}=x^2 \ \vec{i} + yz \ \vec{j} -xy \ \vec{k}$$ I tried to solve it by removing the brackets hence we have B dot the gradient of A.

First, is it right to remove the brackets ? when do we have the right to remove the brackets in the problems of divergence and curl?

Second, if this is right to remove the brackets, then is the following the right solution to the problem?

compute the gradient of the vector function A as follows $$\nabla\vec A = \nabla \cdot \vec A + \nabla \wedge \vec A$$ following this link Gradient of a vector field? then apply dot product with B.

Third, is there any other way to solve this problem? Is there an identity that can be used to solve this problem?

  • $\begingroup$ $$\displaystyle\vec B\cdot\vec\nabla=\frac{\partial(x^2)}{\partial x}+\frac{\partial(yz)}{\partial y}+\frac{\partial(-xy)}{\partial z}.$$ Then I do not know how to interpret $\vec A$ and the $()$. Maybe we have another scalar product? $\endgroup$
    – manooooh
    Oct 23, 2018 at 17:52
  • $\begingroup$ I think you are right. $(\vec{B} \cdot \nabla )\vec{A} = \vec{B} \cdot \nabla \vec{A}$. Why would it be written with parentheses? I don't know, but they may be trying to draw your attention to $(\vec{B} \cdot \nabla)$ as a new operator defined by them, just as $\nabla$ or $\frac{d}{dx}$ are operators. So instead of doing the operation $\frac{d}{dx} f$, you might define the operation $(2 + \frac{d}{dx})f$ which is just the derivative plus 2f, but trying to think of it as it's "own, newly defined, single operator". I can't speak to the other questions $\endgroup$
    – DWade64
    Oct 23, 2018 at 17:57
  • $\begingroup$ This might help too: wikipedia page $\endgroup$
    – DWade64
    Oct 23, 2018 at 18:07
  • $\begingroup$ Here is the answer: (math.stackexchange.com/questions/836041/…). I don't know if my other comments were correct, probably not $\endgroup$
    – DWade64
    Oct 29, 2018 at 11:53
  • $\begingroup$ Do you mean Schur Hadamard product? $\endgroup$ Oct 30, 2018 at 22:13

1 Answer 1


No. What you have is $\nabla\cdot B$.

$\vec{B}\cdot\nabla= x^2\frac{\partial }{\partial x}+ yz\frac{\partial}{\partial y}- xy\frac{\partial}{\partial z}$.

That is a "scalar operator". Applying it to A, like multiplying a scalar by a vector, means applying that to each component of A: $\left[x^2\frac{\partial 2yz}{\partial x}+ yz\frac{\partial 2yz}{\partial y}- xy\frac{\partial 2yz}{\partial z}\right]\vec{i}$$+ \left[x^2\frac{\partial -x^2y}{\partial x}+ yz\frac{\partial -x^2y}{\partial y}- xy\frac{\partial -x^2y}{\partial z}\right] \vec{j}+ \left[x^2\frac{\partial xz^2}{\partial x}+ yz\frac{\partial xz^2}{\partial y}- xy\frac{\partial xz^2}{\partial z}\right]\vec{k}$ $= \left[2yz^2- xy^2\right]\vec{i}- \left[2x^3y- x^2yz\right]\vec{j}+ \left[x^2z^2- 2x^2yz\right]\vec{k}$


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