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?
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}$ $= \left[2yz^2- xy^2\right]\vec{i}- \left[2x^3y- x^2yz\right]\vec{j}+ \left[x^2z^2- 2x^2yz\right]\vec{k}$