How Does the Dot Product Interact with the Gradient in Vector Calculus?

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dot product, and the gradient urgent pls!...

Homework Statement


Δ<-- this be the gradient and B<-- be a vector B X= xi +yj + zk
*<---- be the dot product.
(B*Δ)X=B


Homework Equations


n/a


The Attempt at a Solution



im not sure how to go about this but this is what i did

i did Δ*B so i got Bx + By + Bz
then its multiplyed by X :/ not making any sense to me :cry:

help pls
BA
 
Last edited:
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any one ? :/
 


So you're trying to show that

(\mathbf{B}\cdot\nabla)\mathbf{X} = \mathbf{B}.

Well the scalar product in the brackets is a differential operator. What's its form ?
 


To expand on what bigubau said:
<br /> \mathbf{B}\cdot\nabla =B_{x}\frac{\partial}{\partial x}+B_{y}\frac{\partial}{\partial y}+B_{z}\frac{\partial}{\partial z}<br />
This is applied to each of the components of X, so for example:
<br /> (\mathbf{B}\cdot\nabla )\mathbf{X}|_{x}=B_{x}\frac{\partial X_{x}}{\partial x}+B_{y}\frac{\partial X_{x}}{\partial y}+B_{z}\frac{\partial X_{x}}{\partial z}<br />
 
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