# Vector Calculus HW question

## Homework Statement

http://faculty.fiu.edu/~maxwello/phz3113/probs/set1.pdf

I'm working on problem 2. Trying to prove that the dot product between a vector field and its curl is zero.

## Homework Equations

The basic identities of vector calculus and how scalar fields and vector fields interact

## The Attempt at a Solution

My only real attempt is expanding what was given using an identity. What i have now is that

f(del X A) = A X del(f)

In my head the proof is trivial since by the very definition of the cross product, the new vector while be perpendicular to both vectors that served as the argument. I do have a question though. In the above notation and in the identity when they say f(del X A) do they mean to input the curl vector as an argument for the scalar function? Thanks!

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blue_leaf77
Homework Helper
In my head the proof is trivial since by the very definition of the cross product, the new vector while be perpendicular to both vectors that served as the argument.
I don't precisely know which vectors you were referring to by "both vectors", but I have a feeling that you are concerned with ##\mathbf{A}## being always perpendicular to ##\nabla \times \mathbf{A}## because the cross product in between the last expression makes it perpendicular to ##\mathbf{A}##. I believe this is not necessarily the case. Anyway, the more elegant way is to multiply by dot product the identity you have there with ##\mathbf{A}##.
In the above notation and in the identity when they say f(del X A) do they mean to input the curl vector as an argument for the scalar function? Thanks!
No, they are just vector multiplied with a scalar.

Mario Carcamo
I don't precisely know which vectors you were referring to by "both vectors", but I have a feeling that you are concerned with ##\mathbf{A}## being always perpendicular to ##\nabla \times \mathbf{A}## because the cross product in between the last expression makes it perpendicular to ##\mathbf{A}##. I believe this is not necessarily the case. Anyway, the more elegant way is to multiply by dot product the identity you have there with ##\mathbf{A}##.

No, they are just vector multiplied with a scalar.
Yeah what i mean is if A X B = C then C must be perpendicular to A and B so if your doing Del X A = C then C must be perpendicular to A but maybe that is not the case when your talking about del?

blue_leaf77