Finding a Plane with Zero Circulation for a Given Vector Field

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Homework Statement



Suppose that f is a vector field such that curl f=(1,2,5) at every point in R^3. Find an equation of a plane through the origin with the property that \oint_{C}f dot dX = 0 for any closed curve C lying in the plane.

Homework Equations



http://img187.imageshack.us/img187/291/1fdf437d8e18a23191b63dfnj8.png

The Attempt at a Solution



With Stokes' theorem and a bit of algebra I get: \int\int ( 1,2,5) dot \nablag dy dx) = 0 . So, 1*dx+2*dy+3*dz=0; let dx=1; let dy=1; dz=-1. The resulting plane is x+y-z=0. Is this right?
 
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You want curl(F)=(1,2,5) to be normal to the plane, right? I don't think that gives you x+y-z=0.
 
Er, oops :eek:.

Then I guess x+2y+5z=0 would simply be the answer?
 
bfr said:
Er, oops :eek:.

Then I guess x+2y+5z=0 would simply be the answer?

Seems so to me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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