Unit Tangent Vector in a Scalar Field

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Hello,
I am attempting to calculate unit normal and tangent vectors for a scalar field I have, Φ(x,y). For my unit normal, I simply used:
[itex]\hat{n}=\frac{\nabla \phi}{|\nabla \phi|}[/itex]​
However, I'm struggling with using this approach to calculate the unit tangent. I need to express it in terms of the gradient of the scalar field but I am having a hard time visualizing this.
Thanks!
 

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Take a linear combination of basis vectors with unknown coefficients. Equate its dot product with ## \hat n ## to zero. You should be able to find two independent vectors as solutions. Those will span the tangent space to each point.
 
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Take a linear combination of basis vectors with unknown coefficients. Equate its dot product with ## \hat n ## to zero. You should be able to find two independent vectors as solutions. Those will span the tangent space to each point.
Thanks for your response! I think I tried this: I used [itex] \hat{n} \cdot \hat{T} = 0 [/itex], writing each in terms of an x- and y-component, and then solving for the x-component and y-component of the unit tangent [itex] \hat{T} [/itex]. My [itex] T_x [/itex] and [itex] T_y [/itex] are written only in terms of unit normal components [itex] n_x [/itex] and [itex] n_y [/itex] (by defining [itex] \sqrt{T_x^2 + T_y^2} = 1 [/itex]. When I do this however, and plot my unit tensor field, I don't get clean, tangential vectors. Some are tangential, but they can be a bit of a mess. I am doing this in FEA software, could some of my issues be due to my mesh size? Or am I going about it the wrong way?
 
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One obvious problem I see, is that you're assuming the problem is two dimensional, Its not! You should work in three dimensions. Otherwise ##\hat n\cdot \vec T=0## and ##|\vec T|=1## will completely determine the components and you'll get only one vector as a solution which we know isn't right.
In fact you should've known this from the start because you are considering a function of two variables which can only be a surface in three dimensions.
 

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