# Tangent plane for point on surface mesh

I am working on the generation of hexahedral mesh , for surface construction i need to find the tangent plane for each point on the surface.

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The tangent plane depends on your surface interpolation method. Do you know what method you are using? – sam hocevar Jan 29 '13 at 17:27
"Please help me with the code" is the wrong question. Here you'll get the concepts and understanding, the code is your job. – Byte56 Jan 29 '13 at 19:16
Please rephrase your question to tell us the concepts you're unclear about. – Byte56 Jan 29 '13 at 21:44

If you use OpenGL, you will have to split your hexahedrons into triangles. Each point lies inside a triangle, so find the tangent plane for that triangle.

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Thank you for your response. I am actually looking for a way to find the tangent plane. Could you be able exaplain me the process – AJAY Jan 29 '13 at 15:31
I think you don't understand what tangent plane is. Do you? – Ivan Kuckir Jan 29 '13 at 21:25
Hello Ivan, Thank you for your advice. Well, the problem is I have surface some of the elements are triangles and some of them are quadrilaterals. I know that i can split the quadrilaterals in to triangles use the same formula as triangles. But, is there any method calculate the tangent plane for quadrilaterals. – AJAY Jan 30 '13 at 8:42

(continued from Ivan's answer)

What format do you want the tangent plane in? A combination of (point, normal) already is a unique representation of a tangent plane. For example, if I have a triangle at points A, B, C; I can find the normal via the cross product `N = (A-B)x(A-C)`.

Since (A, N) uniquely defines the plane, I could write it out as the equation

``````Nx(x - Ax) + Ny(y - Ay) + Nz(z - Az) = 0
``````
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Hello Jimmy, Thank you for your advice. Well, the problem is I have surface some of the elements are triangles and some of them are quadrilaterals. I know that i can split the quadrilaterals in to triangles use the same formula as triangles. But, is there any method calculate the tangent plane for quadrilaterals. – AJAY Jan 30 '13 at 8:42