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Calculating Basis of Tangent Plane

  1. Feb 12, 2014 #1
    I've looked at this topic for a while and I have yet to come to any sort of conclusive answer when it comes to calculating the basis of a surface's tangent vector. Do you have a concrete method or know where I can find one for doing this?

    Thank you
  2. jcsd
  3. Feb 12, 2014 #2


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    I'm not sure what you mean by "the" basis. Any vector space has an infinite number of bases. If a surface is given by f(x, y, z)= Constant, then the normal vector to the surface is [itex]\nabla f[/itex] so the tangent plane at [itex](x_0, y_0, z_0)[/itex] is give by [itex]\nabla(x_0, y_0, z_0)\cdot <x- x_0, y- y_0, z- z_0>= 0[/itex]. You can get one vector in that tangent plane by taking y= y_0, x= x_0+ 1 and another, independent vector so they form a basis, by taking x= x_0, y= y_0+ 1.

    For example, if the surface is given by [itex]x^2yz= 1[/itex] then the normal vector at any point is [itex]<2xyz, x^2z, x^2y>. At (1, 1, 1) that would be <2, 1, 1>. The tangent plane there is 2(x- 1)+ y- 1+ z- 1= 0 or 2x+ y+ z= 4. if x= 2, y= 1, then 4+ 1+ z= 1 so z= -1. The point (2, 1, -1) is also in that plane so the vector <2- 1, 1- 1, 1-(-1)>= <1, 0, 2> lies in that tangent plane. If x= 1, y= 2, then 2(0)+ 1+ z= 4 so z= 3. The point (1, 2, 3) is also in that tangent plane so the vector <1- 1, 2- 1, 3-(-1)>= <0, 1, 4> lies in that tangent plane. The two vectors <1, 0, 2> and <0, 1, 4> are two independent vectors in thet tangent plane and so form a basis.
  4. Feb 13, 2014 #3
    HallsofIvy, thanks for the response! Sorry my question was sort of ambivalent. By 'the' basis, I meant 'a' basis. That's a very concise and helpful explanation.

    Thank you!
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