# Calculating Basis of Tangent Plane

1. Feb 12, 2014

### flyinjoe

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. Feb 12, 2014

### HallsofIvy

Staff Emeritus
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 $\nabla f$ so the tangent plane at $(x_0, y_0, z_0)$ is give by $\nabla(x_0, y_0, z_0)\cdot <x- x_0, y- y_0, z- z_0>= 0$. 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 $x^2yz= 1$ 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.

3. Feb 13, 2014

### flyinjoe

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!