# Deriving the equation of a tangent plane

by autodidude
Tags: deriving, equation, plane, tangent
 P: 333 I am trying to derive the equation of a tangent plane at some point $$(x_0, y_0)$$ on a surface using vectors. This is how I have been trying to do it: The tangent line at $$(x_0, y_0)$$ in the x-direction is $$z=z_0+f_x(x-x_0)$$ so the vector parallel to it is $$L_1=<(x-x_0), 0, (z-z_0)>$$. Similarly, the vector parallel to the tangent line with respect to y is $$<(0, (y-y_0), (z-z_0)>$$. Taking the cross product, I got the normal vector $$<-(z-z_0)(y-y_0), -(x-x_0)(z-z_0), (x-x_0)(y-y_0)>$$ Then taking the dot product between the normal vector and a vector in the plane ($$L_2-L_1$$), I got a formula which does equal zero but from which I cannot seem to derive the desired equation. This is what I keep getting: $$(x-x_0)(y-y_0)(z-z_0)-(x-x_0)(y-y_0)(z-z_0)=0$$ I've tried different vector representations of the lines but I keep getting the same result.
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P: 21,280
 Quote by autodidude I am trying to derive the equation of a tangent plane at some point $$(x_0, y_0)$$ on a surface using vectors.
The point would actually be at (x0, y0, f(x0, y0)).
 Quote by autodidude This is how I have been trying to do it: The tangent line at $$(x_0, y_0)$$ in the x-direction is $$z=z_0+f_x(x-x_0)$$ so the vector parallel to it is $$L_1=<(x-x_0), 0, (z-z_0)>$$. Similarly, the vector parallel to the tangent line with respect to y is $$<(0, (y-y_0), (z-z_0)>$$. Taking the cross product, I got the normal vector $$<-(z-z_0)(y-y_0), -(x-x_0)(z-z_0), (x-x_0)(y-y_0)>$$
Let's take a slice of the surface with a cut parallel to the x-z plane. IOW, with y held fixed. At the point P0(x0, y0, f(x0, y0)) the slope of the tangent line along the cut surface is fx(x0, y0). A vector in this direction is <1, 0, fx(x0, y0)>.

Now, make a slice parallel to the y-z plane, so that x is held fixed. At P0, we can use fy(x0, y0) to find a vector in the direction of this new tangent line. A vector with this direction is <0, 1, fy(x0, y0)>. If you cross these two vectors, you'll get a normal to the tangent plane.

I'll let you do the work, but in the meantime, let's call it N = <A, B, C>.

To get the equation of the tangent plane, form a vector from the point of tangency, P0, and an arbitrary point on the plane, P(x, y, z). That vector would be <x - x0, y - y0, z - z>.

This vector is perpendicular to the normal, so their dot product would be zero:
N ##\cdot## <x - x0, y - y0, z - z> = 0.

 Quote by autodidude Then taking the dot product between the normal vector and a vector in the plane ($$L_2-L_1$$), I got a formula which does equal zero but from which I cannot seem to derive the desired equation. This is what I keep getting: $$(x-x_0)(y-y_0)(z-z_0)-(x-x_0)(y-y_0)(z-z_0)=0$$ I've tried different vector representations of the lines but I keep getting the same result.