## Deriving the equation of a tangent plane

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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Mentor
 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.
 Thanks! Got it now! Just curious, could you please point out somet things I did wrong? I notice that I used the same 'change in z' for both of the vectors parallel to the tangent lines which doesn't seem right (in lecture 9 of the MIT's 18.02 multivariable calculus course, the professor used the same z when he wrote down the two equations for the tangent lines).

Mentor

## Deriving the equation of a tangent plane

Some of what you wrote was confusing if not actually wrong. For example, you started with the tangent plane at (x0, y0). The plane is tangent to the surface z = f(x, y) at a point on the surface, so the point of tangency is a point in R3, so has three coordinates.

When you were working on the tangent line in the x direction you wrote
z = z0 + fx(x - x0). That was confusing, because it suggests that the partial fx is a function of one variable, not two. It's also incorrect, because you don't want the partial derivative function -- fx -- you want its value at a particular point -- fx(x0, y0). This is like the difference between √x and √4.

After that, you got an equation for this tangent line L1, but you didn't use the partial at all.

You had L1 = <x - x0, 0, z - z0>. From a previous line you had z = z0 + fx(x - x0). Here fx really means fx(x0, y0). You can use this latter equation to replace z - z0 in your equation for L1.

You really don't need L1, or for that matter, L2. All you need are the two vectors that correspond to the tangent lines in the x direction and y direction, respectively.

A tangent vector in the x direction is <1, 0, fx(x0, y0)>. My vector is just a scaled-down version of yours, where yours is the (x - x0) scalar multiple of mine. When you're working with vectors, it usually doesn't matter which one you work with, as long as it points in the right direction.

Same comments apply to what you did for the other tangent line.

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