Setting a tangent plane parallel to another plane-Cal III

[crims0ned]

At what point on the paraboloid $$y=x^2+z^2$$ is the tangent plane parallel to the plane $$x+2y+3z=1$$?


Tangent plane equation is...
$$Fx(X,Y,Z,)(x-X)+Fy(X,Y,Z)(y-Y)+Fz(X,Y,Z)(z-Z)=0; for x^2+z^2-y=0$$

My attempt at the problem...

First I found the unit normal for the plane I'm trying to match $$x+2y+3z=1$$

so.. $$\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$$

to the unit normal is $$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}},$$

now I set that equal to the tangent plane equation and solve for the the point right? So...

$$2x(X,Y,Z,)(x-X)-1(X,Y,Z)(y-Y)+2z(X,Y,Z)(z-Z)=\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$$

Am I on the right track?

 

[Chaos2009]

You had a good idea, but you have mistakes in there. Your last statement is not correct. You wanted to replace $$F_x \left(x,y,z\right)$$ with $$2x$$, not just the $$F_x$$. The same goes for the rest of the partial derivatives. Also, that gave you the equation for the tangent plane, not the tangent plane's normal vector so you can't just set it equal to the plane's normal vector and solve.

What you want is that you know two planes are parallel if their normal vectors are parallel. This means that you can multiply one of the normal vectors by some scalar to get the other normal vector.

$$n_1 = \lambda n_2$$

Well, know we want to know what the normal vector of the surface is at a given point. I hope you've learned gradients because they make this much easier.

If we have a function, $$f \left(x,y,z\right) = c$$, then the normal vector to the surface at a given point is the gradient of the function evaluated at that point, $$\nabla f \left(x_0,y_0,z_0\right)$$.

Just in case you don't know or forgot, the gradient of a function is defined as follows:
$$\nabla f \left(x,y,z\right) = \left(f_x \left(x,y,z\right),f_y \left(x,y,z\right),f_z \left(x,y,z\right)\right)$$

And we want to know when $$\nabla f_1 \left(x,y,z\right) = \lambda \nabla f_2 \left(x,y,z\right)$$ where $$\lambda$$ can be any number. You would also want to check these values to see if they actually lie on your surface.

 

[crims0ned]

Yeah I think I got it now, I set the gradient equal to the normal of that other plane.

$$\nabla f \left(x_0,y_0,z_0\right)= u=<1,2,3>$$
then I set the partials equal to that normal and I get $$f_x=2x_0; f_y=-1; f_z=2z_0$$

so... $$\nabla f \left<2x_0,-1,2z_0\right>=<1,2,3>$$
then my only problem is my y's don't correspond so I multiplied the normal vector by -1/2 and get
$$\nabla f \left<2x_0,-1,2z_0\right>=<\frac{-1}{2},-1,\frac{-3}{2}>$$

and I find $$x_0$$ and $$z_0$$

$$2x_0=\frac{-1}{2};2z_0=\frac{-3}{2}$$

$$x_0=\frac{-1}{4};z_0=\frac{-3}{4}$$

now I can plug those x's and z's into the original function to get my y.

$$y_0=(\frac{-1}{4})^2+(\frac{-3}{4})^2$$

$$y_0=\frac{1}{16}+\frac{9}{16}$$

$$y_0=\frac{5}{8}$$

So now I have the point $$(\frac{-1}{4},\frac{5}{8},\frac{-3}{4})$$