# Planes perpendicular to vectors

1. Feb 7, 2010

### riskybeats

a)Find the equation of the tangent plane to the graph z = x2 + 4y2 at the point (a, b, a2 + 4b2)

b) For what values of a and b is the tangent plane perpendicular to the vector (3, -4, 2)?

I figured you have to use the tangent plane equation z - zo = zx(a,b)(x-a) + zy(a,b)(y-b). This gave me the equation

Z = 2ax + 8by - a2 - 4b2

For b, I think the dot product could be used if I set it to zero. But how would I go about solving for a and b? I think I may just be getting tired and a bit sloppy, because I know this is already stuff I have gone over, the only new thing is solving for a and b.

Any direction on this would be great, I'm just having trouble visualizing this problem.

Last edited: Feb 7, 2010
2. Feb 8, 2010

### riskybeats

I couldn't get to sleep because I kept on thinking this problem through. As far as I can see, if the plane's norm is the same as the vector <3, -4, 2>, then it must be perpendicular to it. I am not sure how to make the norm on the plane into that though based off of just of values of a and b. Does anyone have any suggestions?

3. Feb 8, 2010

### boboYO

Doesn't have to be the same, just parallel, i.e. costheta=1 or -1.

Find the equation of the norm of the plane first.

4. Feb 8, 2010

### riskybeats

The norm of my plane is influenced by a and b. This is what I have:

2a (X - a) + 8b (y - b) - z + a^2 + 4b^2 = 0

So if the vector is <3, -4, 2>, I could put the norm of my plane as <-3/2, 2, - 1> since that would be parallel? Hence a = -3/4 and b = 1/4 ?

5. Feb 8, 2010

### boboYO

Hm, could you post the equation of the normal?

6. Feb 8, 2010

### riskybeats

Sorry but I'm not sure what you mean by the equation of the normal. Do you mean the vector equation of the tangent plane?

<-3/2, 2, -1>$$\cdot$$<x + 3/4, y - 1/4, z> = 0?

Although I do still have those other numbers a2 + 4b2. Would that influence the normal of my plane still? I am tired and will probably think about this better later. Thanks for the help so far!