Planes perpendicular to vectors

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Homework Help Overview

The discussion revolves around finding the equation of the tangent plane to the surface defined by z = x² + 4y² at a specific point and determining the conditions under which this plane is perpendicular to a given vector (3, -4, 2).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the tangent plane equation and the concept of the dot product to establish perpendicularity with the vector. There are attempts to derive the normal vector of the plane and relate it to the given vector.

Discussion Status

Participants are actively discussing the relationships between the tangent plane's normal and the specified vector. Some have proposed equations and values for a and b, while others seek clarification on the definitions and implications of the normal vector.

Contextual Notes

There is an indication of uncertainty regarding the influence of the terms a² + 4b² on the normal vector, as well as the need for further exploration of the conditions for perpendicularity.

riskybeats
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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.
 
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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?
 
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.
 
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 ?
 
riskybeats said:
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 ?


Hm, could you post the equation of the normal?
 
boboYO said:
Hm, could you post the equation of the normal?

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>[tex]\cdot[/tex]<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!
 

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