Solving Tangent Plane and Perpendicular Plane at (1,1,1)

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Discussion Overview

The discussion revolves around finding the equation of the tangent plane to the surface defined by z=f(x,y)=x^2 + y^2 - 1 at the point (1,1,1), and subsequently determining an equation for a plane that is perpendicular to that tangent plane and also passes through the same point.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents the problem of finding the tangent plane and a perpendicular plane at a specific point.
  • Another participant references the theorem that the gradient of a surface is normal to that surface, suggesting its use in solving the problem.
  • A participant claims to have found the equation of the tangent plane as 2x+2y-z-3=0 and identifies the normal vector as <2,2,-1>, but expresses confusion about finding the perpendicular plane.
  • One participant notes that there are infinitely many planes that can be normal to the tangent plane and suggests that any tangent vector in the tangent plane can serve as the normal for the chosen plane.
  • Another participant reinforces the idea that having "too many variables" allows for the selection of one variable to simplify the problem, emphasizing the infinite options for normal planes through a given point.

Areas of Agreement / Disagreement

Participants generally agree on the existence of infinitely many normal planes to the tangent plane, but there is no consensus on how to proceed with the next steps in solving the problem.

Contextual Notes

Participants express uncertainty regarding the selection of variables and the implications of having multiple solutions, which may depend on the definitions and constraints applied to the problem.

Who May Find This Useful

This discussion may be useful for individuals interested in multivariable calculus, particularly those studying tangent planes and their properties in relation to surfaces.

brads
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Got a problem which should be easy but having trouble...

"Find the equation of the tangent plane to z=f(x,y)=x^2 + y^2 - 1 at point (1,1,1)
"Find an equation to a plane that is perpendicular to that tangent plane and also passes through the point (1,1,1)

Thanks!
 
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Welcome to PF!

Are you familiar with the theorem that the gradient of a surface G(x,y,z)=0 is normal to that surface?
Use that hint (G(x,y,z)=z-f(x,y)).
 
I figured out the first part of the problem the tangent plane to f(x,y) is 2x+2y-z-3=0. The normal vector of that plane is thus <2,2,-1>. I am lost on how to solve the next part, I keep trying different equations but coming up with too many variables.
 
There's an infinity of planes which are normal to the tangent plane.
Pick one of them; the requirement is merely that some tangent vector in the tangent plane should be the normal to the plane you choose.
(the normal of the tangent plane is to be a tangent vector in the plane you choose)
 
Last edited:
"too many variables" only allows you to set one of them to any convenient value. Like arildno said, there are an infinite number of normal planes passing through a given point; so you get to pick anyone you like.
 

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