Finding a Tangent Plane on a 3D Surface

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    3d Point Surface
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Homework Help Overview

The discussion revolves around finding a tangent plane to a 3D surface defined by the equation x² + 2y² + 3z² = 12, in relation to a given parametric line. Participants are exploring the connection between the surface and the tangent plane.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of deriving the tangent plane, including the use of parametric equations and vector derivatives. Questions arise regarding the specific point used in the calculations and the reasoning behind it.

Discussion Status

Some participants express confusion about the steps taken to find the tangent plane, particularly regarding the identification of a specific point on the surface. There is an indication of ongoing exploration and attempts to clarify the process without reaching a definitive conclusion.

Contextual Notes

One participant notes a discrepancy with an answer key, suggesting that there may be additional information or context that is not fully addressed in the discussion.

Spectre32
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Ok I have a surface withthe eqn of x^2 + 2y^2 +3z^2 = 12. The question tells us that there is a perpendicular plane tangent to the the line is as follows:

x = 1 + 2t
y = 3 + 8t
z = 2 - 6t
 
Last edited:
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Whoops I neglected to see something on his answer key... I see what's going on.. Nevermind
 
lol .. ok now i have question. After looking at the real solution my teacher used the Para eqn's to make a vector... athen took the derivite of F(x,y,z) and got <2x,4y,6z>. Then he gets this point (1,2,-1)... I'm unclear as to how he got this point.
 
Yeah I'm been crackin away at this for a while and i sitll don't get it, anyone anyone anyone..
 

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