Finding a Tangent Plane on a 3D Surface

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To find the tangent plane on the surface defined by the equation x^2 + 2y^2 + 3z^2 = 12, the discussion revolves around deriving the necessary vector from the parameterized equations of the line. The derivative of the function F(x, y, z) yields the gradient vector <2x, 4y, 6z>, which is crucial for determining the tangent plane. A specific point (1, 2, -1) is mentioned, but the method of obtaining this point remains unclear to the participant. The conversation highlights the challenge of understanding the transition from the parameterized line to the tangent plane's equation. Overall, the thread reflects a struggle with applying calculus concepts to 3D surfaces and tangent planes.
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
 
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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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