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Find a Point on a 3D surface.

  1. Oct 10, 2004 #1
    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: Oct 10, 2004
  2. jcsd
  3. Oct 10, 2004 #2
    Whoops I neglected to see something on his answer key... I see whats going on.. Nevermind
  4. Oct 10, 2004 #3
    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.
  5. Oct 10, 2004 #4
    Yeah I'm been crackin away at this for a while and i sitll don't get it, any one anyone anyone..
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