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henpen

It is given that at the point \[ [x_0,y_0,z_0] \] the normal vector to the plane is \[ n= [p,r,s] \]. You can deduce from this that the equation of the plane is \[ p(x-x_0)+r(y-y_0)+s(z-z_0) =0\] How?

  • one year ago
  • one year ago

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  1. TuringTest
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    I would walk you through this personally, but my connection sucks right now, so... http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

    • one year ago
  2. Algebraic!
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    think of it as a dot product...

    • one year ago
  3. Algebraic!
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    (x -xo), (y-yo) etc are the components of any vector in the plane....

    • one year ago
  4. Algebraic!
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    if <p,r,s > dot < any vector in the plane>=0, then <p,r,s> is normal to the plane and defines the surface..

    • one year ago
  5. Algebraic!
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    or the orientation of the surface, rather

    • one year ago
  6. henpen
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    Yes, POMN says similarly. I think I've got the intuition, thanks both.

    • one year ago
  7. Algebraic!
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    what's POMN?

    • one year ago
  8. Algebraic!
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    oh, Paul's online math notes... never mind:)

    • one year ago
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