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henpen

  • 3 years ago

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?

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  1. TuringTest
    • 3 years ago
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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

  2. Algebraic!
    • 3 years ago
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    think of it as a dot product...

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

  4. Algebraic!
    • 3 years ago
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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..

  5. Algebraic!
    • 3 years ago
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    or the orientation of the surface, rather

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

  7. Algebraic!
    • 3 years ago
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    what's POMN?

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

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