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What is the purpose of the normal equation of the plane?

  1. May 14, 2012 #1
    Let r and a each be a position vector of a point on a plane. Let n be the normal to the plane.

    The normal equation of the plane would be:

    n.(r-a) = 0
    r.n = a.n

    My question is, the equation doesn't really 'describe' a plane...so what's the purpose?
     
  2. jcsd
  3. May 15, 2012 #2

    quasar987

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    Hi Alshia,

    If you know that a plane passes through a certain point x, and if you know the normal direction to the plane at x, then certainly you know how to describe the plane: it is all the points y such that the vector going from x to y is perpendicular to the normal.

    That's the situation in words. In math language now, to know the normal direction means to know a normal vector n. And the vector going from x to y is y-x. And for two vectors to be perpendicular means their dot product is 0. So what we said in words above is that the plane P is described by P = {y in R³ | (y-x).n=0 }
     
  4. May 16, 2012 #3

    mathwonk

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    there are two kinds of equations, one kind lets you recognize whether a point someone brings up belongs to the plane or not, and the other kimnd lets you produce as many points that are on the plane as you want.


    the normal equation is type one. given any point you plug it into the equation and if you get ero, then that point was a point of the plane.

    a parametrization of form (s,t)--->(p + sv + tw) give you as many points on the plane passing through p nd parallel to the independent vectors v and w as you want.

    i.e. any two numbers s and t will produce another point on that plane.

    i.e. there are two kinds of equations "implicit" and "explicit". the implicit kind only gives a way to check whether candidate points are or are not on the plane. explicit ones give a way to produce as many points as desired that are on the plane.
     
  5. May 17, 2012 #4
    @quasar987:

    Hmm, that description certainly isn't very visual, unlike the description of a vector or a straight line. But I guess it's acceptable.

    @mathwonk:

    So for example, the linear equation would be Type 1 and Type 2? What about the quadratic equation?
     
  6. May 17, 2012 #5

    quasar987

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    This description of a plane is certainly the most visual and understandable one. It is just me who failed to convey to you the image I have of it in my mind. But you can watch this guy talk about it and draw pictures for 13 minutes:

     
    Last edited by a moderator: Sep 25, 2014
  7. May 17, 2012 #6
    @quasar987:

    OK, so is the premise like this?:

    If n.(r-a)=0, then there exist a plane on which r and a lies.



    Why is the cartesian equation of a plane ax + by + cz = d? Does this require some kind of advanced math to prove? I'm asking this because the book uses the normal equation to show that that's cartesian equation, but in the video the cartesian equation is given first. I prefer the video's sequence though.
     
    Last edited: May 17, 2012
  8. May 17, 2012 #7

    quasar987

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    No, that is not the premise. In fact, given any two points r and a in space, there are infinitely many planes on which both lie.

    No advanced math is required to see that a plane can also be described by an equation of the form ax + by + cz = d, but it certainly is not obvious to me.

    Much more understandable is the description in terms of a point and a normal vector.

    Look, the basic thing to realize is this. Take a pen in your fingers and hold it before you. If I tell you that there is a plane passing through the point at the base of the pen and which is perpendicular to the pen, can you tell me where that plane is? Experiment with your free hand playing the role of the plane and realize that there is only one way to position your hand (representing a plane) at the base of the pen, so that it make an angle of 90° with the pen.

    This shows that to specify a plane in space, it is sufficient to specify
    a) a point lying on it, and
    b) a line (or vector) perpendicular to it
     
  9. May 18, 2012 #8
    Another way to think about it: if I gave you 3 points that lie in a plane (and are not collinear), do you believe that you could then uniquely determine the plane? How about if I gave you two (non-parallel) lines that line in the plane?

    If you believe that either of those descriptions uniquely determines the plane, it's not too hard to get to the description via a normal vector and one point. Given two non-parallel lines that lie in the plane, their point of intersection will give you a point a lying in the plane and taking the cross product of their direction vectors will give you a normal vector n.
     
  10. May 19, 2012 #9
    Let your normal vector be (a,b,c), r be (x,y,z), d= -n.a . The cartesian falls into your lap.
     
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