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Vector equation of a plane.

  1. May 21, 2013 #1
    Just curious about a certain facet of the vector description of a plane. My query is as to why it is defined as n • (r-r0) = 0. Great, that's because any two vectors with a dot product of 0 must be orthogonal to each other and if we have a point on a infinite plane with an associated vector we can define the plane perfectly. BUT here is my problem (Which will be resolved after a few minutes on here, I'm sure) If we can define the vector (r-r0) as a single vector, say a, then why it seems to me that a multitude of different n vectors could be orthogonal to vector a, and thus it seems like a poor definition of a plane.

    Just wondering if anyone can clear this up for me.


  2. jcsd
  3. May 21, 2013 #2


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    That's NOT a "definition" of a plane but it is a property of any plane. Every plane is two dimensional- given any point in the plane there exist an infinite number of lines through that point lying in the plane. Further, there exist a unique line (normal to the plane) through that poit and perpendicular to all those lines.
  4. May 21, 2013 #3
    Right, BUT my book says that it is a definition (Or at least says that it determines a single plane) which is weird.

    Also, how can we derive the standard cartesian equation a(x-x0) + b(y-y0) + c(z-z0) from the vector equation?
  5. May 21, 2013 #4


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    That "multitude" of vectors is the set
    $$P=\left\{\mathbf r\in\mathbb R^3:\mathbf n\cdot\mathbf r=\mathbf 0\right\}.$$ This is a plane that contains ##\mathbf 0##. Now let ##\mathbf r_0\in\mathbb R^3## be arbitrary and consider the set
    $$Q=\left\{\mathbf r\in\mathbb R^3:\mathbf n\cdot(\mathbf{r-r_0})=\mathbf 0\right\}.$$ For all ##\mathbf r\in Q##, ##\mathbf r-\mathbf{r_0}## is in the plane Q. This observation should make it easier to see that Q is a plane that contains ##\mathbf r_0##. Q is what you get if you take P and add ##\mathbf r_0## to every vector in it.

    &\mathbf r =(x,y,z)\\
    &\mathbf r_0 =(x_0,y_0,z_0)\\
    &\mathbf n =(a,b,c)
  6. May 21, 2013 #5
    Oh, I see the mistake in my logic, I was looking at the cartesian equation being ambiguous as a result of the vector equation, I suppose that the vector equation is ambiguous until we add the actual vector, then we get a unique equation for the plane.

    Don't know what I was thinking. There's only one plane orthogonal to a particular 'normal' vector, even if there are several vectors orthogonal to a single plane.
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