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Homework Help: Euclidean geometry-2 planes

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Three-dimensional, euclidean space. We've got 2 non-parallel planes:
    [tex] \vec{OX} \cdot \vec{b_1}=\mu_1[/tex] and [tex] \vec{OX} \cdot \vec{b_2}=\mu_2[/tex]. Find all the points Y such that Y lies on the first plane and Y+[tex]\vec{a}[/tex] lies on the 2nd one. What did you get?

    2. Relevant equations
    Come in (3.)

    3. The attempt at a solution
    So we start with such equations:
    [tex]\vec{OY} \cdot \vec{b_1}=\mu_1 [/tex] AND [tex]\vec{OY} \cdot \vec{b_2} + \vec{a} \cdot \vec{b_2}=\mu_2 [/tex].
    Simple manipulation and we get:
    [tex] \vec{OY} \cdot (\vec{b_1} - \vec{b_2} )= \mu_1 - \mu_2 +\vec{a} \cdot \vec{b_2} [/tex] (substracted 2nd equation from the 1st one).
    And this is where I get stuck, I have no idea how to have [tex]\vec{OY} [/tex] on one side and the rest on the other.

    Thx in advance for your help!

    E: I'm not even sure that what I've got so far is correct-I believe that those points should all lie on one line, and I've got a plane again. But what I believe in geometry is not necessarily true :)
    Last edited: Jun 9, 2010
  2. jcsd
  3. Jun 9, 2010 #2


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    Homework Helper

    first consider all the points given by Y + a, this will be a plane parallel to the first offset by the vector a.

    The points you want will be the intersection of the offset plane with the 2nd plane
  4. Jun 9, 2010 #3
    I still don't catch it. I mean, I understand and see what you wrote, but have no idea, how to proceed with the exercise. If only it was possible to get a plane equation in the form of OX=a+td1+sd2...
  5. Jun 9, 2010 #4


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    ok, i agree the equation of teh plane is takes a little getting used to

    To simplify notation lets call the plane X and any vector form the origin to a point on the plane x = OX, then we have
    [tex] x \cdot b_1=\mu_1[/tex]

    now say [itex] |b_1|^2 = \beta_1^2 [/itex]

    take consider the vector [itex] b_1' = \frac{\mu_1}{\beta_1^2} b_1 [/itex]

    clearly that vector is on X, as
    [tex] b_1' \cdot b_1 = \frac{\mu_1}{\beta_1^2} b_1 \cdot b_1 = \mu_1[/tex]

    now say you have another vector x on the plane, and consider [itex] x' = x-b_1' [/itex], then:

    [tex] x' \cdot b_1 = (x-b_1') \cdot b_1 = (x \cdot b_1)- (b_1' \cdot b_1) =0[/tex]

    so now you know b_1 is normal to the plane X, and b_1' is a vector on X, which should allow you to manipulate the equation of the plane
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