1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook