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Find the equation of the plane in the canonical basis

  1. Sep 21, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    Hi!
    Can help me with this problem with my exercise?I don´t know if i did it okay or i have to do anymore
    Is there another form to do it?
    Be the π plane, whose equation with the base B (with coordinates (x ', y', z')) is z'= 0.
    B={(1,1,0),(0,1,1),(1,-1,1)}.
    a)Find the equation of the plane in the canonical basis. Prove analytically, how to get the same result.
    b)Could you give a base B´(coordinate (x'', y'', z'')) in which the equation of the plane is y''= 0?

    So, i did this:
    a)
    B = { u1, u2, u3} and canonic base K = { i, j, k }
    u1 = i + j, u2 = j + k
    u1 x u2 = u3
    the equation of plane π is z ' = 0
    therefore : this plane passes through origin and is perpendicular to u3 = i - j + k
    the equation of plane π in K is :
    OM . (i - j - k) = 0
    giving :
    x - y + z = 0

    b)the equation of plane π is : third coordinate = 0
    it is enough to keep the same vectors as those in U provided the SECOND is u3 :
    in basis :
    V = { u1, u3, u2 }
    the equation of plane π is : second coordinate : y" = 0
     
    Last edited by a moderator: Sep 21, 2015
  2. jcsd
  3. Sep 21, 2015 #2

    andrewkirk

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    Hello and welcome to physicsforums!

    Your answers look correct to me.
     
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