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Perpendicular planes

  1. Jul 3, 2009 #1
    How to find (equations of) two perpendicular planes intersecting to a plane (say 3x+y+z=10) in a point (say 1,2,5 ).
    All the three planes are perpendicular to each other and intersecting at a single point (say (1,2,5) in this example)
    If possible explain it in vector form and non-vector forms.

    Thanks in advance.
     
  2. jcsd
  3. Jul 3, 2009 #2
    So you got three perpendicular planes:

    Ax+By+Cz+D=0
    Ex+Fy+Gz+H=0
    Ix+Jy+Kz+W=0

    If two planes are perpendicular to each other then, then their normal vectors are also perpendicular to each other.

    The normal vectors are (A,B,C), (E,F,G) and (I,J,K). This means that the scalar product (A,B,C) o (E,F,G) = 0 , (A,B,C) o (I,J,K) =0 , (E,F,G) o (I,J,K)=0

    Also the vector product of two of them gives us the third vector.
    For ex. (A,B,C) x (E,F,G) = (I,J,K)

    The intersection point is standing on all of the planes. So that if it is [itex](x_1,y_1,z_1)[/itex] then:
    [tex]Ax_1+By_1+Cz_1+D=0 [/tex]
    [tex]Ex_1+Fy_1+Gz_1+H=0 [/tex]
    [tex]Ix_1+Jy_1+Kz_1+W=0[/tex]


    Regards.
     
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