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Point at which a line intersects a plane

  1. Nov 14, 2012 #1
    So i know the equation of a plane.

    Ax + By +Cz = D

    Normal is the normal vector to the plane.

    A = normal.x
    B = normal.y
    C = normal.z
    p1 and p2 are 2 points on the line (which will intercept a plane at some point)
    the .x and .y and .z refer to there respective components of the vector.

    X = (p2.x - p1.x)* T + p1.x
    Y = (p2.y - p1.y)* T + p1.y
    Z = (p2.z - p1.z)* T + p1.z

    I also know what D equals I solved for that by moving stuff around

    The problem is a need a general solution for T.

    It should be something like

    A ((p2.x - p1.x) * T + p1.x)) + B ((p2.y - p1.y) * T + p1.y)) + C ((p2.z - p1.z) * T + p1.z)) = D

    Except isolated for T (I believe)

    In case your curious this is for a programming function. Thats why i'm using nothing but variables.

    I can solve for every equation but T, while I can solve for T by myself given specific numbers im not sure how to isolate it even if I expand out the equation to stuff like

    Ap2.x - Ap1.x * AT + Ap1.x ...

    I'm thinking maybe im going down the wrong path here or something.

    Any help would be much appreciated. :)
     
  2. jcsd
  3. Nov 14, 2012 #2

    Simon Bridge

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    You have p1 p2 and n?
    Why not group terms in T ... then the equation has form: [itex]\lambda T + \mu = D[/itex]
     
    Last edited by a moderator: Nov 14, 2012
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