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Point at which a line intersects a plane |
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| Nov14-12, 12:07 AM | #1 |
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Point at which a line intersects a plane
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. :) |
| Nov14-12, 01:37 AM | #2 |
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Recognitions:
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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] |
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