Point at which a line intersects a plane

  • #1

Main Question or Discussion Point

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. :)
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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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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