Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A simple camera problem

  1. Dec 23, 2012 #1
    Hello, I have a simple(or not?) math problem:

    I have equations of 3 lines in R3 trought the origin:

    line l1:
    1*x+β1*y+γ1*z=0
    2*x+β2*y+γ2*z=0

    line l2:
    3*x+β2*y+γ3*z=0
    4*x+β3*y+γ4*z=0

    line l3:
    5*x+β4*y+γ5*z=0
    6*x+β5*y+γ6*z=0

    I know every λ, β and γ - they are real constants.

    I also have a plane δ and :

    l1 intersects δ in point p1,
    l2 intersects δ in point p2,
    l3 intersects δ in point p3

    I also know:
    ....the distance between p1 and p2 = h
    ....the distance between p2 and p3 = w
    ....p1, p2 and p3 form a right triangle with right angle at p2 (h2+w2=(p1p3)2)

    I want to find the equation of δ in terms of λ, β, γ, h and w

    _______________________________

    it should be easy to find the equation of the plane if I find the points p1, p2 and p3

    I think I should compose a system, containing:
    ....the first six equations(which are linear, so it should be easy to solve with matrices),
    ....the equation for right angle in R3: w2+h2=(p1p2)2,
    ....the two equations for distances w and h: h2=(p1x-p2x)2+(p1y-p2y)2+(p1z-p2z)2 and w2=(p2x-p3x)2+(p2y-p3y)2+(p2z-p3z)2

    There are 9 equations with 9 unknowns (3 points, each with 3 coordinates) - it should be solvable.

    Have you got any ideas how to solve this? If there weren't 3 quadratic equations it would be easy.


    *I know that there should be 2 planes, matching the conditions - one that I've drawn(see the image) and the other is at the opposite side of the origin. I think the two roots of the quadratic equations should find them.
     

    Attached Files:

  2. jcsd
  3. Dec 23, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey xzardaz.

    Try using the intersections to get the normal of the plane and use one of the points as a point on the plane.

    Basically a line is represented by L(t) = a*(1-t) + (b-a)*t where the line passes through points a and b.

    Now if you solve for t for all the lines you build a normal by considering n = (p2-p1) X (p3-p1) which you normalize and then use the relationship n . (r - r0) = 0 where n is what you solved, r0 = p1, and r is a general point on the plane.
     
  4. Dec 25, 2012 #3
    I don't think I fully understand you. Can you give me some references ?

    If I solve L(t) = a*(1-t) + (b-a)*t for all the lines, I can't get the points, because there are infinite number of points matching the equations. How do I use w and h ?
     
  5. Dec 25, 2012 #4

    chiro

    User Avatar
    Science Advisor

    You can solve for t by using the fact that L(t) = point on plane for some t and then re-arrange to get the value of t.
     
  6. Jan 1, 2013 #5
    Happy new year,

    L(t)=a(1-t)+(b-a)t\\

    L(t)=p1x(1-t)+(p2x-p1x)t
    L(t)=p1y(1-t)+(p2y-p1y)t
    L(t)=p1z(1-t)+(p2z-p1z)t

    L(t)=p2x(1-t)+(p3x-p2x)t
    L(t)=p2y(1-t)+(p3y-p2y)t
    L(t)=p2z(1-t)+(p3z-p2z)t

    Is that what you mean ? If so, how do I find p1, p2 and p3 ?
     
    Last edited: Jan 1, 2013
  7. Jan 1, 2013 #6

    chiro

    User Avatar
    Science Advisor

    What you have is basically L_1(t) = p1 = a1*(1-t1) + (b1-a1)*t1, L_2 = a2*(1-t2) + (b2-a2)*t2 and so on.

    You extrapolate your variables in terms of the t's by relating the different forms of the line equations together.

    The normal is calculated in terms of (p3-p1) X (p2-p1) and this will give the normal to the plane with point p1 on the plane.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A simple camera problem
Loading...