A simple camera problem

xzardaz

Hello, I have a simple(or not?) math problem:

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

line l₁:
|λ₁*x+β₁*y+γ₁*z=0
|λ₂*x+β₂*y+γ₂*z=0

line l₂:
|λ₃*x+β₂*y+γ₃*z=0
|λ₄*x+β₃*y+γ₄*z=0

line l₃:
|λ₅*x+β₄*y+γ₅*z=0
|λ₆*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=(p_1x-p_2x)²+(p_1y-p_2y)²+(p_1z-p_2z)² and w²=(p_2x-p_3x)²+(p_2y-p_3y)²+(p_2z-p_3z)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 Thumbnails

 

chiro

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.

xzardaz

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 ?

chiro

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.

xzardaz

Happy new year,

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

L(t)=p_1x(1-t)+(p_2x-p_1x)t
L(t)=p_1y(1-t)+(p_2y-p_1y)t
L(t)=p_1z(1-t)+(p_2z-p_1z)t

L(t)=p_2x(1-t)+(p_3x-p_2x)t
L(t)=p_2y(1-t)+(p_3y-p_2y)t
L(t)=p_2z(1-t)+(p_3z-p_2z)t

Is that what you mean ? If so, how do I find p₁, p₂ and p₃ ?

chiro

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.