Finding the Equation of a Plane in R3 Using Intersection Points and Distances

[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.

 

[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]

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.

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.