
#1
Dec2312, 01:10 PM

P: 10

Hello, I have a simple(or not?) math problem:
I have equations of 3 lines in R3 trought the origin: line l_{1}: λ_{1}*x+β_{1}*y+γ_{1}*z=0 λ_{2}*x+β_{2}*y+γ_{2}*z=0 line l_{2}: λ_{3}*x+β_{2}*y+γ_{3}*z=0 λ_{4}*x+β_{3}*y+γ_{4}*z=0 line l_{3}: λ_{5}*x+β_{4}*y+γ_{5}*z=0 λ_{6}*x+β_{5}*y+γ_{6}*z=0 I know every λ, β and γ  they are real constants. I also have a plane δ and : l_{1} intersects δ in point p_{1}, l_{2} intersects δ in point p_{2}, l_{3} intersects δ in point p_{3} I also know: ....the distance between p_{1} and p_{2} = h ....the distance between p_{2} and p_{3} = w ....p_{1}, p_{2} and p_{3} form a right triangle with right angle at p_{2} (h^{2}+w^{2}=(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 p_{1}, p_{2} and p_{3} 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: w^{2}+h^{2}=(p_{1}p_{2})^{2}, ....the two equations for distances w and h: h^{2}=(p_{1x}p_{2x})^{2}+(p_{1y}p_{2y})^{2}+(p_{1z}p_{2z})^{2} and w^{2}=(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. 



#2
Dec2312, 04:08 PM

P: 4,570

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*(1t) + (ba)*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 = (p2p1) X (p3p1) 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. 



#3
Dec2512, 04:22 AM

P: 10

If I solve L(t) = a*(1t) + (ba)*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 ? 



#4
Dec2512, 05:30 PM

P: 4,570

A simple camera problem
You can solve for t by using the fact that L(t) = point on plane for some t and then rearrange to get the value of t.




#5
Jan113, 08:03 AM

P: 10

Happy new year,
L(t)=a(1t)+(ba)t\\ L(t)=p_{1x}(1t)+(p_{2x}p_{1x})t L(t)=p_{1y}(1t)+(p_{2y}p_{1y})t L(t)=p_{1z}(1t)+(p_{2z}p_{1z})t L(t)=p_{2x}(1t)+(p_{3x}p_{2x})t L(t)=p_{2y}(1t)+(p_{3y}p_{2y})t L(t)=p_{2z}(1t)+(p_{3z}p_{2z})t Is that what you mean ? If so, how do I find p_{1}, p_{2} and p_{3} ? 



#6
Jan113, 08:12 PM

P: 4,570

What you have is basically L_1(t) = p1 = a1*(1t1) + (b1a1)*t1, L_2 = a2*(1t2) + (b2a2)*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 (p3p1) X (p2p1) and this will give the normal to the plane with point p1 on the plane. 


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