How can I obtain UV coordinates of rectangle in 3D with raytracing method?

Question

I am currently working on raytracing. I have problem with view Ray collisions. I cant figure out how to get intersection point of ray and plane, to be more precise, my problem is not figure out intersection point of ray vs plane, problem is to convert this coordinate into uv coordinate(this rectangle can be rotated anyhow in world) for texture mapping. I know One point on this rectangle, its normal and bounds.

Answer 1

We have 4 vertices of a rectangle lying on a sphere:

A - top left
B - top right
C - bottom right
D - bottom left

Center of the sphere:

O

And intersection point on the sphere inside rectangle ABCD:

I

The idea is to identify all sides of the triangle AID, because it will allow us to know the coordinates of the point I on the plane. So if we move the rectangle on the plane with A(0, rect.height) and D(0, 0) then point I could be found by solving the following system of equations:

x^2+y^2=DI^2               - circle equation with center in point D and radius DI
x^2+(y-rect.height)^2=AI^2 - circle equation with center in point A and radius AI

from which it follows that:

y = (DI^2-AI^2+rect.height) / (2*rect.height)

and x could have 2 values (positive and negative), however we are interested only in positive value, because only it will be inside the rect.

x = sqrt(DI^2-(DI^2-AI^2+rect.height)/(2*rect.height))

Then UV could be calculated the following way uv(x/rect.width, y/rect.height)

However length of AI and DI still not known, but could be calculated using formula of Great-circle distance

AI = (Radius of the Sphere) * (Angular orthodromy length must be in radians)
Radius of the Sphere = sqrt((O.x - A.x)^2+(O.y - A.y)^2+(O.z - A.z)^2)
Angular orthodromy length = arccos(sin(a1)*sin(a2)+cos(a1)*cos(a2)*cos(b2-b1))

a1 is angle AOA1,  where A1(A.x, O.y, A.z)
b1 is angle O1OA1, where O1(O.x, O.y, A.z)
a2 is angle IOI1,  where I1(I1.x, O.y, I.z)
b2 is angle O2OI1, where O2(O.x, O.y, I.z)