I am confused that for the picture at the bottom right corner, why the vertical axis is M^(-1) * (o - p0)? Why this vector is orthogonal to both vector u and v? And if u and v are both less than 1, that means the point is in triangle?
kalebm
How exactly does M^-1 transform the ray's direction to be orthogonal to the plane? Isn't the ray defined as o + td, and not o-p_0?
Jessica
The ray now is represented as M^(-1) * (o - p0), and is shown in the (u,v,t) coordinate.
ecohen2
In reference to my previous question, I tried rewatching this part of the lecture but am still not fully clear, can someone walk through exactly what is happening here? From my understanding, we change the triangle to be on the plane defined by u,v, then set it equal to the equation of the ray. From there we shift to 3d (u,v,t) coordinates .... but then where do we go from here? I know M^-1 bring it back to 2d, but where does that get us? How does it help us find the intersection?
sc2019
When you transform it back to 2d you can then use the Barycentric coordinates test to find the intersection. Essentially, we cannot do the test in 3d coordinates (or it's possible but more complex) so we transform it to 2d first.
adampahlavan
Shouldn't M be a 3x3 matrix here?
yueli96
still confused about how this M^(-1) transformation happened.
I am confused that for the picture at the bottom right corner, why the vertical axis is M^(-1) * (o - p0)? Why this vector is orthogonal to both vector u and v? And if u and v are both less than 1, that means the point is in triangle?
How exactly does M^-1 transform the ray's direction to be orthogonal to the plane? Isn't the ray defined as o + td, and not o-p_0?
The ray now is represented as M^(-1) * (o - p0), and is shown in the (u,v,t) coordinate.
In reference to my previous question, I tried rewatching this part of the lecture but am still not fully clear, can someone walk through exactly what is happening here? From my understanding, we change the triangle to be on the plane defined by u,v, then set it equal to the equation of the ray. From there we shift to 3d (u,v,t) coordinates .... but then where do we go from here? I know M^-1 bring it back to 2d, but where does that get us? How does it help us find the intersection?
When you transform it back to 2d you can then use the Barycentric coordinates test to find the intersection. Essentially, we cannot do the test in 3d coordinates (or it's possible but more complex) so we transform it to 2d first.
Shouldn't M be a 3x3 matrix here?
still confused about how this M^(-1) transformation happened.