Lecture 10: Ray Tracing (18)
evan1997123

Can someone explain how they got the last step for solving t. I know Professor Ren said it's just moving parts, but I thought you couldn't do that if you had dot products.

ellinzhao

(op)N=tdN(\mathbf{o} - \mathbf{p}')\cdot \mathbf{N} = -t\mathbf{d}\cdot \mathbf{N}

From here you can just divide out dN\mathbf{d}\cdot \mathbf{N} because the result of the dot product is just a scalar

ziyaointl

Algebraically, pN\mathbf{p}' \cdot \mathbf{N} should be cancelled on both sides, leaving the final equation as t=(po)NdNt = \frac{(\mathbf{p}-\mathbf{o}) \cdot \mathbf{N}}{\mathbf{d} \cdot \mathbf{N}}. However, since p\mathbf{p} and p\mathbf{p}' are just any two points on the plane, so I guess it doesn't matter.

aninrusimha

A cool side note: If d is perpendicular with N, then the dot product of d and N is going to be 0. This means the result is undefined, which makes sense as it that case there are either 0 solutions (o is not on the plane) or infinite solutions

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