# How to raytrace triangular faces when camera is not at origin?

I've watched the Pixar In A Box, Maths of Rendering Tutorial on Khan Academy. I've started writing a raytracing program and have come to the point where I must solve for t (the parameter for the parametric ray through the camera and a point on the view plane).

In the tutorial, this is done through simply substituting each the parametric formula for each x, y, and z into the equation of the plane that the triangle lies on. This leaves a single unknown (t).

However, in all the examples and exercises that are given by the tutorial the camera is located at [0,0,0]. As I'm sure you'll understand, this is in many cases unrealistic. Because the camera is at 0, they simplify as follows:

$i=\frac{1}{t}C+tP$

becomes simply:

$tP$

Which makes for an easy solve. My question is when the camera is not at [0,0,0] should I:

1. Calculate the coordinates of the plane relative to the camera. (So the camera acts as if it is [0,0,0]
2. Solve the equation as a quadratic. I'm still confused a bit about why this is a quadratic in the first place. Could someone explain this? And which solution do I use?

Of the two options, which is more acceptable for use in this community? Why? Is one more computationally efficient?

Edit: I think I have found a solution as follows: The parametric equation (which is effectively just a weighted average) is represented like this in the tutorial:

$x(t)&space;=&space;\frac{1}{t}C_{x}+tP_{x}$

But I realize now that is the same as:

$x(t)&space;=C_{x}&space;+t(P_{x}-C_{x})$

Which is dramatically easier to solve.

So my final question is, which of the three is the most acceptable way to solve for t? I'm assuming the third one.

• Have you been misreading $1-t$ as $1/t$ this whole time? – Rahul Oct 3 '18 at 6:44
• Yeah for like 5 days – PolymorphismPrince Oct 3 '18 at 21:49

Wow I'm stupid, the whole problem was I thought it was 1/t and it is actually 1-t in parametric form. Hope this helps someone anyway.