Ray Tracing Box Intersection: (3,8) to (6,-1), (-2,3), and i-5j

[haddow64]

Ok, so this is probably a really basic question, and I've got a feeling that my brain is just too 'fuzzy' just now and that all my problem is is simple arithmetic, but here it goes...

1. Determine whether or not the box with vertices at (2,1), (5,1), (5,5), and (2,5) is intersected by the ray originating at (3,8) and:

(i) ends at (6,-1)
(ii) ends at (-2,3)
(iii) has direction i-5j



Box:
AB: (2+3u / 1) -(1)
CD: (5-3u / 5) -(2)
BC: (5 / 1+4u) -(3)
AD: (2 / 1+4u) -(4)

(i) Ray (3,8) to (6,-1)
(3+3t / 8-9t)

(1) & (i) -> (2+3u / 1) = (3+3t / 8-9t)


How do I get this down to give t and u to show that intersection occurs? I'm sorry that this is such a basic question, I've been working all day and don't seem to be able to think properly just now.

Thanks in advance for any help.

 

[Dick]

Take your ray for each of the problems and intersect with the lines that form the boundaries of the box, look at the intersection points and draw conclusions from there. Even a simple graphical sketch of the problem might tell you what which lines you have to consider and which you don't.

 

[Architeuthis Dux]

Hi there,

First of all, don't worry about your brain feeling fuzzy. It's completely normal to feel mentally exhausted after working all day. Let's take a step back and break down the problem.

Ray tracing is a technique used in computer graphics to simulate the path of light rays in a virtual environment. In this case, we are dealing with a ray starting at (3,8) and ending at (6,-1). We also have a box with four vertices at (2,1), (5,1), (5,5), and (2,5).

To determine whether or not the ray intersects with the box, we need to find the point of intersection between the ray and the box. This point will have coordinates (x,y) that satisfy both the equation of the ray and the equations of the lines that make up the sides of the box.

Let's start with the equation of the ray. We can write it as (3+3t, 8-9t), where t is a parameter that represents the distance along the ray. Now, let's look at the equations of the lines that make up the sides of the box.

AB: x = 2+3u, y = 1-1u
CD: x = 5-3u, y = 5+0u
BC: x = 5+0u, y = 1+4u
AD: x = 2+0u, y = 1+4u

Notice that the equations for AB and CD only have one variable (u), while the equations for BC and AD only have one variable (v). This is because these lines are either horizontal or vertical, so they can be represented by a single variable.

Now, let's plug in the coordinates for the endpoints of the ray into its equation. For (6,-1), we have x = 6 and y = -1. Plugging these values in, we get:

6 = 3+3t
-1 = 8-9t

Solving for t, we get t = 1. Plugging t = 1 into the equation for the ray, we get the point of intersection as (6, -1).

Next, we need to plug in the coordinates for this point into the equations for the lines that make up the sides of the box. If the point satisfies all four equations, then we know that