Complex Math Problem - Any takers?

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AltF8

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AltF8
I've got a slight dilemna, and I thought I'd put it out to the big brains that sometimes hover around GTP. I've got a little math problem for you, that if you solve - I'll be really grateful. ;)

What I need is, how to programmatically find the formula that defines the line between two points in a three-dimensional space.

*ducks*

Basically, I know what x, y, and z are for both points, and I need to know what other points it intersects between those two.

Thoughts?
 
Hmmm...looking deeper, I find that this is extremely complicated and requires knowledge of vectors. :(

Unless someone else has any ideas...
 
Originally posted by AltF8
I've got a slight dilemna, and I thought I'd put it out to the big brains that sometimes hover around GTP. I've got a little math problem for you, that if you solve - I'll be really grateful. ;)
Click to expand...

Key word there.
 
Let's try this:
http://www.wcsscience.com/pythag3d/page.html

It's fairly straight forward, but you have to know what the x,y,&z coordinants are at both ends to arive at the answer. By define, do you mean how long it is and where it's placed within the 3D space?
 
No, what I need is the equation that defines the line. The Pythagorean Theorem gives the distance between the points.
 
Have you found the "Stargate"?




Actually, I have a son that is a math whiz. When he gets home from work I'll check and see if he has an answer to your question.
I only do gas mileage, drug conversion, and IV drip rates.:lol:
 
Originally posted by AltF8
No, what I need is the equation that defines the line. The Pythagorean Theorem gives the distance between the points.
Click to expand...

Well, by reverse-engineering an equation from only two points, you're only ever going to get a straight line. What if your line is not straight?
 
When do you need the answer for this? I think we'll be studying the Z dimension a little bit in Algebra II the coming semester. :D
 
y = mx + b in a given plane. Define the plane with the 'z' coordinates.
Y and X are the variables and m is the slope and b is the Y intercept.
I don't understand it but am told this is the formula you seek.
 
Originally posted by Gil
y = mx + b in a given plane. Define the plane with the 'z' coordinates.
Y and X are the variables and m is the slope and b is the Y intercept.
I don't understand it but am told this is the formula you seek.
Click to expand...
Well, problem is, isn't y=mx+b only aplicable in the x and y planes? The z plane doesn't really figure into that equation...

Bah, let Pako figure it out! :lol:
 
I believe that the formula you want is ax+by+cz+d=0. It's called a Cartesian equation of the plane.

If you need more than that, I'll type up an explanation or find a good website link.
 
Originally posted by youth_cycler
Well, problem is, isn't y=mx+b only aplicable in the x and y planes? The z plane doesn't really figure into that equation...

Bah, let Pako figure it out! :lol:
Click to expand...
Please see the last line of my last post.:D
I'm quite serious. I DON'T understand it. My son gave me the formula.
 
Originally posted by GilesGuthrie
Well, by reverse-engineering an equation from only two points, you're only ever going to get a straight line. What if your line is not straight?
Click to expand...

The line will always be straight.
 
Originally posted by Tom M
I believe that the formula you want is ax+by+cz+d=0. It's called a Cartesian equation of the plane.

If you need more than that, I'll type up an explanation or find a good website link.
Click to expand...

It's the best thing I've seen yet, but the key is how to define the plane that the line sits on.
 
the formula is OP1 + t(P1P2) = 0

OP1 is a vector from the origin to the first point

P1P2 is a vector from the first point to the second point

t is a variable (time)
 
:lol:

I think I'll leave the 4th dimension out of this one, thanks. ;)

(Of course, the application that would be using this would entail objects travelling at supraluminal speeds...)
 

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