Cartesian and Parametric

  • Thread starter prasannapakkiam
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  • #1
prasannapakkiam

Main Question or Discussion Point

Okay, I was doing 3D modelling. To save space I used vector functions to render terrain. Anyway, I came up with 3 parametric equations - each a function of an axis: e.g.: x=4t, y=5t+6, z=7t-9. How can you convert this into a Cartesian Equation?:confused:
 

Answers and Replies

  • #2
590
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What do x, y and z stand for?
 
  • #3
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Okay, I was doing 3D modelling. To save space I used vector functions to render terrain. Anyway, I came up with 3 parametric equations - each a function of an axis: e.g.: x=4t, y=5t+6, z=7t-9. How can you convert this into a Cartesian Equation?:confused:
Not sure what you mean by a "Cartesian Equation".
In your case, your are looking for a nonhomogeneous linear system of 2 equations in 3 unknowns.

EDIT.

Basic theorem from linear algebra:
Every linear manifold is a solution set of a nonhomogeneous linear system of equations.

In your example, the function-locus parametrically defined by f(t) = (4t,5t+6,7t-9) can be considered a linear manifold.
It is the range of f.
It is the line of intersection of the two planes determined by the system.
 
Last edited:
  • #4
2,063
2
z = 5t + 6 + 2t - 15
 
  • #5
prasannapakkiam
z(t)=7t-9
x(t)=4t
y(t)=5t+6
i.e. it is like a vector
I was wondering if there is a way to transfer this into a Cartesian equation. I mean if the z did not exist, I would just solve the 2 equations for t and then equate them: x/4=(y-6)5. With that extra function, how would one go about creating a single cartesian equation?
 
  • #6
90
0
z(t)=7t-9
x(t)=4t
y(t)=5t+6
i.e. it is like a vector
I was wondering if there is a way to transfer this into a Cartesian equation. I mean if the z did not exist, I would just solve the 2 equations for t and then equate them: x/4=(y-6)5. With that extra function, how would one go about creating a single cartesian equation?

Read the edit in post 3.
 
  • #7
1,425
1
There is no cartesian equation for 3d vectors. Think about it: if a single relationship included three variables, x, y, z, then any of these variable could not be determined by the value of simply 1 other variable. In fact, such a relationship represents a surface, not a curve or line thereof.
 
Last edited:
  • #8
prasannapakkiam
Exactly what I wished to confirm...
 
  • #9
Actually, there is a Cartesian form.

Given a position vector A <a1,a2,a3>
And a direction vector B <b1,b2,b3>

The Cartesian form of a line in 3 dimensional space is:

(x - a1) / b1 = (y - a2) / b2 = (z - a3) / b3

This can be determined from two points (C,D) by generating the direction vector like so:
A = C - D

And then picking one of the points as the position vector.
See:
http://www.netcomuk.co.uk/~jenolive/vect17.html

And I believe you can generate a Cartesian form of a line in N dimensions..
 
  • #10
192
3
Some people refer to the Cartesian Equation Form as the
Symmetric Form

JFYI
 

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