Converting cartesian to parametric equation

[votr]

converting cartesian to parametric equation R3

hi,

I can't convert cartesian to parametric equation this equation 3x-y+4z-6=0
In example is given only 3x-y+4z-6=0 and says to convert it to parametric form ?

how this can be done ?

 

[Mark44]

So that I don't completely answer your question for you, let me show you how it works for a different problem. Suppose your equation was x + 2y - z + 2 = 0. This equation represents a plane in R³, as does your equation.

Solving for x gives
x = -2y + z - 2
We're going to need a set of equations for x, y, and z, so here are two more:
y = y
z = z
The last two equations are obviously and trivially true.

Here's what we have:

x = -2y + z - 2
y = y
z =      z

From this, we can see that any point (x, y, z) in the plane can be written in terms of two parameters y and z. Namely, (x, y, z) = y*(-2, 1, 0) + z*(1, 0, 1) + (-2, 0, 0).

If you like, you can use different letters for the parameters, say r and s, so that you have 
(x, y, z) = r*(-2, 1, 0) + s*(1, 0, 1) + (-2, 0, 0).

If you look at this in terms of vectors, the vector (-2, 0, 0) takes you from the origin to the point (-2, 0, 0) in the plane, and the other two vectors take you from that point to any other in the plane. 

You can apply the same thinking to your problem. Hope that helps.

 

[HallsofIvy]

Note that this is a plane. That is, it is two dimensional and so will require two parameters (unlike a line or curve that is one dimensional and so requires only one parameter). What Mark44 did was use y and z as parameters. If you don't like that, that is, if you prefer to use, say, s and t, as parameters, just replace y and z by that.

That is, if x= -2y+ z- 2 then x= -2s+ t- 2, y= s, z= t are parametric equations for the plane.

In general, if you can solve an equation, in x, y, and z, representing a surface, for anyone of the variables, you can use the other two as parameters.

 

[votr]

thank you for your answers,
now I understand how to express it with parameters
but this part, (x, y, z) = r*(-2, 1, 0) + s*(1, 0, 1) + (-2, 0, 0)
i don't understand
how (-2,1,0) and (1,0,1) vectors are acquired !

thanks again

 

[Mark44]

From here:

x = -2y + 1z - 2
y = 1y + 0z
z = 0y + 1z