# Converting from Cartesian to Parametric form

1. Sep 20, 2007

### JFonseka

[SOLVED] Converting from Cartesian to Parametric form

1. The problem statement, all variables and given/known data

Find a parametric vector equation of for the plane in R^3 having cartesian equation

4y + 5z = - 6

2. Relevant equations

None

3. The attempt at a solution

What I did was, first I turned the equation into 4x + 5y = -6, cause I'm more comfortable just treating equations like that.

Then I solved for y and got, y = (-4x -6)/5 = t

Hence x = (4t + 6)/5

and now y = t.

Then in parametric form it becomes:

(x, y) = (6/5, 0) + (4/5, 1)t

Is this right or wrong?

2. Sep 20, 2007

### genneth

As it's a plane, you need two parameters to describe it. This is a standard transform -- the equations for a plane in cartesian form and parametric form are well known, and have easy to understand geometric ties. Understand how they both relate to the geometry, and it will be easy to swap between them.

3. Sep 20, 2007

### HallsofIvy

Staff Emeritus
Unfortunately, it's not true now! Your original equation had y and z, not x and y! You can't just arbitrarily change it!

?? If y= (-4x-6)/5, then x is NOT (4y+6)/5. From 4x+ 5y= -6 (which is incorrect itself), solving for x gives x= (-5y-6)/4.

What happened to z?? This is a plane in three dimensions. Since a plane is a two-dimensional object, parametric equations must involve two independent parameters.
You can do this: solve 4y+ 5z= -6 for either y or z: say, z= (-6-4y)/5. Now you can let y be the parameter: y= t, z= (-6-4t)/5.

Since your equation tells you nothing about x, x can be any number for all y and t: let x be the other parameter. Now what do you have?

4. Sep 20, 2007

### JFonseka

I still don't get it, so Iwe have z = (-6 -4t)/5 now and I let x be equal to anything???

5. Sep 20, 2007

### Dick

As genneth and Halls said, you should expect to have two parameters. y= t, z= (-6-4t)/5, x=s is one way to write it.

6. Sep 21, 2007

### JFonseka

Yea I managed to solve it a few hours ago, thanks for the replies guys!

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