# Parametric Representation of a Plane

1. Sep 30, 2007

### Alw

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

Give a parametric representation of the plane x + y + z = 5.

2. Relevant equations

I am really not sure, I've been over the chapters we've covered for a little over an hour now, and the only mention i can find of a parametric representation of a plane is in passing once, merely stating that such a thing exists. All examples and explanations relate to

0 = a(x-x1) + b(x-x1) + c(z-z1)

where <a, b, c> is a vector normal to the plane, and (x1,y1,z1) is a point on the plane.

3. The attempt at a solution

well, I am going to assume that 0 = a(x-x1) + b(x-x1) + c(z-z1) is the standard form for planes, so I started by putting x + y + z = 5 in that form.

x + y + z = 5
x + y + z -5 = 0

i picked an arbitrary point on the plane, (2,2,1)

a(x-2) + b(y-2) + c(z-1) = 0, and therefore the coefficents must all be 1, giving me

(x-2) + (y-2) + (z-1) = 0, along with <1,1,1> being a vector normal to this plane.

i am really not sure where to go after this...
i know how to find the parametric representation of the intersection of two planes, but of the plane itself. . .

I am sorry i don't have much work to show for this, but I really have no idea where to start.

2. Sep 30, 2007

### Dick

The equation you came up with is really the same on you started with, and it's not 'parametric'. Take your point P0=(2,2,1) on the plane and find two more points on the plane P1 and P2 such that P0,P1 and P2 don't all lie on the same line. Then you can write a parametric representation of the plane as P0+s*(P1-P0)+t*(P2-P0). Do you see why this lies on the plane for any s and t, and do you see why any point on the plane can be written in this form?

3. Sep 30, 2007

### HallsofIvy

Staff Emeritus
A plane is two dimensional. "Parametric equation" for a two-dimensional figure must be of the form x= f(u,v), y= g(u,v), z= h(u,v).

Since you are told that " the plane x + y + z = 5", and so z= 5- x-y, there is nothing at all wrong with taking x and y themselves as parameters.

4. Sep 30, 2007

### Dick

Good point. Much simpler.

5. Sep 30, 2007

### Alw

ok, i understand how you can use those 3 points to define the plane since they arent in the same line, and i think i understand why s and t can be any value. Regardless of what value they have, they are are just a coefficient on the (position vector?) kind of like how with the equation y = x, (1,1) is on that line, along with a*(1,1), where a is any number. Any point on the plane can be used becuase, any point on the plane is still in the plane :P.

i'm not sure i fully understand. are you saying that i can use x and y as parameters for z, x and z as parameters for y, y and z for parameters for x? how would i continue on with the problem, or can i answer it by saying

z = 5 - x - y
x = 5 - y - z
y = 5 - x - z

? I dont think I fully understand what you are saying

Thank you very much for the help :)

Last edited: Sep 30, 2007
6. Sep 30, 2007

### Dick

I'm glad you get the vectorial way of thinking about it. Hall's is thinking about it algebraically. x(s,t)=s, y(s,t)=t, then z(s,t)=5-s-t is on the plane. It's the same as the other approach if P0=(0,0,5), P1=(1,0,4) and P2=(0,1,4).

7. Sep 30, 2007

### HallsofIvy

Staff Emeritus
No, I meant:
x= u
y= v
z= 5- u- v.