# Curve contained in a Plane?

1. Oct 17, 2009

### hungryhippo

Given a parametric equation of a curve, how would you show that it is contained in a plane?

In 3-D space, a plane is defined by a point, and a vector that is orthogonal to the plane. For this question type, you're only given the parametric equation.

How I would appraoch this problem, would use t=0 to find a point on the curve that exists. Then, would I just find another point, in which it doesn't exist on the curve, since two nonparallel vectors are needed. Then, I would find the normal (n) by taking the cross product of those points...Would this be the way to go? If so, would I then just conclude that since the curve contains the point with parameter t=0, it is contained in a plane?

In desparate need of help on this

2. Oct 17, 2009

### Dick

Well, yeah. Just as you say. But you need to work with tangent vectors. If you can find two nonparallel tangents then you can compute the normal. Then just show any tangent is perpendicular to that normal. If you are just working with two points on the curve then you need to find a third point to define the plane and the normal.

Last edited: Oct 17, 2009
3. Oct 25, 2009

### kiwilava

Hi, regarding in this question of how to prove the curve is contained in a plane. You said i need to work with tangent vectors, so i find the tangent vector for the curve, then do i just take any t (t=0, t=1) to find two nonparallel tangents to compute for the normal?

4. Oct 25, 2009

### HallsofIvy

Staff Emeritus
I don't see why you need to work with tangent vectors at all.

I presume you are given parametric equations for the curve, x= f(t), y= g(t), z= h(t) and an equation for the surface, F(x,y,z)= 0. Replace x, y, and z by f(t), g(t), h(t) in the equation for the surface, F(f(t), g(t), h(t))= 0. If that is true for all t, then the curve lies in the plane.

5. Oct 25, 2009

### kiwilava

no, we were giving the parametric equations for the curve only, and need to show that the curve is contained in a plane.
so what i did was let t=0,-1,1 so obtain 3 points on the curve, and then subtracting the points, i found 3 vectors, then i cross product two of the vectors, and dot product the third vector, and i get 0, which means the vectors are coplanar, so they lie on the same plane, so the curve is contained in a plane. is that correct?

6. Oct 25, 2009

### Office_Shredder

Staff Emeritus
All that tells you is that those three points are contained in a plane (which is trivially true). You could do the same thing for any curve, it doesn't prove anything.

What you need to do is show that, given those three points lie on a specific plane, every other point also lies on that plane

7. Oct 26, 2009

### phizz

I'm not sure exactly how you would show that every other point in the curve lies on this plane. I was thinking:

Find three points on the curve, setting t=-1,0,1. Then using those points to come up with 2 vectors v1 and v2 and taking v1 X v2. This gives us a normal vector, telling us that we have a plane (defined as a point (t=0) and a normal vector). is this sufficient enough to prove that because we have a plane in which 3 of the points of the curve lie, the rest of the points also lie in this plane?

8. Oct 26, 2009

### Dick

Nooo. Let the curve r(t). If you find a normal using n=(r(1)-r(0))x(r(-1)-r(0)) now you have to show n.(r(t)-r(0))=0 (where '.' is dot product) for all t to show it's in the same plane.

9. Oct 26, 2009

### kiwilava

does it matter if i find the normal using n=(r(1)-r(0))x(r(-1)-r(1))?

10. Oct 26, 2009

### Dick

No, the normal you get by taking the cross product of any two of the differences will be parallel. This whole thing might be a little easier to get a handle on if you had a specific example of an r(t) you want to show is a plane curve. Do you?

11. Oct 26, 2009

### kiwilava

since that doesnt matter, then i've proved n.(r(t)-r(0))=0 for all t , thanks for your help once again

12. Oct 26, 2009

### phizz

My example is:

x = 1+3t+2t^2
y = 2-2t+5t^2
z = 1-t^2

13. Oct 26, 2009

### Dick

Ok, you seem to be on the right track. That is a plane curve. Do understand why dotting the normal with r(0)-r(t) being zero proves that?

14. Oct 26, 2009

### phizz

because that is the vector equation of a plane and you are showing that the normal vector N is perpendicular to r(t) which is in this plane?

15. Oct 26, 2009

### Dick

Yeah, basically. Except N isn't perpendicular to r(t). It's perpendicular to r(t)-r(0). The plane might not pass through (0,0,0). That's the case for your example.

16. Oct 26, 2009

### phizz

okay im actually not getting 0 when i take the dot product...i don't know if you actually did it out but if you did is this correct? or is my math wrong?

17. Oct 27, 2009

### Dick

Yes. I worked it out. You'd better show us what you are doing is you aren't getting 0 for the dot product. Remember you don't want r(t).n=0, you want (r(t)-r(0)).n=0.