General equation for plane through the origin

[issisoccer10]

[SOLVED] General equation for plane through the origin

Homework Statement

Show that for any funcions f(t), g(t), and h(t), the graph of

r(t) = [f(t)-g(t)]i + [g(t)-h(t)]j + [h(t)-f(t)]k

lies on a plane passing through the origin.

Homework Equations

The Attempt at a Solution

I have tried various equations for f(t), g(t), and h(t) and the equations obviously holds true. However, that doesn't bring me any closer as to why it works. It seems pitifully simple, I know, but the solution escapes me. Any help would be greatly appreciated...Thanks a lot

 

[Defennder]

This question appears to be quite tedious, but if it lies in a plane you should be able to show that the torsion of r(t) is 0 everywhere. The resulting mathematical expression appears very complicated. Anyone else here has a better approach?

 

[ozymandias]

I'd break up the proof to two parts:
1. That these points all lie on a single plane.
2. That the plane passes through the origin.

I'd solve #1 by computing the derivative r'(t) and guessing a normal n to the plane, showing that n dot r'(t) = 0 regardless of t.
Once the normal has been found, one can solve #2 by substituting in one of the points on the plane and finding out exactly whether the plane passes through the origin or not (obviously it does, or they wouldn't have asked you to prove this :) ).

Assaf
"www.physicallyincorrect.com"[/URL]

 

[coomast]

I don't think you need the derivative of f(r). Consider two points at locations of the parameter t at t1 and t2 of the curve. The position vector of these points are now:

\vec{V}_1=(f_1-g_1)\vec{i}+(g_1-h_1)\vec{j}+(h_1-f_1)\vec{k}
\vec{V}_2=(f_2-g_2)\vec{i}+(g_2-h_2)\vec{j}+(h_2-f_2)\vec{k}

The normal vector of these two position vectors, meaning the normal vector of the plane formed by the legs of these vectors, can be obtained as the cross product, thus:

\vec{N}=\vec{V}_1 \times \vec{V}_2

After this has been obtained, you have an equation of the normal as follows:

\vec{N}=n_x\vec{i}+n_y\vec{j}+n_z\vec{k}

With the terms each a function of the two points. Now in order to show that any point of the curve is in the plane pasing through the two position vectors, you must have:

\vec{N} \cdot \vec{p}=0

This is nothing more than:
[Edit] error in next formula corrected
n_x(f-g)+n_y(g-h)+n_z(h-f)=0

Or:

(n_x-n_z)f+(n_y-n_x)g+(n_z-n_y)h=0

This is not difficult to show and therefore the proof is finished. Because the position vectors start in the origin, the plane passes through it as well. It's one and a half page long, not so tedious at all.

 

[HallsofIvy]

By the way, your title, "General equation for plane through the origin" is a little misleading. My first reaction was that since this function depends on a single variable, t, it must be a one dimensional curve and not a plane! Of course you were asking about a curve lying on a single plane.

 

[ozymandias]

coomast,

Your V1 and V2 ARE derivatives of sorts :).
(finite differences, to be exact)
Regarding the condition N dot p, what is p? Is it p=r(t0) for some t0? If so, then I beg to differ.
Similarly, your assertion

Because the position vectors start in the origin, the plane passes through it as well

is incorrect. Consider:

r(t) = (sin(t),cos(t), 1)

Obviously r(t) starts at the origin, but the shape traced by it (a circle at z=1) belongs to a plane which does NOT pass through the origin.

Assaf
"www.physicallyincorrect.com"[/URL]

 

[coomast]

@ozymandias: It is indeed a kind of derivative. You are right on this.

The point p is nothing more than a random t value inserted into original equation and then written with its position vector from the origin. I should have been a bit more clear, again you're right.

Regarding the example you gave, I don't see that this is valid for the assumptions in the original post made.

f-g=sin(t)
g-h=cos(t)
h-f=1

Does not have a solution for f, g and h, so it can't be used here.

If you state that:

f=sin(t)
g=cos(t)
h=1

It is a valid one.

In case this is incorrect, please post.