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View Full Version : Calc 3: Proof of coplanar vectors

SeeDRankLou
02-13-2009, 09:55 AM
So there's a proof in calc 3 that I can't get. Coplanar points are just on the same plane, but coplanar vectors are defined as any 3 vectors a, b and c, and any scalars s,t and u where s,t or u != 0 that satisfy

sa + tb + uc = 0.

There is a theorem that I'm having trouble proving (I like having proofs for theorems). The theorem goes:

a = PA, b = PB, c = PC are coplanar if and only if the points P, A, B, C all lie on the same plane.

I can prove half of this theorm, If the vectors are coplanar then the 4 points lie on the same plane.

Proof:
If the vectors are coplanar, then sPA + tPB + uPC = 0. Assuming s != 0, then PA = (-t/s)PB - (u/s)PC. By setting up the dot product:

(PB x PC) dot PA = (PB x PC) dot [-(t/s)PB - (u/s)PC]
= (-t/s) (PB x PC) dot PB - (u/s) (PB x PC) dot PC
The cross product is perpendicular to both of its components, so:
= (-t/s) (0) - (u/s) (0) = 0

However, (PB x PC) dot PA is suppose the be the volume of the parallelepiped (or tetrahedral, whatever you want to call it) formed by those vectors. Since the volume = 0, then there is only a plane. Therefore, P, A, B, C are all on the same plane.

I am however having trouble proving the other half: If P, A, B, C are on the same plane, then PA, PB, PC are coplanar.

It's not enough to say that since the 4 points are on the same plane then their 3 vectors are as well, I have to somehow derive that sPA + tPB + uPC = 0. I'm having trouble figuring out how.

Can anyone help me?

Aerith's Knight
02-13-2009, 09:30 PM
Well, I'm probably misunderstanding the question, but if you have three vectors with the same point of origin in the same plane, then this equation would be something like:

sa +tb +uc = 0

Where PA = a = A-P = (Ax - Px, Ay - Py, Az - Pz) etc etc.

If s, t and u are scalars that you can define yourself it seems to me that it would boil down to these equations:

s * ax + t * bx + u * cx = 0
s * ay + t * by + u * cy = 0
s * az + t * bz + u * cz = 0

Where (x, y, z) = (ax, ay, az), (bx, by, bz) and (cx, cy, cz)

These are three equations with three unknowns, so you can solve this by the standard method of reduction. (im assuming you do know how seeing the complexity of the previous question (and this one) on this board, if not, just shout and I'll give an example).

SeeDRankLou
02-13-2009, 10:56 PM
That's how one would show that the 3 vectors are coplanar. What I'm trying to do is derive the formula sa + tb + uc = 0 based on the fact that the 4 points are on the same plane. I can't just say that sa + tb + uc (unless I can), I have to somehow derive it.

What I've got to far is this, but I'm stuck.

The normal to this plane could be PA x PB, which would be perpendicular to all lines on the plane. So PA x PB is perpendicular to PA, PB and PC. So the dot product of the normal with each of those vectors = 0. If I take 3 scalars s, t and u, at least one != 0, then I can say:

(PA x PB) dot sPA = 0
(PA x PB) dot tPB = 0
(PA x PB) dot uPC = 0

If I add these 3 equations and factor out the cross product, I get:

(PA x PB) dot (sPA + tPB + uPC) = 0

But I can't figure out how to get rid of the cross product, or to somehow say that it is perpendicular to an arbitrary vector, which would make it the zero vector, or anything. Maybe I'm not thinking of this correctly, or maybe I'm missing something simple, I don't know.

Aerith's Knight
02-14-2009, 12:33 AM
You can't say it is perpendicular to anything more, because that is impossible in a field.(Unless you have some 4th dimension up your sleave).

Another way would be to formulate the expression so that the only way it can be true is that (sPA + tPB + uPC) = zero vector, but that doesn't seem to apply here.

I'll have to think about it.

SeeDRankLou
02-17-2009, 09:45 PM
Ok, so how does this sound:

If P, A, B, C are on the same plane, then the vectors PA, PB, PC are on the same plane. Since the all share a common point P, the vectors would look something like so:

http://i47.photobucket.com/albums/f155/maddylouis/EoFF/PABC.jpg

Since they are on the same plane, the Parallelogram Law of adding vectors applies. If this is true, then there exists 2 scalars b & c such that:

PA = bPB + cPC

By picking an arbitrary scalar s, any vector parllel to PA would = sPA. So,

sPA = sbPB + scPC

sPA - sbPB - scPC = 0.

And by letting t=-sb and u=-sc,

sPA + tPB + uPC = 0.

Would that work?

Aerith's Knight
02-17-2009, 09:51 PM
I don't think you can, seeing as we are calculating the volume of a parrallellapiped, you can't make those factors dependant on each other. If instead t or u you would take 3(or any real number), the volume would still be 0.

to put it simple, the statement sPA + tPB + uPB = 0 only says that whatever the sizes of the vectors are, if the angle of PA is 0, then the volume is zero.

SeeDRankLou
02-17-2009, 11:35 PM
Dammit, you're right.

The proof of this part of the theorem is actually one of the problems at the end of the section in the book. It's an odd question, so the answer is in the back of the book. The question goes:

"Suppose that points P, A, B and C all lie on the same plane. Show that vectors a=PA, b=PB and c=PC are coplanar vectors."

And the only thing the back of the book says for this answer is

"a dot (b x c) = 0"

I looked up the book online, and it gives the answer to all questions in the book in more detail than in the back of the book. And still, for this question all it says is

"a dot (b x c) = 0"

Is it really simple enough to just say that because the points all lie on the same plane, then PA dot (PB x PC) = 0, so a, b and c are all on one plane and thus coplanar?

Aerith's Knight
02-18-2009, 12:31 AM
Dammit, you're right.

The proof of this part of the theorem is actually one of the problems at the end of the section in the book. It's an odd question, so the answer is in the back of the book. The question goes:

"Suppose that points P, A, B and C all lie on the same plane. Show that vectors a=PA, b=PB and c=PC are coplanar vectors."

And the only thing the back of the book says for this answer is

"a dot (b x c) = 0"

I looked up the book online, and it gives the answer to all questions in the book in more detail than in the back of the book. And still, for this question all it says is

"a dot (b x c) = 0"

Is it really simple enough to just say that because the points all lie on the same plane, then PA dot (PB x PC) = 0, so a, b and c are all on one plane and thus coplanar?

I think it's supposed to be an insight question. They can only be in the same plane when their volume is zero, which brings it down to that equation.

I don't think you're supposed to derive it, but that it derives itself from the situation. Because I found the same lack of explination in my notes, and if it was an important theorem, I would've gotten it, seeing as my teacher loved to teach us abstract things we don't care about. :p

SeeDRankLou
02-18-2009, 01:07 AM
I dislike insight theorems, but you're probably right.