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?

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?