# Equation of a plane in R4 from three points

1. Jan 17, 2012

### Easy_as_Pi

1. The problem statement, all variables and given/known data
Find a parametric equation of each of the following planes:
d) The plane in R4 containing the points P: (1,1,-1,2), Q: (2,3,0,1), and R: (1,2,2,3)

3. The attempt at a solution
I found vector PQ <1,2,1,-1> and vector PR <0,1,3,1>
My next thought was to find a vector orthogonal to those two (cross product PQ x PR) and then use that as my normal vector, and choose a point. Then my answer would be: <x,y,z,m> + t*normal vector.

The problem is that I don't know how to take the cross product in 4 dimensions. I have read that it can only be taken in 3D and 7D ....but I'm rather lost as to how to find this plane's equation. Any help is thoroughly appreciated.

2. Jan 17, 2012

### Staff: Mentor

The cross product isn't applicable here, as you mention below. What you show as your answer would be a vector sum that determines a line in R4. To determine a plane, you need a point and two vectors that aren't parallel.

If u and v are vectors in R4 and OP is a vector from the origin to point P, and if u and v aren't parallel, then the vector equation of the plane containing the vectors is r(s, t) = OP + su + tv.

Geometrically, we need a vector from the origin to the plane (OP). Then any point in the plane is some linear combination of the two vectors u and v for some scalars s and t.

3. Jan 17, 2012

### Dick

<x,y,z,m> + t*normal vector wouldn't be the equation of a plane even it the vectors were three dimensional. It would be the equation of a line, since there is only one parameter t. The equation of a plane should have two parameters. That should make it pretty easy.

4. Jan 18, 2012

### Easy_as_Pi

So, to make sure I'm following you, Mark44, I would end up with a function r of s and t. OP= <1,1,-1,2>, u=PQ=<1,2,1,-1> , and v=PR=<0,1,3,1>. r(s, t) = OP + su + tv
r(s, t) = <1,1,-1,2> + s<1,2,1,-1> + t<0,1,3,1>

I know I need to have a point in the plane and two vectors from that point which are not parallel. In this case I chose point P, and had vectors PQ and PR, which are not parallel. I could have selected any of the three points and made 2 vectors from the selected point, one to each of the other points, and, as long as the vectors were not parallel, arrive at a correct parametric equation, right? Also, out of curiosity, how would I find the normal vector for this plane? I know it needs to be perpendicular to both PQ and PR, but that I cant use the cross product. Would I just solve a very drawn out dot product set equal to 0?

Thanks for the help!

5. Jan 18, 2012

### Dick

Sure, that's it. And you've got the right procedure to find normal vectors as well, solve n.PQ=0 and n.PR=0. You'll find there is a two dimensional subspace of normal vectors. In R3 it would be one dimensional, which is why there you can talk about 'the normal vector' direction.
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