Equation of a plane in R4 from three points

[Easy_as_Pi]

Homework Statement

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)

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.

 

[Mark44]

Homework Statement

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)

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 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.

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.

 

[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.

 

[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 can't use the cross product. Would I just solve a very drawn out dot product set equal to 0?

Thanks for the help!

 

[Dick]

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 can't use the cross product. Would I just solve a very drawn out dot product set equal to 0?

Thanks for the help!

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.