# Intersection of two planes in R4

1. Dec 20, 2012

### Gauss M.D.

1. The problem statement, all variables and given/known data

I have two planes in R4, namely {[2, 0, 0, 1], [1, 1, 2, 0]} and {[-2, 0, 0, 1], [0, 1, -1, 0]}.

2. Relevant equations

3. The attempt at a solution

Tried to row eliminate, didn't work. Tried figuring out a normal equation, but clearly that won't work in R4. Don't know what to do really.

2. Dec 20, 2012

### HallsofIvy

Staff Emeritus
Two vectors do not define a plane if R4. I suspect you mean the subspaces that are spanned by the two vectors, planes that include the origin. It looks to me like the only point of intersection is the origin.

Last edited: Dec 20, 2012
3. Dec 20, 2012

### Staff: Mentor

What does your notation mean? All points a*(2, 0, 0, 1)+b*(1, 1, 2, 0), for example?

You can set this equal to a corresponding expression for the second plane, and find all points which can satisfy both conditions at the same time (there are not many points in the intersection...)

4. Dec 20, 2012

### Gauss M.D.

"What does your notation mean? All points a*(2, 0, 0, 1)+b*(1, 1, 2, 0), for example?

You can set this equal to a corresponding expression for the second plane, and find all points which can satisfy both conditions at the same time (there are not many points in the intersection...)"

Yes, my notation meant the span of the vectors in brackets.

Your method is what I would do with two planes in R3. But in R3, you have normal equations for planes. In R4, you don't. So how do you set the corresponding plane equations equal to each other?

5. Dec 20, 2012

### Staff: Mentor

You don't need normal equations for those planes. But if you like them, you can use them in R^4, too - you get two equations per plane, and the intersection has to satisfy them all.