All possible planes, given two points

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1. Mar 14, 2016

The Subject

1. The problem statement, all variables and given/known data
Find the equation of all planes containing the points P(2, -1, 1) and Q(1, 0, 0)

2. Relevant equations

3. The attempt at a solution
I use PQ to get a vector, (-1, -1, 1). I some how need to use another vector so I can use the cross product to find the planes.

So i let another point A(x, y, z) and find a vector PA, (x+1, y+1, z-1)?
If i use the cross product i get (-z-y, z+x, 2-y+x) with a scalar equation of a(-z-y)+b(z+x)+c(2-y+x) = 0

I don't feel confident with my conclusion, is there a concept I'm misunderstanding about finding the planes?

2. Mar 14, 2016

SteamKing

Staff Emeritus
How many points determine one plane? What if you have fewer points than this number?

If you cross vector PQ with some other vector, how many planes does that determine?

3. Mar 14, 2016

The Subject

Thanks for the insightful question, I actually had to check my understanding.

3 points, anything less contains no planes?

PQ cross with another vector gives me 1 plane.

Since PQ cross another vector (x y z), this should give me any possible planes for some arbitrary vector, right?

4. Mar 14, 2016

SteamKing

Staff Emeritus
3 points uniquely determine 1 plane. If you have one or two points, how many different planes can pass thru these points?
Try it and see.

5. Mar 14, 2016

The Subject

Oh okay, it contains infinite planes!

So I let the arbitrary vector (1, 1, 1) the I get - 2a+2b+2c=for some d.

I just realize, how do I ensure this scalar equation is correct?

Since I have points P and Q that needs to go through, it makes sense to sub abc with points P and Q. It seems I'm getting different numbers for d

6. Mar 14, 2016

Staff: Mentor

Check your arithmetic. $PQ \ne <-1, -1, 1>$.

7. Mar 14, 2016

The Subject

Oh wow how did that happen! Thanks

8. Mar 14, 2016

SteamKing

Staff Emeritus
Maybe you should review the equation of the plane in 3 dimensions:

https://en.wikipedia.org/wiki/Plane_(geometry)