All possible planes, given two points

[The Subject]

Homework Statement

Find the equation of all planes containing the points P(2, -1, 1) and Q(1, 0, 0)

Homework Equations

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?

 

[SteamKing]

Homework Statement

Find the equation of all planes containing the points P(2, -1, 1) and Q(1, 0, 0)

Homework Equations

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?

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?

 

[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?

 

[SteamKing]

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

3 points, anything less contains no planes?

3 points uniquely determine 1 plane. If you have one or two points, how many different planes can pass thru these points?

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?

Try it and see.

 

[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

 

[Mark44]

Homework Statement

Find the equation of all planes containing the points P(2, -1, 1) and Q(1, 0, 0)

Homework Equations

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.

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

 

[The Subject]

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

Oh wow how did that happen! Thanks

 

[SteamKing]

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

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

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