# Prove plane is subspace of R3

1. Sep 19, 2006

### Nothing000

I have a question that states:
Let a, b, and c be scalers such that abc is not equal to 0.
Prove that the plane $$ax + by + cz = 0$$ is a subspace of $$R^3$$.

Last edited: Sep 19, 2006
2. Sep 19, 2006

### Nothing000

Here is what I am thinking:
The plane would intersect the origin when (a,b,c) = (0,0,0).
But since abc does not equal zero, niether a, b, or c can equal zero. So there must be a hole in the plane at the origin.
Since the plane does not pass through the origin the zero vector does not lie within the plane, and therefore the plane is NOT a subset of $$R^3$$.

Last edited: Sep 19, 2006
3. Sep 19, 2006

### Nothing000

So I guess I disproved it. But is that correct?

4. Sep 19, 2006

### Data

No. If it had a hole in it, it wouldn't be a plane. (0,0,0) is on the plane since 0a+0b+0c=0.

So can you tell me what other things do you need to test in order to show that something is a subspace?

5. Sep 19, 2006

### Nothing000

Tell me if I am defining the set correctly. I am going to call it W.
$$W = \{[x,y,z] | x,y,z \in R3\}$$

1. So for each vector u and vin the set W, the sum u+v must be in W.

2. For each vector u in W and each scaler c, the vector cu is in W.

Last edited: Sep 19, 2006
6. Sep 19, 2006

### Nothing000

It fits these criteria, because any vector that lies in the plane will still lie in the plane after any vetor addition with another vecor in the plane, and after multiplication with a scaler.

So it is closed under Vector Addition, and Scaler Multiplication.

Last edited: Sep 19, 2006
7. Sep 19, 2006

### Nothing000

So would I just prove it to be true by saing something like:
let $$\vec{u} = [a_{1},b_{1},c_{1}]$$ and $$\vec{v} = [a_{2},b_{2},c_{2}]$$.

$$\vec{u}+\vec{v} = [(a_{1} + a_{2}) ,(b_{1} + b_{2}) ,(c_{1}+ c_{2})]$$

Therefore $$\vec{u} + \vec{v}$$ is in $$W$$ and is closed by vector addition.

And then do something similiar for scaler multiplication?

Last edited: Sep 19, 2006
8. Sep 20, 2006

### HallsofIvy

Staff Emeritus
The origin is NOT at (a,b,c)= (0,0,0), it is at (x,y,z)= (0,0,0). a, b, c are fixed constants for this problem.
To prove that {(x,y,z)| ax+ by+ cz= 0} is a subspace, show that it satisfies the properties of a subspace.