What are the properties to prove a plane is a subspace of R^3?

In summary, the plane ax+by+cz=0 is not a subspace of R^3 because there is a hole in it at the origin.
  • #1
Nothing000
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0
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 [tex]ax + by + cz = 0[/tex] is a subspace of [tex]R^3[/tex].
 
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  • #2
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 [tex]R^3[/tex].
 
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  • #3
So I guess I disproved it. But is that correct?
 
  • #4
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
Tell me if I am defining the set correctly. I am going to call it W.
[tex]W = \{[x,y,z] | x,y,z \in R3\}[/tex]





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.
 
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  • #6
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.
 
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  • #7
So would I just prove it to be true by saing something like:
let [tex] \vec{u} = [a_{1},b_{1},c_{1}] [/tex] and [tex] \vec{v} = [a_{2},b_{2},c_{2}] [/tex].

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

Therefore [tex] \vec{u} + \vec{v} [/tex] is in [tex]W[/tex] and is closed by vector addition.

And then do something similar for scaler multiplication?
 
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  • #8
Nothing000 said:
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.

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.
 

1. What is a subspace?

A subspace is a subset of a vector space that satisfies all the properties of a vector space, including closure under addition and scalar multiplication.

2. How can you prove that a plane is a subspace of R3?

To prove that a plane is a subspace of R3, we need to show that it satisfies the three properties of a subspace: it contains the zero vector, is closed under addition, and is closed under scalar multiplication.

3. What is the zero vector in R3?

The zero vector in R3 is (0, 0, 0). It is the vector with all components equal to zero and is the identity element for vector addition.

4. Why is it important for a subspace to be closed under addition and scalar multiplication?

Closure under addition and scalar multiplication ensures that the operations of adding vectors and multiplying a vector by a scalar will always result in a vector that is also in the subspace. This is necessary for the subspace to maintain its structure and properties.

5. Can a plane in R3 be a subspace if it does not pass through the origin?

No, a plane in R3 must pass through the origin in order to be a subspace. This is because the zero vector must be contained in the subspace, and it lies at the origin.

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