Finding the basis of a subspace

[HizzleT]

Homework Statement

How do I find a basis for:
the subspace of R^3 consisting of all vectors x such that x ⋅ (1,2,3) = 0.

Homework Equations

I believe this is performed through setting x = x,y,z, setting each parameter sequentially equal to 1 while the others are set to o, putting into a matrix where a^i --> v(subscript i), and transforming into row reduced echelon form.

The Attempt at a Solution

I believe I have done this correctly, but please tell me if I have not.

x ⋅ (1,2,3) = 0.
(x,y,z) ⋅ (1,2,3) = 0.
x + 2y + 3z = o
x = -2y -3z
that is,
x = -2s - 3t
y = s
z = t

Giving: (-2s-3t,s,t)

Set s = 1, t = 0
(-2,1,0)
Set s = 0, t = 1
(-3,0,1)

-2 -3
1 0
0 1

ROW REDUCTION
yields:

1 0
0 1
0 0

Transforming v1 and v2 into the elementary columns. Thus,
{(-2,1,0),(-3,0,1)} form a basis for the subspace of R^3 consisting of all vectors x such that x ⋅ (1,2,3) = 0.
DimS = 2

_____

If it is of any assistance, the answer in the back of the textbook is:
{(-2,1,0),(-3,0,1)} with dimS = 2

 

[HizzleT]

I apologize for the redundancy setting y = s and z = t. That was clearly not needed.

 

[BiGyElLoWhAt]

Your subspace is going to be a plane in r3. That plane will consist of all your vectors x, which are orthogonal to 1,2,3. Your method looks alright, if you doubt the answer, find a way to check it. *cough* cross some vectors *uncough*

Also, if this were any more complex, using x for a vector equal to <x,y,z> could be confusing. I wouldn't recommend making a habit of it.

 

[HizzleT]

Your subspace is going to be a plane in r3. That plane will consist of all your vectors x, which are orthogonal to 1,2,3. Your method looks alright, if you doubt the answer, find a way to check it. *cough* cross some vectors *uncough*

Also, if this were any more complex, using x for a vector equal to <x,y,z> could be confusing. I wouldn't recommend making a habit of it.

Thank you for the reply!

Do you mean to say I would be better saying x = <x1,x2,x3>?
Perhaps I've misunderstood you completely.

 

[BiGyElLoWhAt]

Something along those lines, yea. or x = <e1,e2,e3>

Just setting x = <x,*,*> can be confusing, you know, since its just x on the left side, and it's just x in one of the components on the right side. However you want to differentiate between those is up to you; but I suggest you do, otherwise you'll get lost in your own work no doubt about it.

 

[HizzleT]

Something along those lines, yea. or x = <e1,e2,e3>

Just setting x = <x,*,*> can be confusing, you know, since its just x on the left side, and it's just x in one of the components on the right side. However you want to differentiate between those is up to you; but I suggest you do, otherwise you'll get lost in your own work no doubt about it.

Thank you!
It's been a while since I've done this.

 

[BiGyElLoWhAt]

No problem, glad to help =]