Subspace / basis problem

[MarcL]

Homework Statement

Let S, a subspace of ℝ³ be the set of vectors orthogonal to vector (1,2,3)
a)describe Set S
b) find a basis for Set S

2. Relevant Equations

That a basis has to be linearly independent and span R^3

The Attempt at a Solution

[/B]
I would do this:
I know that vector (1,2,3) is the cross product of 2 vectors v1xv2
so I could put it in a matrix (where v1=a,b,c and v2=d,e,f)

a b c
d e f

But I am lost as to describe set S... Wouldn't I need to row reduce to see which variable is free, and then I could say whether or not it is a line or a plane ( well the dimension)

 

[Mark44]

Homework Statement

Let S, a subspace of ℝ³ be the set of vectors orthogonal to vector (1,2,3)
a)describe Set S
b) find a basis for Set S

2. Relevant Equations

That a basis has to be linearly independent and span R^3

The Attempt at a Solution

[/B]
I would do this:
I know that vector (1,2,3) is the cross product of 2 vectors v1xv2
so I could put it in a matrix (where v1=a,b,c and v2=d,e,f)

a b c
d e f

What's another way to show that an arbitrary vector <x, y, z> is orthogonal to a given vector <1, 2, 3>?

But I am lost as to describe set S... Wouldn't I need to row reduce to see which variable is free, and then I could say whether or not it is a line or a plane ( well the dimension)

If you take constant multiples of the given vector <1, 2, 3>, what sort of geometric object do you get? What's the dimension of this subspace?

 

[BvU]

What you attempt is to find a,b,c,d,e,f such that (a,b,c) x (d,e,f) = (1,2,3)
Might seem reasonable, but it's a bit tedious: you want to find six variables from three equations. And you don't really need the lengths (that comes in part b).

Orthogonal to a vector $\vec v$ is anything that has $\vec a \cdot \vec v = 0$ with $|\vec a| > 0\;$. That's only one equation with three unknowns.
Pick a vector v1 that satisfies that equation.
Pick a different one and call that v2.
I think then you have a basis already according to this link (exercise doesn't ask for orthogonal or orthonormal basis !)

But if you do want them orthonormal you can do v1 x (1,2,3) to get a v2 that is perpendicular to both.
Then normalize v1 and v2.

Re describing S: Why do you think there is the possibility that S is a line ?

 

[MarcL]

Ah I see what you're getting at, once I've found two indepedent solutions i can just put it in a matrix and solve it. I see what you mean by not spanning R^3 too. it will span R^2 if I understand correctly.

What's another way to show that an arbitrary vector <x, y, z> is orthogonal to a given vector <1, 2, 3>?

I could also find the dot product, apart from that I am not too sure what I could use to prove the orthogonality. Geometrically speaking it is a plane, so the vector would be the normal.

As for the post above me, I meant in any case ( wasn't solved and I wasn't entirely sure). However, I expected it to be a plane.

 

[Mark44]

The dot product is the primary tool to show orthogonality.

Yes, set S is a plane.