How can two vectors with three components define a plane?

[nobahar]

Hello!
I was wondering, if I have two vectors each with three components, this is not sufficient to define all of R3, but it defines a subspace.
Does this subspace have to be a plane?
I think it does, but I am having difficulty visualising it or describing it. I think I can ascertain that if there are two vectors with three components, then once two of the "outputs" are given values, the third is also determined and is not free to vary. (This is complicated by collinear vectors, which I am trying not to consider.)
So if I have two vectors and three components, how do I determine if it is a plane that is formed?
This is frsutrating to no end!
Thanks in advance.

 

[HallsofIvy]

Even more- a (two- dimensional) subspace of a three dimensional space is a plane containing the origin. A one-dimensional subspace of a three dimensional space is a line through the origin.

I don't know what you mean by "if I have two vectors and three components". Do you mean you are given two vectors in a three dimensional vector space and asked to determine the subspace they span?

For example, if you are given <1, 0, 1> and <1, 2, 0> any vector in their span is of the form s<1, 0, 1>+ t<1, 2, 0>= <s+ t, 2s, t>. If we think of the vector as <x, y, z> that is saying x= s+ t, y= 2s, z= t. From y= 2s, s= (1/2)y. From z= t, of course, t= z. Then x= s+ t= (1/2)y+ z or 2x= y+ 2z, 2x- y- 2z= 0, the equations of a plane.

Of course, if the two vectors are not independent, that is, if one is a multiple of the other, then they both point along the same line and so the subspace is a line throught the origin.

For example, if the two vectors are <1, 0, 1> and <3, 0, 3>, any vector in their span is of the form s<1, 0, 1>+ t<3, 0, 3>= <s+ 3t, 0, s+ 3t>. Now we have x= s+ 3t, y= 0, z= s+ 3t. We have immediately x= z which "looks" like a line but y is always 0 whereas just "x= z" implies y can be anything. We would have to write that as x= t, y= 0, z= t for parameter t.

 

[nobahar]

Thanks for the response Halls, I hope the following is not to long winded. The bit I think is most important is in bold.

I don't know what you mean by "if I have two vectors and three components".

Sorry, I thought components were the little 'bits'/'parts' that make up a vector (e.g. a₁ in the vector <a₁, a₂, a₃>). I think they may be called elements (as suggested by wiki).

Do you mean you are given two vectors in a three dimensional vector space and asked to determine the subspace they span?.

Pretty much, that is what I was trying to establish, but in a generalised form. There isn't a specific question, as it's for conceptual understanding. I figured that the number of vectors in a set will determine the type of subspace: If I have two vectors, regardless of how many elements they have (e.g. 2: <a,b> or 3: <a,b,c>), the maximum subspace they can give is a plane; even if they have three elements, the two vectors can only form a plane. If I have three vectors with four elements (e.g. <a, b, c, d> and <e, f, g, h> and <k, m, n, p>), then I can only obtain a three dimensional subspace in four-dimensions. So the maximum extent of the subspace is dependent on the number of vectors in the set (as well as the number of ‘parts’).

Is this true? I have determined (with varying levels of success!) the span of a number of specific vectors, and I find I can’t represent all of R3 using two vectors.

If I could establish that ANY two linearly independent vectors each with two elements can span all of R2, then I know that a set containing three 2-tuple (containing two elements?) vectors will be linearly dependent, one will be a linear combination of the others. IF, I don’t believe it’s true, but IF I could represent the whole of R3 using ANY two linearly independent, 3-tuple vectors, then I know that I cannot have three linearly independent vectors; but, by the same reasoning, if I have three 3-tuple vectors, that are linearly independent, then the previous case cannot be true, and two linearly independent 3-tuple vectors will be insufficient to span all of R3. But this is all rests on the assumption that if a set of n linearly independent vectors span some space, then ALL the sets of n vectors that are linearly independent vectors also span the same space. (e.g. if two linearly independent 2-tuple vectors span R2, then ALL sets of two linearly independent 2-tuple vectors span R2).

Of course, if the two vectors are not independent, that is, if one is a multiple of the other, then they both point along the same line and so the subspace is a line through the origin.

For example, if the two vectors are <1, 0, 1> and <3, 0, 3>, any vector in their span is of the form s<1, 0, 1>+ t<3, 0, 3>= <s+ 3t, 0, s+ 3t>. Now we have x= s+ 3t, y= 0, z= s+ 3t. We have immediately x= z which "looks" like a line but y is always 0 whereas just "x= z" implies y can be anything. We would have to write that as x= t, y= 0, z= t for parameter t.

Just to clarify, this IS a line, but it could be mistaken for a plane, as y=anything, allows no restriction in this direction: it would be of the form x + 0y – z = 0, if it was a plane?

Many thanks, I hope this is not too long, it was difficult to make it more succint.