Describing span(x) geometrically

[Cottontails]

Homework Statement

There is a vector space with real entries, in ℝ³ with the subset X = <br /> \begin{pmatrix}<br /> 2\\<br /> -1\\<br /> -3<br /> \end{pmatrix}\\<br /> ,<br /> \begin{pmatrix}<br /> 4\\<br /> 0\\<br /> 1<br /> \end{pmatrix}\\<br /> ,<br /> \begin{pmatrix}<br /> 0\\<br /> 2\\<br /> 7<br /> \end{pmatrix}<br /> <br />
and you have to describe span(x) geometrically.

Homework Equations

In answering this question, I found the span of the subset through a<br /> \begin{pmatrix}<br /> 2\\<br /> -1\\<br /> -3<br /> \end{pmatrix}\\<br /> + b<br /> \begin{pmatrix}<br /> 4\\<br /> 0\\<br /> 1<br /> \end{pmatrix}\\<br /> + c\begin{pmatrix}<br /> 0\\<br /> 2\\<br /> 7<br /> \end{pmatrix}=<br /> \begin{pmatrix}<br /> x\\<br /> y\\<br /> z<br /> \end{pmatrix}<br />

The Attempt at a Solution

This formed the matrix:
<br /> \begin{pmatrix}<br /> 2 &amp; 4 &amp; 0 &amp; | &amp; x\\<br /> -1 &amp; 0 &amp; 2 &amp; | &amp; y\\<br /> -3 &amp; 1 &amp; 7 &amp; | &amp; z<br /> \end{pmatrix}<br />
Using row operations, I then made the matrix into reduced row echelon form and it was non-trivial with the final result being:
<br /> \begin{pmatrix}<br /> 1 &amp; 2 &amp; 0 &amp; | &amp; x/2\\<br /> 0 &amp; 1 &amp; 1 &amp; | &amp; x/4+y/2\\<br /> 0 &amp; 0 &amp; 0 &amp; | &amp; -x/4-7y/2+z<br /> \end{pmatrix}<br />
So, we can then interpret the span geometrically as the plane in ℝ³ with the equation -1/4x - 7y/2 + z=0
Is this right?

 

[vela]

You made a mistake somewhere. Those three vectors are linearly independent, so you shouldn't end up with a row of zeros. But if that were the correct matrix, then yes, the span would be the plane you said.

 

[Cottontails]

Sorry, I didn't realize but I put the first entry as 3 instead of -3 and now I have corrected it. Is that correct now?

 

[vela]

Yes, that matches what I get.