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For instance, can the 3 vectors $\vec a=[1, \ 0, \ 1]^T, \vec b=[2, \ 7, \ -2]^T, \vec c=[3, \ 1,\ 5]^T$ lie on the same plane in $\mathbb R^3$?

My understanding is that the span of 2 linearly independent vectors in $\mathbb R^3$ such as $\vec a$ and $\vec b$ is a plane that passes through $\vec 0$. Yet $\vec c$ is not in the span of $\vec a$ and $\vec b$. What's happening?

I can see $a^Tb=0$. Is that relevant? According to Wikipedia, "Note that v and w can be perpendicular, but cannot be parallel." I take to understand that $\vec c$ would be the $\vec r$ and $\vec 0$ would be the $\vec r_0$ in the equation that was given $\vec r-\vec r_0=s\vec v+t\vec w$

(This is an edit to add a new idea): Do the vectors span $\mathbb R^3$ because they are all linearly independent? Then it that case, what does this mean? What's the plane that is spanned by 3 linearly independent vectors?

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    $\begingroup$ I assume you mean $[1,0,1]$, not $[101]$, and $[2,7,-2]$, not $[27-2]=[25]$. $\endgroup$ – mr_e_man Oct 6 '18 at 23:57
  • $\begingroup$ @mr_e_man Yes. I thought spacing would helping. Anyway, I put commas. Thanks. $\endgroup$ – user198044 Oct 6 '18 at 23:57
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    $\begingroup$ I think you should look at the distinction between linear subspaces and affine subspaces. A plane is determined by 3 points. The 3 points $\vec0,\vec a,\vec b$ determine a plane, which is a linear subspace. The 3 points $\vec a,\vec b,\vec c$ determine a plane, which is an affine subspace. The 4 points $\vec0,\vec a,\vec b,\vec c$ determine the entire 3D space. $\endgroup$ – mr_e_man Oct 7 '18 at 0:11
  • $\begingroup$ @mr_e_man Can you help me here? How is a plane a particular solution plus the span of linearly independent vectors? $\endgroup$ – user198044 Oct 7 '18 at 0:12
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    $\begingroup$ An affine subspace is a linear subspace displaced from the origin. The "particular solution" is that displacement. $\endgroup$ – mr_e_man Oct 7 '18 at 0:25
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There is nothing wrong with a vector in $\mathbb R^3$ not being in the span of two linearly independent vectors, since $\mathbb R^3$ is $3$ dimensional, whereas the span of the two vectors will be only $2$ dimensional, i.e. a plane.

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  • $\begingroup$ Thank you! I just figured out that the vectors span $\mathbb R^3$. So what is the "plane" formed by 3 vectors than span $\mathbb R^3$? $\endgroup$ – user198044 Oct 7 '18 at 0:06
  • $\begingroup$ $\mathbb R^3$ itself; though it is not usually called a "plane". $\endgroup$ – Chris Custer Oct 7 '18 at 0:08
  • $\begingroup$ Chris Custer, can you help me here? How is a plane a particular solution plus the span of linearly independent vectors? $\endgroup$ – user198044 Oct 7 '18 at 0:12
  • $\begingroup$ Sure. This can always be done. The basic idea is that any plane can be considered a translation of a plane through the origin. $\endgroup$ – Chris Custer Oct 7 '18 at 0:30
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Chris Custer Oct 7 '18 at 1:55

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