# linear dependence and coplanar vectors

I am confused about the coplanarity of vectors, and the relation of coplanarity to linear dependence.

If I have real vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$, with $\mathbf{w}$ a linear combination of $\mathbf{u}$ and $\mathbf{v}$, the three are then linearly dependent. (This much is clear.)

However, I'm not clear on the following:

• What does "coplanar" mean? (No, seriously; when I think of three points - vectors - I imagine them as determining a plane a priori; I do not think of two vectors as determining a plane. I imagine that my definition of coplanar is somewhat off here.)

• How does linear dependence/independence relate to coplanarity (however it is actually defined)?

Googling (I have Strang's book, which doesn't introduce the notion of linear independence) led me to believe that coplanarity $\iff$ linear dependence (for three vectors), but I do not understand this.

(Also: if you are so inclined, a nice linear algebra reference would be appreciated....)

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Your problem in understanding coplanarity is of distinguishing "points" from "vectors," and of distinguishing 2-dimensional linear subspaces from planes in general. Every three points determine a plane, as you say, but in general this plane doesn't pass through the origin. In linear algebra we single out the planes that pass through the origin, since they're subspaces of $\mathbb{R}^3$. Then we have a different definition of the plane determined by just two vectors, which is their linear span $\{au+bv: a, b \in \mathbb{R}\}$. This necessarily includes the origin. You can also think of the span of $u$ and $v$ in terms of points, as the plane determined by $u, v$, and the origin.
As you've said, if $w$ is in the span of $u$ and $v$, i.e. is a linear combination of them, then $u,v,w$ are linearly dependent. What this means is precisely that $u,v,w$ are coplanar, in the sense that $w$ is in the plane determined by $u,v,$ and the origin.
First, I ought to correct for the fact that I phrased the last comment badly. We can only define linear functions on vector spaces, so that there's no good way to speak about the function $v\mapsto 2v$ on a plane like $z=1$. This is because an important fact about that function is that $2(u+v)=2u+2v$; but for $u+v$ isn't even in $z=1$ when $u$ and $v$ are, so we don't get to use that fact! The point is that functions $f$ with $f(a+b)=f(a)+f(b), f(ca)=cf(a)$ are very nice, and they're only definable on subspaces. –  Kevin Carlson Sep 10 '12 at 4:06