# 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....)

• Vectors are collinear in the sense that they can be "ported" onto the same line (plainly put: are parallel). And vectors are coplanar in the sense that they can be "ported"onto the same plane. Commented Mar 27, 2021 at 13:13

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.

• This brings up a number of questions in my mind. One, if it's not too broad, is "why are we so concerned with subspaces?" Commented Sep 10, 2012 at 3:42
• Could you offer a quantitative example? Commented Sep 10, 2012 at 3:48
• 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. Commented Sep 10, 2012 at 4:06

First of all you should know that two vectors(Forces) are always coplanar. Now, when it comes to the coplanarity of three vectors the linear independence and dependence comes into picture. Since two vectors are always colplanar, the third will be coplanar or not depends on the linear relationship the third vector has with the first two. FOR EXAMPLE: Consider two vectors 'a' and 'b' and some third vector be 'r'.

1. Case 1: If the third vector is linearly dependent on the other two i.e. r= ax+by (where x and y are scalars and belongs to real numbers.), then r is coplanar with the first two.Or we can say that the three vectors are linearly dependent aka coplanar.

2. Case 2: If the third vector is independent of the first two vectors i.e. r!=ax+by (where x and y are scalars and belongs to real numbers. '!=" is read as not equal to), then that suggests that the third vector is not in the plane of first two and thus the system formed by the three vectors is non-coplanar.