Densities of planes I have  misunderstanding about   what does  mean  density of planes. For   example I was trying to figure out  definition of co-vector, and  while browsing   in internet  definition of it, I found following   pdf file: http://profstewart.org/pm2/ln11.pdf.
While there is  given some terminology, like  comparing vectors and covectors, the magnitude of a vector is given by its length,
while the magnitude of a covector is given by the density of its planes. The direction of
a vector is along its length (intrinsically oriented), while the direction of a covector is
normal to its planes (extrinsically oriented) in the sense that $n\cdot v = 0$ for any vector $v$ lying in the plane of the covector $n$ (sorry using Latex  I can't  show superscripts and  subscripts).
My actual question is  what is meant by  density of planes, of course more or less I did not understand definition of  co-vector, but maybe after clarify  first problem, this will be more  clear. Thanks  very much.
 A: For covectors:
A covector is a vector in the dual space. The important thing here is that vector spaces always have a homomorphism to and from the addition on the ground field. If you construct all of those homomorphisms, you can construct an addition and scalar multiplication essentially by adding the lists of values of the homomorphism pointwise. (try this on a vector space over a finite field)
Now, what isomorphisms of vector spaces do, is they perform changes of basis. But what is a change of basis? On one hand, vectors can be expressed as linear combinations of basis vectors. But the linear combination you need can be expressed as a series of coordinates. So you have a meter stick for each basis vector, saying how much it goes along each vector. This is what the dot product does, for real vector spaces.
But consider what a change of basis like rotation does to your axes, if you think of the axes on your eyes. You can't tell whether a vector was rotated in the plane spanned by the axes, or the axes rotated the other way and the vector stayed put. So the meter sticks are apparently a different kind of object from the vectors, in that they transform differently (see contravariance and covariance of vectors). Furthermore, you can apply the meter stick to the vector to get a scalar, which is not what you're usually looking for in an algebra over a vector space. But if you think of the scalar result as an element of the ground field as a vector space, and taking a product as being a linear map from the direct product of the original vector space and its dual space, you get the concept of a bilinear map.
A: Given any vector space $V$, its elements $x$ are called vectors.  Ordinary $3$-space $V:={\mathbb R}^3$ is an example in question. We think of the vectors $x\in{\mathbb R}^3$  as triples $(x_1,x_2,x_3)$ of real numbers; but at the same time we draw them as arrows attached at the origin. We add them as triples, but also in a geometric way using the "parallelogram of forces" idea. 
A linear map $\phi:\ V\to{\mathbb R}$ is called a functional or a covector. In terms of coordinates such a $\phi$ appears as $\phi(x)=a_1x_1+a_2x_2+a_3x_3$ with certain coefficients $a_k$. Now, while it is intuitively clear how to draw vectors and interpret them in a geometric  (or physical) sense, it is definitely not so with covectors. As such covectors turn up in all sorts of situations (e.g., as forces in mechanics, or as heat flow vectors) we are nevertheless interested in some sort of graphical representation in our figures.
One way out of this dilemma is using the scalar product, when available: Each functional $\phi$ can be represented by a vector $a$ (that can be drawn in the same figure) in the following way:
$$\phi(x)=a\cdot x\qquad(x\in V)\ .\tag{1}$$
This vector $a$ is determined uniquely by $\phi$; furthermore length and direction of the vector $a$ can be interpreted  in terms of properties of the function $\phi$.
An example: When $f: V\to{\mathbb R}$ is a differentiable function and $p$ is a given point then
$$f(p+X)-f(p)=df(p)(X) + o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{2}$$
Here $df(p)$, the differential of $f$ at $p$, is a linear functional on the tangent space $T_p$; in other words: it is a covector, or an element of the cotangent space $T_p^*$. The vector in $T_p$ which represents $df(p)$ in the sense of $(1)$  is the gradient of $f$ at $p$, denoted by $\nabla f(p)$. So instead of $(2)$ we can write
$$f(p+X)-f(p)=\nabla f(p)\cdot X + o\bigl(|X|\bigr)\qquad(X\to0)\ .$$
Now the "density of planes", in my view a somewhat outmoded notion: A given functional $\phi$, being a function $\phi:\ \ V\to{\mathbb R}$,  possesses  level planes $\phi(x)\equiv c$ for given values  $c\in{\mathbb R}$. These planes fill our drawing space as the  leaves of paper fill a book, and they are orthogonal to the vector $a$ associated to $\phi$ via $(1)$. Assume now that we choose a step size $\delta$ (maybe related to some physical unity), and that we draw only the planes $\phi(x)\equiv k\>\delta$ for $k\in{\mathbb Z}$. Then we get a discrete set $\Sigma$ of parallel planes with a  distance $d={\delta\over|a|}$ between neighbouring planes. It follows that this distance decreases, or $\Sigma$ becomes more dense, when the norm $|a|$ of $\phi$ increases. In the same vein: The density of these planes is the number of such planes per unit length in the orthogonal direction. The latter number is ${|a|\over\delta}$, and is directly proportional to $|a|$.
A: I guess I got a correct interpretation:
A covector could be also defined as a family $(H_t)_{t\in\Bbb R}$ of parallel planes (within the 3d space), such that $0\in H_0$, and the indexing is 'linear' in the sense that $H_t=H_0+t\cdot v\ $ for some $v$ vector $\notin H_0$, for all $t\in\Bbb R$. So, the density of the planes can be also thought as the speed of $H_t$ if $t$ is considered as time, and, for example for a 'covector' $(H_t)$ we can define its double: $(H_{2t})_t$, it contains 'twice' as much planes e.g. for $t\in (0,1)$ than $(H_t)_t$, that is its density is bigger.
If we require of the $v\notin H_0$ that it be orthogonal to $H_0$ (and hence to all $H_t$), then this is unique. In other words, if $\underline\sigma=(H_t)$ is given, then there is a uniqe $n\in H_1$ such that $n\perp H_0$, and thus, this covector corresponds to $n$ (and $(H_{2t})_t$ to $2n$, etc), and the dot product of a covector and a vector can be rewritten
$$\underline\sigma\cdot v = n\cdot v\,.$$
