Sets of vectors with a particular property Suppose I have two sets, $A$ and $B$, each of which contains a number of $n$-dimensional vectors. Each of the elements of these vectors are real and non-negative. I want these sets to have the property that $\mathbf{a}\cdot \mathbf{b} = 1$ whenever $\mathbf{a}\in A$ and $\mathbf{b}\in B$. Here, $\mathbf{a}\cdot \mathbf{b}$ is the usual dot product, i.e. $\mathbf{a}\cdot \mathbf{b} = \sum_i a_i b_i$.
There are a few easy ways to construct examples of this, but my question is about how to properly characterise the set of all possible solutions.
One example is where $A=\{(1/2,1/2)\}$ and $B = \{(1,1),(1/2,3/2)\}$, but this isn't a very interesting example because $A$ only has one element. A slightly more interesting example is $A=\{(1,0,0),(1,1,0)\}$, $B=\{(1,0,0),(1,0,1)\}$, but that relies on elements of the vectors being zero.
My first question is, is there an example where both $A$ and $B$ have more than one element, and all of the elements of each vector are positive?
My second question is, is there a straightforward way to construct the set of all possible sets $A$ and $B$, given $n$ (the dimensionality of the vector space) and the number of elements in $A$ and $B$? 
 A: What follows is in $\mathbb{R^3}$ (with a 2D interpretation).
Let us begin by a solution :
$A=\{(0.5,2,1),(0.5,1,2)\}$ with for example $B=\{(0.8,0.2,0.2),(1.4,0.1,0.1)\}, $
and more generally, keeping the same set $A$,  $B$ is the (infinite) set of all vectors of the form $(2-6t,t,t)$ for $0 < t < 1/3$.
Here is a general method that explains the crucial aspect of the choice of $A$ (that I have used to find the particular case presented upwards).

Being given the set $A=\{(a,b,c),(a'b'c')\}$ in the positive orthant $\Omega:=\mathbb{R_+^3}$, find under which conditions on $a,b,c,a',b',c'$ does the following system
$$\tag{1}\begin{cases}ax+by+cz&=&1\\a'x+b'y+c'z&=&1\\\end{cases}$$
possess solutions $(x,y,z) \in \Omega$.

(1) can be written under the following equivalent form:
$$\tag{2} \text{Find} \ \ x,y,z >0 \ \ \text{such that} \ \ x\binom{a}{a'}+y\binom{b}{b'}+z\binom{c}{c'}=\binom{1}{1}.$$
A necessary and sufficient condition for (2) to hold is as follows:

Let $P\binom{a}{a'}, Q\binom{b}{b'}, R\binom{c}{c'}$. Let $\Delta$ be the bissector line of the positive quadrant in $\mathbb{R^2}$, and $\Delta_-$, resp. $\Delta_+$ the open regions delimited by $\Delta$ (polar angles $<\pi/4$, resp. $>\pi/4$ ; see figure).
There is a solution to (1) iff triangle $PQR$ is a non-flat triangle crossed by $\Delta$.

(i.e., not all $P,Q,R$ in $\Delta_+$ and not all in $\Delta_-).$

Proof:
If $P\binom{a}{a'},Q\binom{b}{b'},R\binom{c}{c'}$ are all in $\Delta_+$ or all in $\Delta_-$,  any linear combination of them with positive coefficients will still be in the same region, leading to an impossibility to have the RHS of (2) as their sum.
In a reverse manner (see figure), let us assume that e.g., $P \in \Delta_-$ and $Q, R \in \Delta_+$, line $\Delta$ crosses triangle $PQR$ in two points $S$ and $T$. Let $I$ be the midpoint of $[ST]$, with coordinates $(\alpha,\alpha)$ for a certain $\alpha$. $I$ has what is called barycentrical coordinates $(x,y,z)=(x_0,y_0,z_0)$ with $x_0,y_0,z_0>0$ and $x_0+y_0+z_0=1$ such that:
$$x_0\binom{a}{a'}+y_0\binom{b}{b'}+z_0\binom{c}{c'}=\binom{\alpha}{\alpha}.$$
It suffices then to take $x=x_0/\alpha, \ y=y_0/\alpha, z=z_0/\alpha $ in order that (2) is fullfilled.
Now, why are there an infinite number of solutions for (2) ? Two different reasons can be given

*

*(i) Any point $I$ between $S$ and $T$ (not necessarily their midpoint) could have been taken, and thus could have generate a solution.


*(ii) Because (1) (unless $(a,b,c)$ and $(a',b',c')$ are proportional) is the intersection of 2 planes, i.e., represents a straight line passing by $(x_0/\alpha,y_0/\alpha,z_0/\alpha)$, i.e, of the form $(x,y,z)=(x_0/\alpha+pt,y_0/\alpha+qt,z_0/\alpha+rt), \ t \in \mathbb{R}$. Simple continuity arguments show that such a line has an infinite number of points in $\Omega$.
Remark : $(p,q,r)$ introduced above can clearly be taken as the cross product of $(a,b,c)$ and $(a',b',c')$.
Appendix: If $A$ has three elements, one can evidently extend this reasoning. But the  additional constraint $a''x+b''y+c''z=1$ in relationship (1) gives (if the equivalent condition of being in a certain triangle is fulfilled) a unique solution $(x,y,z) \in \Omega$ in the general case. Thus $A$ has three elements but $B$ has only one. For $n>3$, there are no solutions in the general case ((1) turns into an overdetermined system).
In conclusion : one can say in a certain sense that $A$ is represented by triangle $PQR$ and $B$ is represented by line segment $[ST]$.
A: In terms of linear algebra, and without looking at the signs of the coordinates of the involved vectors, you can say the following: Each $B$-vector defines a hyperplane $H_i\!:\ {\bf b}_i\cdot{\bf x}=1$ in ${\mathbb R}^n$. The intersection $S:=\bigcap_{i=1}^r H_i$ of these hyperplanes is an affine subspace of ${\mathbb R}^n$ of dimension $n-r$ or greater. The configuration $(A,B)$ is admissible if all vectors ${\bf a}_k\in A$ are lying in $S$.
The constraint which is difficult to handle is that all involved coordinates should be $\geq0$.
A: To get towards a classification, let us first make some simplifying observations.  Suppose you have a good set $(A,B)$.  Given any vector $(a_1, a_2, ..., a_n) \in A$ whose coordinates are all positive, define $A'$ by rescaling $A$ by a factor $a_i^{-1}$ in the $i$th direction.  To preserve inner products, let $B'$ be obatined by scaling $B$ by a factor $a_i$ in the $i$th direction.  These scaling operations preserve positivity and inner products between vectors in $A$ and $B$, so the scaled pair $(A', B')$ is still good, and now we have $(1, 1, ..., 1) \in A'$.
In particular, this means that every vector in $B'$ must satisfy $\sum_i b_i = 1$ as well as $b_i > 0$ for all $i$.  This means $B'$ lies in the interior of the convex of the set of basis vectors, $\{ e_i : 1 \le i \le n \}$ (where $e_i$ is $1$ in the $i$th coordinate and $0$ everywhere else).  Let us denote this interior by $F^0$.
Now to generate a good pair, fix any subset $B' \subseteq F^0$.  As has been mentioned in the previous answers, every point $\vec{b} \in B'$ defines a hyperplane $H_{\vec{b}} = \{\vec{x} \in \mathbb{R}^n : \vec{x} \cdot \vec{b} = 1 \}$.  Note that, since $\sum_i b_i = 1$, we must have $\vec{1} = (1, 1, ..., 1) \in H_{\vec{b}}$.  Hence, setting $\vec{y} = \vec{x} - \vec{1}$, we can translate by $\vec{1}$ and work with the subspace $V_{\vec{b}} = \{ \vec{y} \in \mathbb{R}^n : \vec{y} \cdot \vec{b} = 0 \} = \{ \vec{b} \}^{\perp}$.
We want this condition to hold for every vector in $B'$, so we consider the intersection of all of these subspaces, $V = \cap_{\vec{b} \in B'} V_{\vec{b}} = (B')^{\perp}$, the orthogonal space to $B'$.  Note that since $V$ is a subspace, it is always non-empty (since $\vec{0} \in V$).  Moreover, its dimension is equal to $n - \dim \operatorname{span}(B')$.  Hence if $B'$ is full-dimensional, then $V$ will only consist of $\vec{0}$; otherwise $V$ will have infinitely many points.
Finally, we must take care of the positivity requirement.  We want all vectors $\vec{x} \in A'$ to have positive coordinates.  However, recall that $V$ was obtained by translating by $\vec{1}$.  Thus we are interested in the set $S = \{ \vec{y} \in V : \forall i, y_i > -1 \}$.  Note that this contains an open ball around the origin, and so is non-empty (and in fact has the same dimension as $V$).  Now we can get the set $A'$ by taking any subset of $S$ and translating back by $\vec{1}$; that is, any subset $A' \subseteq S'$ will do, where $S' = \{ \vec{y} + \vec{1} : \vec{y} \in S \}$.
This gives us a good pair $(A', B')$ with $\vec{1} \in A'$.  For full generality, we can rescale the coordinates; given any positive reals $\lambda_i$, $1 \le i \le n$, define the set $A$ by rescaling $A'$ by a factor of $\lambda_i$ in the $i$th direction, and the set $B$ by rescaling $B'$ by a factor $\lambda_i^{-1}$ in the $i$th direction.  This gives a fully general pair $(A,B)$ satisfying the requirements.
TLDR: Let $B'$ be a set of vectors with positive coordinates that sum up to $1$; let $S$ be the vectors in the orthogonal complement of the span of $B'$ whose coordinates are all bigger than $1$, and get $A'$ by adding $\vec{1}$ to every vector in some subset of $S$.  Now, for some positive $\lambda_i \in \mathbb{R}$, get the pair $(A,B)$ by scaling $A'$ by $\lambda_i$ and $B'$ by $\lambda_i^{-1}$ in the $i$th direction for each $i$.
A: Working in $n$ dimensions, let's suppose $A$ contains some positive number of vectors. Choose a vector in $A$ and call this vector $a_1$. The set of all vectors $b$ such that $a_1 \cdot b = 1$ is the hyperplane of $n - 1$ dimensions perpendicular to $a_1$ passing through the point at displacement $\frac{a_1}{\|a_1\|^2}$ from the origin on the same side of the origin as $a_1$. So $B$ must be a subset of that hyperplane. 
If we suppose there is another element of $A$, call it $a_2$, where $a_2\neq a_1$, it follows that $B$ must be a subset of the hyperplane of $n - 1$ dimensions perpendicular to $a_2$ passing through the point at displacement $\frac{a_2}{\|a_2\|^2}$ from the origin on the same side of the origin as $a_2$. In fact, $B$ must be a subset of the intersection of the two hyperplanes already described. If $a_2$ is a scalar multiple of $a_1$, the intersection is empty; otherwise it is a hyperplane of dimension $n-2$. 
If there is a third vector in $A$, it could be linearly independent of $a_1$ and $a_2$, in which case $B$ is a subset of a hyperplane of dimension $n-3$ that is the intersection of the three hyperplanes determined by the three vectors; or the third vector could be a linear combination of the form $ra_1 + (1-r)a_2$ where $r$ is any real number, in which case it does not make any new restriction on $B$. That is, once we have two vectors in $A$ an entire one-dimensional line within the vector space (or any subset of that line) can be included in $A$ "free of charge."
Continuing in this fashion, if we identify $k$ linearly independent vectors in $A$ then these vectors determine an $(n-k)$-dimensional hyperplane of which $B$ must be a subset, as well as a $(k-1)$-dimensional hyperplane of which $A$ must be a subset. 
In three dimensions, if we want $A$ and $B$ each to contain more than one vector, each of the sets $A$ and $B$ must be a subset of a line in the vector space. Setting aside (until later) the requirement that coordinates be positive, an example is when $A=\{(x,0,1) \mid x\in\mathbb R\}$ and 
$B=\{(0,y,1) \mid y\in\mathbb R\}$. We can also have $A=\{(x,0,a) \mid x\in\mathbb R\}$ and $B=\{(0,y,\frac1a) \mid y\in\mathbb R\}$ where $a$ is a non-zero real constant, and we can apply a rotation about the origin simultaneously to these two lines and get two new lines that are suitable to contain $A$ and $B$. Note that any rotation that takes $(0,0,1)$ to a point with only positive coordinates will cause the two lines also to contain segments with only positive coordinates.
For example, let $A$ be any subset of the line through $(\frac23,\frac53,2)$ and $(\frac32,\frac54,\frac34)$ and let $B$ be any subset of the line through $(\frac38,\frac14,\frac16)$ and $(\frac3{10},\frac25,\frac1{15})$.
In case you're wondering how I found those particular points, I guessed the line $(2,1,0)+t(-2,1,3)$ (in parametric form) would contain some points that avoided zeros and duplicated coordinates, let that be the line containing $A$, solved for the line containing $B$, which is $(0,1,-\frac13)+u(1,-2,\frac43)$, and picked two suitable points on each line.
