Orthogonal Vectors in a 2D Lattice with minimum area I came across an interesting problem in my research (not a mathematician). Here it goes:
Suppose, there is a 2D lattice $\Lambda$ in the X-Y plane with basis vectors $\vec{a}$ and $\vec{b}$, which are not orthogonal to each other. I want to find a rectangle in this lattice, whose area is the minimum of all possible rectangles. So far, this is what I have:
Consider two vectors in the lattice:
$\vec{v_1} = m_1 \vec{a} + n_1 \vec{b}$
$\vec{v_2} = m_2 \vec{a} + n_2 \vec{b}$
If they are orthogonal, then $\vec{v}_1 \cdot \vec{v}_2 = m_1 m_2 \left| \vec{a} \right|^2 + n_1 n_2 \left| \vec{b} \right|^2 +  (m_1 n_2 + n_2 m_1) \vec{a} \cdot \vec{b} =0 $
The area subtended by the two vectors is the norm of the cross product, i.e. $\left| \vec{v}_1 \times \vec{v}_2 \right| = \left| \vec{a}_1 \times \vec{b}_2 \right| \left|m_1n_2 - m_2n_1 \right|$
Therefore, the area is minimized if $\left|m_1n_2 - m_2n_1 \right|$ is minimized and greater than zero. And the constraint is:
$m_1 m_2 \left| \vec{a} \right|^2 + n_1 n_2 \left| \vec{b} \right|^2 +  (m_1 n_2 + n_2 m_1) \vec{a} \cdot \vec{b} =0$
I am stuck after this! May be some kind of optimization with a constraint using Lagrange multipliers? But there are too many variables and $m_1, m_2, n_1, n_2 \in \mathbb{Z}$. Any suggestions will be greatly appreciated. Thank you for your time.

A second answer. This is getting pretty good.  Given a lattice with Gram matrix
$$
\left(
\begin{array}{rr}
A & B \\
B & C
\end{array}
\right),
$$
in order for there to be an orthogonal pair of lattice vectors, it is necessary that $A,B,C$ be linearly dependent over $\mathbb Q,$ that is, there are ordinary integers $i,j,k$ such that
$$ iA + j B + k C = 0.  $$
This is still not quite enough. It is also necessary that there be another ordinary integer $w,$ which is allowed to be zero or nonzero, with
$$  j^2 - 4 i k = w^2.  $$ 
This is a pretty strong restriction. However, as mentioned before, it is satisfied by 
$$
\left(
\begin{array}{rr}
1 & 0 \\
0 & \pi
\end{array}
\right),
$$
in which we are given orthogonal basis vectors but we cannot scale this to an integral lattice; here we can take $i=0, j=1,k=0.$
PROOF, the $i,j,k$ thing:
$$
\left(
\begin{array}{rr}
\alpha & \gamma \\
\beta & \delta
\end{array}
\right)
\left(
\begin{array}{rr}
A & B \\
B & C
\end{array}
\right)
\left(
\begin{array}{rr}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right) =
\left(
\begin{array}{cc}
D & 0 \\
0 & E
\end{array}
\right)
$$
The zero entries in the product give the equation
$$ A \alpha \beta + B (\alpha \delta + \beta \gamma) + C \gamma \delta.  $$
So need to have $iA + jB + kC = 0$ with some multiplier $n$ and
$$n i = \alpha \beta, \; \; nj = \alpha \delta + \beta \gamma, \; \; nk =  \gamma \delta.  $$
If you multiply it out, 
$$  n^2 j^2 - 4nink = n^2(j^2 - 4ik)= (\alpha \delta - \beta \gamma)^2 $$
General comment, given just two irrational numbers, we can find out whether their ratio is rational by getting giant decimal accuracy and writing the continued fraction for the ratio. To find integers $i,j,k$ that solve $iA + jB + kC = 0,$ we use PSLQ with which I have no experience.
