What are the four-vector analogues for the geometrical scalar and vector product of two three-vectors? In Euclidean three-space, we have the usual geometric definitions for the scalar and vector product of two vectors, give by
$\vec A \cdot \vec {B} =|A||B|\cos\theta$ and $A\times B= |A||B|\sin\theta$. Perhaps there's an equivalent for Minkowski-space where $\sin$ and $\cos$ are replaced by $\cosh$ and $\sinh$?
 A: To higher dimensional analog of the scalar product is rather straight forward. In any number of dimensions $A \cdots B = A_1 B_1 + A_2 B_2 + \ldots + A_n B_n$. This is indeed always equal to $A \cdots B = |A||B| \cos \theta$, where $\theta$ is now the angle between the two vectors in the higher dimensional space. 
The four-dimensional analog to the cross product (or vector product) is less straightforward. In three dimensions the cross product takes in two vectors and returns a vector in the direction perpendicular to these two vectors. In a higher number of dimensions there is no longer one unique direction perpendicular to the two input vectors and the operation $A \times B$ becomes ill-defined in that case. 
What we can define is the wedge product the Hodge dual of a vector and another vector $\star A \wedge B$. This more complicated operation will in 3-dimensions reduce to our familiar cross product.
The Hodge dual of a vector $\star A = \varepsilon_{ij\ldots n}A^i = A_{j\ldots n}$ is the (n-1)-dimensional hyperplane, perpendicular to that vector. Here $\varepsilon_{ij...n}$ is the fully antisymmetric Levi-Cevita tensor. The wedge product is defined as $A \wedge B = A_i B_j\varepsilon^{ij\ldots n}$, so the higher dimensional generalization of the cross product will be $\star A \wedge B = (\epsilon_{ab\ldots n}A^a) \varepsilon^{ib\ldots n} B^i$
When you use the above index notation you can use the above result in both Euclidean and Minkowski spacetime. The only difference is that the $A_0$ component of a vector in Minkowski spacetime is negative.
A: The Euclidean dot product generalizes to the Minkowski inner product $A_\mu B^\mu = {\bf A} \cdot {\bf B} - A^0 B^0$.  There's no single natural way to generalize the cross product, which is unique to three and seven dimensions.  For three-vectors, ${\bf a} \times {\bf b} = {\bf c}$ in components becomes $\epsilon_{ijk} a^i b^j = c_k$, so one can e.g. define a generalized "cross product" between three four-vectors via $D_\sigma = \epsilon_{\mu \nu \rho \sigma} A^\mu B^\nu C^\rho$.
A: If you intend your $\theta$ to be the [real-valued] rapidity (arc-length of a unit hyperbola in Minkowski spacetime), then your 4-vectors should be both future-timelike [or both past]. I think it can work with spacelike vectors as long as they are coplanar with the future-timelike vectors, correspondingly orthogonal to the timelike-vectors, and probably must both be in the forward [or backward] spatial direction (so that their unit vectors lie on a common hyperbola... in other words, on a 2D spacetime diagram with those timelike vectors). Then, you can use the hyperbolic trig function of the rapidity.
For the generalization of the cross-product, think of a "bivector".
Its area is related to the $\sinh$-function. In 3D, the Hodge dual (with $\epsilon_{abc}$) would map that to a [pseudo-]vector... not so in 4D (since the $\epsilon_{abcd}$ has four indices).
