Given $n$ points (no two identical, no three colinear, no four coplanar, etc.), I'd like to find a formula for the normal vector to the unique hyperplane that intersects each of these points.
In three dimensions, we use a cross product: given $x_1, x_2, x_3$, the normal vector is given by $(x_1 - x_2) \times (x_1 - x_3)$. How does this generalize?
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Another way to do this, which I (re)discovered when helping son learn about vectors, is to generalize the representation of the cross-product of two 3-vectors as a 3 by 3 matrix with the first row being the three unit vectors and the other two rows being the components of the two 3-vectors.
In $R^n$, if we have $n-1$ n-vectors, form the n by n matrix with the first row being the n unit vectors, and the next n-1 rows being the components of the vectors. the resulting vector is orthogonal to each of the n-1 vectors.
You want a normal vector rather than "tangent vector." You make an $(n-1)$ by $n$ matrix with row $i$ given by $x_i - x_n.$ Then the normal vector is the null space of this matrix. The fast way to do this is Gauss elimination, but it can also be done as a recipe by pretending the matrix is square, and find the entries of the proper column of the adjoint matrix, given by cofactors with certain $\pm$ signs. The traditional cross product is often taught this way in physics classes.