# How do you compute the normal vector to a hyperplane in $\mathbb{R}^n$ given $n$ representative points?

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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Do you mean "normal vector"? – Ben Oct 12 '12 at 1:32
Oops, yes, I totally do. I'll edit that in. – GMB Oct 12 '12 at 1:33

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.
Thanks. When you say "proper column" of the adjoint matrix, what do you mean? When we "pretend the matrix is square," are we filling in the missing row with $0$s? If so, does this tactic generalize to computing nullspaces of, say, $(n-k) \times n$ matrices? – GMB Oct 12 '12 at 1:54
@LevDub, see en.wikipedia.org/wiki/Cross_product#Matrix_notation and consider doing the analogous thing with a 4 by 4 matrix. For the $n-k$ idea I'm not sure. – Will Jagy Oct 12 '12 at 3:04
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.