We note here that f(u, v; w) = (u•v)w − 2(w•u)v is a linear function in each of its three variables and that F(u, v) = λw.f(u, v; w) thus is a bilinear function with values in V→V. Also F(n, n) is the reflection about the plane orthogonal to n when n is constrained to be length 1. The scheme here is to extend the domain of F from Vk to Ck so that F(n, n) is a congruence when n is somehow limited to some set of Clifford numbers deemed to represent rotations.

The most general bilinear form is f(∑imiei, ∑iniei; ∑iwiei) = ∑iφiei
where φi = ∑jklcijklmjnkwl.
We seek the c's. There are many but there is also much symmetry.

Here is how to recover a bilinear from b from a quadratic form q. First if b(x, y) is bilinear then so is (b(x, y) + b(y, x))/2 which is symmetric and produces the same quadratic form. We shall try for a symmetric bilinear form. The function f above is not symmetric but we could symmetrize it.
q(x) = b(x, x)
q(x+y) = b(x+y, x+y) = b(x+y, x) + b(x+y, y)
= b(x, x) + b(x, y)+b(y, x) + b(y, y) = q(x) + q(y) + 2b(x, y)

b(x, y) = (q(x+y) −q(x) − q(y))/2
Voilà! and it is symmetric.