But there is a commutative ring available, along the lines of what Mariano says. If $k$ is a field and $V$ is a vector space, then $k \oplus V$ is a commutative ring by the rule that a scalar times a scalar, or a scalar times a vector, or a vector times a scalar, are all what you think they are. The only missing part is a vector times a vector, and you can just set that to zero. The dot product is then a special bilinear form on the algebra. In the formalism, I think that everything that you wrote makes sense.
Theo says in a comment that "even better", one should work over $\Lambda^*(V)$, the exterior algebra over $V$. The motivation is that this algebra is supercommutative. I considered mentioning this solution, and supposed that I really should have, because it arises in important formulas. For example, the Gauss formula for the linking number between two knots $K_1, K_2 \subseteq \mathbb{R}^3$ is: $$\mathrm{lk}(K_1,K_2) = \int_{K_1 \times K_2} \frac{\det \begin{bmatrix} \vec{x} - \vec{y} \\ d\vec{x} \\ d\vec{y} \end{bmatrix}}{4\pi |\vec{x} - \vec{y}|^3}$$ $$= \int_{K_1 \times K_2} \frac{\det \begin{bmatrix} x_1 - y_1 & x_2 - y_2 & x_3 - y_3 \\ dx_1 & dx_2 & dx_3 \\ dy_1 & dy_2 & dy_3 \end{bmatrix}}{4\pi |\vec{x} - \vec{y}|^3}.$$ The right way to write and interpret this formula is indeed as a determinant in the exterior algebra of differential forms. For one reason, it makes it easy to generalize Gauss' formula to higher dimensions.
However, supercommutative is not the same as commutative, and this type of determinant has fewer properties than a determinant over a commutative ring. And different properties. Such a determinant has a broken symmetry: you get a different answer if you order the factors in each term by rows than by columns. (I am using row ordering.) Indeed, the row-ordered determinant can be non-zero even if it has repeated rows. To give two examples, the determinant in the generalized Gauss formula has repeated rows, and the standard volume form in $\mathbb{R}^n$ is $$\omega = \frac{\det ( d\vec{x}, d\vec{x}, \ldots, d\vec{x} )}{n!}.$$
Happily, for Dick's question, you can truncate the exterior algebra at degree 1, which is exactly what I did. This truncation is both supercommutative and commutative.