I read that mean curvature $H=\frac{En+Gl-2Fm}{2(EG-F^2)}$, where $E=\mathbf{x}_u\cdot\mathbf{x}_v$, $F=\mathbf{x}_u\cdot\mathbf{x}_v$, and $G=\mathbf{x}_v\cdot\mathbf{x}_v$ are the coefficients of the first fundamental form, and $l$, $m$, and $n$ are the coefficients of the second fundamental form. But for this formula to make sense, we should always have $EG=(\mathbf{x}_u\cdot\mathbf{x}_v)(\mathbf{x}_v\cdot\mathbf{x}_v)\neq(\mathbf{x}_u\cdot\mathbf{x}_v)(\mathbf{x}_u\cdot\mathbf{x}_v)={F}^2$. But since the dot product is associative and commutative, don't we actually have equality here, implying that the denominator of the formula is 0? I'm very confused by this.
EDIT: I now realize the dot product is not associative. But still, do we always have that $EG\neq{F}^2$? If so, then why?
Because dot product, inherited from $\Bbb R^3$, is a positive definite bilinear form, so $EG-F^2>0$. Alternatively, $EG-F^2$ is the square of the area of the parallelogram spanned by $\mathbf x_u$ and $\mathbf x_v$, and these vectors are linearly independent.
By the way, it makes no sense to say dot product is associative, as dot product of two vectors is a scalar, not another vector.