$X_1\times_h...\times_hX_{n-1}=\sqrt{\det(H)}H^{-1}[\det([E_1\:X])...\det([E_{n-1}\:X]]^t$ derivation 
$H$ is a positive square symmetric matrix of order $n$. The map  defined as $h:\mathbb{R}^{n\times 1}\times\mathbb{R}^{n\times 1}\to\mathbb{R},(X,Y)\to X^t HY$, defines the scalar product in $\mathbb{R}^{n\times 1}$.
Suppose that $V$ is hyper plane in $\mathbb{R}^{n\times1}$, with base $X_1,...,X_{n-1}$. As $Y$ is defined in $V$ iff $\det(Y,X_1,...,X_{n-1})=0$, the hyper plane $V$ is determined by the equation $[\det(E_1,X)...\det(E_n,X)]Y=0$. Therefore, a h-orthogonal vector to $V$ is the vector $Z=X_1\times_h...\times_hX_{n-1}$ defined as:
$X_1\times_h...\times_hX_{n-1}=\sqrt{\det(H)}H^{-1}[\det([E_1\:X])...\det([E_{n-1}\:X]]^t$

In the material I am studying this concepts are introduced this way. I have been looking for derivation but found nothing.
Question:
Where does $\sqrt{\det(H)}H^{-1}[\det([E_1\:X])...\det([E_{n-1}\:X]]^t$ come from? How is it derived? Can someone provide me sources where can find more on  these issues?
Thanks in advance!
 A: Since $V$ is one dimension short of being a hyperplane I will take the liberty of  adding one more vector to the collection of the vectors $X_i$'s.
For completeness, the equality $det(Y,X_1,...,X_{n−1}) = det(E_1,X)...det(E_n,X)]Y$, where $X = [X_1,..., X_n]$ is the matrix with columns $X_1, ... X_n \in \mathbb{R}^{n+1}$, follows from a simple cofactor expansion, just walk down the first column of the matrix
$$\begin{bmatrix} y_1 & x_1^1& x_1^2& ... & x_1^n \\ & & ... &  & \\ y_{n+1} & x_{n+1}^1 & x_{n+1}^2 & .... & x_{n+1}^n  \end{bmatrix}$$
and you get
\begin{align}
det[Y X] &= det \left(\begin{bmatrix} y_1 & x_1^1& x_1^2& ... & x_1^n \\ & & ... &  & \\ y_{n+1} & x_{n+1}^1 & x_{n+1}^2 & .... & x_{n+1}^n  \end{bmatrix} \right)\\
&= y_1 det \left(\begin{bmatrix}  x_2^1 & x_2^2 & ... & x_2^n \\ & ... &  & \\   x_{n+1}^1 & x_{n+1}^2 & .... & x_{n+1}^n  \end{bmatrix} \right) + ... + y_{n+1} det \left(\begin{bmatrix}  x_1^1 & x_1^2 & ... & x_1^n \\ & ... &  & \\   x_n^1 & x_n^2 & .... & x_{n}^n  \end{bmatrix} \right)\\
&= D^t Y
\end{align}
where $D$ is the vector determined by 
$D^t = \left[  det \left(\begin{bmatrix}  x_2^1 & x_2^2 & ... & x_2^n \\ & ... &  & \\   x_{n+1}^1 & x_{n+1}^2 & .... & x_{n+1}^n  \end{bmatrix} \right), ..., det \left(\begin{bmatrix}  x_2^1 & x_2^2 & ... & x_2^n \\ & ... &  & \\   x_{n}^1 & x_{n}^2 & .... & x_{n}^n  \end{bmatrix} \right)\right]$
Now
$$ det \left(\begin{bmatrix}  x_2^1 & x_2^2 & ... & x_2^n \\ & ... &  & \\   x_{n+1}^1 & x_{n+1}^2 & .... & x_{n+1}^n  \end{bmatrix} \right) = det \left(\begin{bmatrix} 1 & x_1^1 & x_1^2 & ... & x_1^n \\ 0 & x_2^1 & x_2^2 & ... & x_2^n\\ &&...&& \\  0 & x_{n+1}^1 & x_{n+1}^2 & .... & x_{n+1}^n  \end{bmatrix} \right) = det[E_1, X]$$
which finally yields the stated equality. Now we have constructed an expression for the orthogonal complement of $V$, what remains is to translate this into an expression for the orthogonal complement with respect to the H-inner product. To this end the author just adds a $H^{-1}$ term to negate the effect of the operator $H$.
As mentioned in the comments, the term $\frac{1}{\sqrt{det(H)}}$ is just a scaling factor used to normalize the vector, and changes nothing. The vector is orthogonal with respect to the H-innerproduct (by inspection), and since this is a hyperplane we know that it completely determines the orthogonal complement, (by the linear span of this vector). 
A: Although it is not stated above, the vectors $X_1,...,X_{n-1}$ are presumably  $H$-orthonormal. In that case the above is a perfect natural way to construct the (up to sign) unique  vector $Z$ so that $(Z,X_1,...,X_{n-1})$ forms an $H$-orthonormal base for ${\Bbb R}^n$. 
It is essentially a (rather nice) explicit implementation of the Hodge-star operator   from Riemannian geometry. 
Denote by 
$\langle X,Y\rangle_H = X^t H Y$ the $H$-scalar product with corresponding norm
$\|\cdot\|_H$.
The determinant is $n$-multilinear so if $Y=y_1 E_1+\cdots y_n E_n$ and ${\bf X}$ is an abbreviation for  $X_1,...,X_{n-1}$ then 
\begin{equation} \det (Y,{\bf X}) = \sum_i y_i \det(E_i,{\bf X}) = y^t \alpha = y^t H H^{-1} \alpha= \langle y, H^{-1} \alpha \rangle_H  
\end{equation}
where $\alpha_i= \det(E_i,{\bf X})$. By anti-symmetry of the determinant, $Y$ belongs to $V$ iff the determinant is zero, i.e. iff $Y$ is $H$-orthogonal to $H^{-1}\alpha$. So indeed $H^{-1} \alpha$ is an $H$-orthogonal vector to $V$.
In order to derive an expression for the normalization factor, let $$c=\frac{1}{\|H^{-1} \alpha\|_H}=\frac{1}{\sqrt{\langle H^{-1}\alpha, H^{-1}\alpha \rangle_H}}.$$ We write $Z=X_0=c  \ H^{-1}\alpha=$ for the $H$-normalized vector in the direction $H^{-1}\alpha$. Writing $M_{ij}= E_i^t X_i $ for the coordinates in the Euclidean base we have:
 $$ \det(M) =c \det(H^{-1}\alpha,{\bf X}) =c \langle H^{-1}\alpha, H^{-1}\alpha \rangle_H
= \frac{1}{c}$$
On the other hand  $(X_0,X_1,...,X_{n-1})$ is an $H$-orthonormal base so we also have
 (summation over repeated indices):
$$1 = \det (\langle X_i,X_j \rangle_H) = \det\left(\ (X_i^t E_k) \ (E_k^t H E_l) \ (E_l^t X_j)\ \right)) = \det(M^t H M)=c^{-2} \det(H)$$
which gives us the normalization $c=(\det(H))^{1/2}$ as we wanted to show.
