Let $r_{1},\dots,r_{n-1}\in\mathbb{R}^{n}$ be linearly independent vectors. Let $v_{1},\dots,v_{n}\in\mathbb{R}^{n}$ be an orthonormal basis for $\mathbb{R} ^{n}$. Define
$$r_{n}=\sum_{i=1}^{n}\det\left(v_{i},r_{1},\dots,r_{n-1}\right)v_{i}$$
show that $r_{n}$ is orthogonal to the space spanned by $r_{1},\dots,r_{n-1}$. What is the geometric meaning of the length of $r_{n}$?
I'm pretty lost.. I know this is supposed to generalize the vector product in $\mathbb{R}^3$ in some way, but I'm not sure how to show the orthogonality. Even for the (hopefully) easier case of $v_i=(0,\dots,0,1,0,\dots,0)$ (where the $1$ is on the $i$th variable) I have no idea...
\begin{align} \left<r_n,r_j\right>&=\sum_{i=1}^n\left<v_i,r_j\right>\det(v_i,r_1,\cdots,r_{n-1})\\ &=\sum_{i=1}^n\det(\left<v_i,r_j\right>v_i,r_1,\cdots,r_{n-1})\\ &=\det(\sum_{i=1}^n\left<v_i,r_j\right>v_i,r_1,\cdots,r_{n-1})\\ &=\det(r_j,r_1,\cdots,r_{n-1})\\ &=0\quad \text{ (for $j\not=n$)} \end{align} Where the second and the third equalities follow from the linearity of determinant, and the fourth one holds since $v_i$ forms an orthonormal basis.