For the hyperplane passing through the origin and orthogonal to the ones vector,$(\underset{n\ \mbox{times}}{\underbrace{1,\dots,1})}$ in $\mathbb{R}^{n}$, what are the $n-1$ remaining orthonormal basis vectors that span the plane in terms of the Cartesian coordinates, $(x_1,\dots,x_{n-1})$ of $\mathbb{R}^{n-1}$?
I started by thinking about the $(n-1)$-product of rotation matrices of 2D subspaces applied to the hyperplane, $(x_1,\dots,x_{n-1},0)$. Then I realized that the angles are not trivial and I thought there may be a better/more standard way?
If $\vec{a} \cdot \vec{y}=0$, $\vec{a},\vec{y}\in\mathbb{R}^n$, is the equation of the hyperplane (with the $a_i$s not-yet-determined), and the constraint of being orthogonal to the ones vector is $\sum_{i=1}^n{y_i}=0$. I would like to know the set of $n-1$ orthonormal vectors, $e_i=f_i(x_1,\dots,x_{n-1})$, $i=1,\dots,n-1$ that along with $e_n=(\underset{n\ \mbox{times}}{\underbrace{1,\dots,1})}$ span the space.
Let $H$ denote the hyperplane in question. If you really want a linear parametrization of $H$ (instead of an implicit description, which can be found immediately from Aretino's answer), you need a basis $(v_{j})_{j=1}^{n-1}$ of $H$, such as \begin{align*} v_{1} &= (-1, 0, \dots, 0, 1) = -e_{1} + e_{n}, \\ v_{2} &= (0, -1, \dots, 0, 1) = -e_{2} + e_{n},\quad \dots \\ v_{n-1} &= (0, 0, \dots, -1, 1) = -e_{n-1} + e_{n}. \end{align*} If instead you want a parametrization in which the coordinates are "Cartesian", your basis needs to be orthonormal, e.g., obtained from $(v_{j})$ by Gram-Schmidt.
In either case, you'd send $(x_{1}, \dots, x_{n-1}, 0)$ to $$ \sum_{j=1}^{n-1} x_{j} v_{j} = x_{1}v_{1} + \dots + x_{n-1} v_{n-1}. $$