Finding perpendicular unit vector in $\mathbb{R}^n$ to hyperplane Suppose I have $n-1$ linearly independent vectors $(v_1, ..., v_{n-1})$ in $\mathbb{R}^n$ that together form a basis of a hyperplane. I'm looking to find a last vector $v$ that is normal to the hyperplane (i.e., perpendicular to all $n-1$ basis vectors), with the extra restrictions that $v$ is a unit vector and that $det([v\quad v_1 \quad ...\quad  v_{n-1}])>0$.
Is there a formula to find such a $v$ given $v_1, ..., v_{n-1}$, that is smooth in $v_1, ..., v_{n-1}$? I was considering the cross-product but I'm not sure how it generalizes to $n-1$ vectors and to $\mathbb{R}^n$ and how to get the determinant right.
 A: Your idea of generalizing the concept of cross-product is a good one; you can read about it here. And it will indeed give you a smooth function (a rational function, in fact).
A: Let $\mathbf{e}=(e_1, \dots e_n)$ be the canonical basis of $\mathbb{R}^n$. The vector you're looking for is $v=\frac{\sum_{i=1}^n\lambda_ie_i}{\sqrt{\sum_{i=1}^n\lambda_i^2}}$ where $\lambda_i=\det([e_i\ v_1\ \dots\ v_{n-1}])$. To see that it is indeed the vector you're looking for, let $\varphi:\mathbb{R}^n\to\mathbb{R}$ be the linear form defined by $\varphi(w)=\det([w\ v_1\ \dots\ v_{n-1}])$. Since the inner product $\langle.,.\rangle$ on $\mathbb{R}^n$ is non-degenerate, there is one and only one vector $v'\in \mathbb{R}^n$ such that for every $w\in \mathbb{R}^n$, $\varphi(w)=\langle w,v'\rangle$. This vector is not zero because $\varphi$ is not zero. Since $\mathbf{e}$ is an orthonormal basis, we have $v'=\sum_{i=1}^n\langle e_i,v'\rangle e_i$ that is $v'=\sum_{i=1}^n\varphi(e_i)e_i=\sum_{i=1}^n\lambda_ie_i$. Then we can see that, for all $1\leqslant i \leqslant n-1 $, $\langle v_i,v'\rangle=\varphi(v_i)=\det([v_i\ v_1\ \dots\ v_{n-1}])=0$ and $\det([v'\ v_1\ \dots\ v_{n-1}])=\varphi(v')=\Vert v'\Vert^2>0$. Finally, we see that $v=\frac{v'}{\Vert v'\Vert}$ is a unit vector orthogonal to the $v_i$'s and $\det([v\ v_1\ \dots\ v_{n-1}])=\Vert v'\Vert>0$. 
