When can one continuously prescribe a unit vector orthogonal to a given orthonormal system? Let $1 \leq k < n$ be natural numbers.  Given orthonormal vectors $u_1,\dots,u_k$ in ${\bf R}^n$, one can always find an additional unit vector $v \in {\bf R}^n$ that is orthogonal to the preceding $k$.  My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,\dots,u_k$, as the tuple $(u_1,\dots,u_k)$ ranges over all possible orthonormal systems?  (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)
When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,\dots,u_k$ that is consistent with a chosen orientation on ${\bf R}^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case).  But I don't know what is going on in lower dimension.  Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.
It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections.
 A: $\def\RR{\mathbb{R}}$ This problem was solved by 
Whitehead, G. W., Note on cross-sections in Stiefel manifolds, Comment. Math. Helv. 37, 239-240 (1963). ZBL0118.18702.
Such sections exist only in the cases $(k,n) = (1,2m)$, $(n-1, n)$, $(2,7)$ and $(3,8)$. 
All sections can be given by antisymmetric multilinear maps (and thus, in particular, can be taken to be smooth). The $(2,7)$ product is the seven dimensional cross product, which is octonion multiplication restricted to the octonions of trace $0$. 
The $(3,8)$ product was computed by 
Zvengrowski, P., A 3-fold vector product in $R^8$, Comment. Math. Helv. 40, 149-152 (1966). ZBL0134.38401
to be given by the formula 
$$X(a,b,c) = -a (\overline{b} c) + a (b \cdot c) - b (c \cdot a) + c (a \cdot b)$$
where $\cdot$ is dot product while multiplication with no symbol and $\overline{b }$ have their standard octonion meanings. Note that, if $(a,b,c)$ are orthogonal, the last $3$ terms are all $0$, so the expression simplifies to $- a (\overline{b} c)$; writing in the formula in the given manner has the advantage that $X(a,b,c)$ is antisymmetric in its arguments and perpendicular to the span of $a$, $b$ and $c$ for all $(a,b,c)$.
A: Unless I'm missing something, I think that the hairy ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
A: The space of orthonormal $k$-frames in $\mathbb{R}^n$ is the Stiefel manifold $V(k, n) = SO(n)/SO(n - k)$. There is a natural $SO(k)$ action on $V(k, n)$ and the quotient is the oriented grassmannian $\operatorname{Gr}^+(k, n) = SO(n)/(SO(k)\times SO(n-k))$. Let $\gamma_k \to \operatorname{Gr}^+(k, n)$ denote the tautological bundle and let $\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$ denote its orthogonal complement. As you indicated, a nowhere-zero section of $\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$ would give rise to a map that you desire. In fact, such a map arises this way if and only if it is $SO(k)$-invariant.
The inner product on $\mathbb{R}^n$ allows us to define the map $P \mapsto P^{\perp}$ which induces a diffeomorphism $f : \operatorname{Gr}^+(k, n) \to \operatorname{Gr}^+(n-k, n)$. Under this diffeomorphism we have $f^*\gamma_{n-k} \cong \gamma_k^{\perp}$, so $\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$ admits a nowhere-zero section if and only if $\gamma_{n-k} \to \operatorname{Gr}^+(n-k, n)$ does. Therefore, we would like to know the answer to the following question: 

For which values of $k$ and $n$ does $\gamma_{n-k} \to \operatorname{Gr}^+(n-k, n)$ admit a nowhere-zero section? 

One necessary condition is that $w_{n-k}(\gamma_{n-k}) = 0$. Said another way, if $w_{n-k}(\gamma_{n-k}) \neq 0$, then there is no $SO(k)$-invariant map.

In a previous version of this answer, I stated what I thought was the $\mathbb{Z}_2$ cohomology ring of $\operatorname{Gr}^+(k, n)$ - I was incorrect. From this mistake, it followed that for $1 < k < n - 1$, $w_{n-k}(\gamma_{n-k}) \neq 0$ and hence there were no $SO(k)$-invariant maps for these values of $k$. This conclusion is false; there is a counterexample when $k = 2$ and $n = 7$ as David E Speyer pointed out in the comments below. 
Somewhat surprisingly, the $\mathbb{Z}_2$ cohomology ring of $\operatorname{Gr}^+(k, n)$ is not known in general, see this question. The values of $k$ and $n$ for which $w_{n-k}(\gamma_{n-k}) \neq 0$ also seems to be unknown in general. However, if $n - k \leq k$, then $w_{n-k}(\gamma_{n-k}) \neq 0$, so for values of $k$ and $n$ with $2k \leq n$, there are no $SO(k)$-invariant such maps.

When $k = n - 1$, you described such a map which is in fact $SO(n-1)$-invariant. By the above correspondence, such maps exist because $\gamma_1 \to \operatorname{Gr}^+(1, n) = S^{n-1}$ is trivial as it is an orientable line bundle (alternatively, $\gamma_1$ is trivialised by the Euler vector field).
When $k = 1$, first note that $\gamma_{n-1} \to \operatorname{Gr}^+(n - 1, n) = S^{n-1}$ is isomorphic to the tangent bundle of $S^{n-1}$: 
\begin{align*}
TS^{n-1} &\cong T\operatorname{Gr}^+(n-1, n)\\ 
&\cong \operatorname{Hom}(\gamma_{n-1}, \gamma_{n-1}^{\perp})\\ 
&\cong \gamma_{n-1}^*\otimes\gamma_{n-1}^{\perp}\\ 
&\cong \gamma_{n-1}\otimes f^*\gamma_1\\ 
&\cong \gamma_{n-1}
\end{align*}
where the last isomorphism uses the fact that $\gamma_1$, and hence $f^*\gamma_1$, is trivial. By Poincaré-Hopf, $TS^{n-1}$ admits a section if and only if $n$ is even. In this case, the map can be written down explicitly: $(v_1, v_2, \dots, v_{n-1}, v_n) \mapsto (-v_2, v_1, \dots, -v_n, v_{n-1})$. Identifying $\mathbb{R}^n$ and $\mathbb{C}^{n/2}$ via $(v_1, v_2, \dots, v_{n-1}, v_n) \mapsto (v_1 + iv_2, \dots, v_{n-1} + iv_n)$, the aforementioned map is nothing but multiplication by $i$.
Note, requiring $SO(k)$-invariance for $k = 1$ is not a restriction as $SO(1)$ is the trivial group.
A: Denoting the Stiefel manifold of orthonormal $k$-frames in $\mathbb{R}^n$ by $V(k,n)$ as in Michael Albanese's answer, what you are asking for is a section of the sphere bundle
$$
S^{n-k-1}\to V(k+1,n)\to V(k,n),
$$
where the projection takes a $(k+1)$-frame to its first $k$ vectors.
According to the paper
Čadek, Martin; Mimura, Mamoru; Vanžura, Jiří, The cohomology rings of real Stiefel manifolds with integer coefficients, J. Math. Kyoto Univ. 43, No. 2, 411-428 (2003). ZBL1061.55015,
the Euler class of this bundle is zero when $n-k$ is odd, and nonzero when $n-k$ is even. 
This extends Michael Albanese's answer slightly, showing that the  maps cannot exist when $n-k$ is even with $1<k<n$, even without requiring $SO(k)$-invariance. I don't know if the bundle admits a section when $n-k$ is odd. The primary obstruction vanishes in these cases, but note that there may be higher obstructions in the groups $H^{i+1}(V(k,n);\pi_i(S^{n-k-1}))$.
