Avoid the planes - the geometry of grassmannians

Suppose we have $n$ planes $H_1, \ldots, H_n$ in $\mathbb{R}^m$ of codimension $q$, or equivalently of dimension $d=m-q$. I want to choose a vector which does not belong to the planes in a continuous way. There are two versions of this problem, depending on how we parameterize the planes, and the answer can be actually different.

1. Unframed version. Let $Gr_q(m) $ be the grassmannian, that is the space of codimension q planes in $\mathbb{R}^m$. Does there exist a function

$$c : Gr_q(m) ^n \to \mathbb{R}^m$$

Such that $c(H_1, \ldots, H_n) \not \in H_i$ for all $i$?

2. Framed version. Let $V_{d,m}$ be the Stiefel manifold, that is the space of orthonormal systems in $\mathbb{R}^m$ of cardinality $d$. Does there exist a function

$$c : V_{d,m} ^n \to \mathbb{R}^m$$

Such that $c(H_1, \ldots, H_n) \not \in H_i$ for all $i$?

Note. I slightly changed the notation to agree with Chris one; now $d$ denote the dimension of the planes and $q$ the codimension.

This is a partial answer, though its not particularly elementary. (EDIT: I am using d to denote the dimension of the planes, rather than the codimension as stated in the question, so things need to be re-indexed). If $2d+1\geq m$, $d\neq m$ then for any $n$, we cannot find such a map. Assume such a map exists, and consider the real steifel manifolds $V_{m,d}$. These are the space of orthonormal $d$ tuples in $\mathbb{R}^n$, and we have a natural surjective map $V_{m,d}\xrightarrow{\pi} V_{m,d-1}$ given by forgetting the last vector in our list. By restricting a hypothetical $c$ map to the first coordinate (with other coordinates some fixed spaces), it suffices to show that no map can exist when $n=1$.

We claim that a map of the form described will yield a section of the map $$V_{m,d+1}\xrightarrow{\pi}V_{m,d}$$

To see this, first normalise the map $c$ to land in unit one vectors in $\mathbb{R}^n$. Then for a given orthonormal $d$ frame, map it to the orthonormal $d+1$ frame with new vector $c(Span(v_1,..v_{d}))$. This is clearly a section, so now consider mod $2$ cohomology, and in particular the cup product structure.

By results of Borel we know this graded ring, which I read at https://www.researchgate.net/publication/254207624_The_cohomology_rings_of_real_Stiefel_manifolds_with_integer_coefficients

The cohomology ring of $V_{m,d}$ is generated by classes of generators $z_i$ of degree $i$ for $m-d\leq i\leq m-1$, subject to the relation $z_i^2=z_{2i}$ when $2i\leq m-1$, and $z_i^2=0$ else. The map $\pi$ induces the natural inclusion of these rings, so consider $z_{m-d-1}^2=z_{2m-2d-2}$ in $H^*(V_{m,d+1},\mathbb{Z}/2\mathbb{Z})$. On one hand, our section is the identity on this element, since $z_{2m-2d-2}$ is in the image of the natural inclusion, but on the other, the minimal positive degree nonzero cohomology group of $V_{m,d}$ is in degree $m-d$, so $z_{m-d-1}$ must be in the kernel. So since the induced map on cohomology rings is a map of graded rings, no such map $c$ can exist in this case.

On the other side, the case of $d=1$, $m> 2$ is also not possible. Similarly, reduce to the $n=1$ case, and thus we obtain a map $\mathbb{RP}^{m-1}\rightarrow S^{d-1}$ such that when we compose with the natural quotient $S^{d-1}\rightarrow \mathbb{RP}^{m-1}$, the induced endomorphism of $\mathbb{RP}^{m-1}$ has no fixed points. Then consider the trace of this endomorphism on mod $2$ cohomology. Since it has no fixed points, this trace is $0$, and thus this endomorphism is nonzero on some nonzero cohomology group (since its an isomorphism on $H^0$), so induces an isomorphism on $H^{m-1}(\mathbb{RP}^{m-1},\mathbb{Z}/2\mathbb{Z})$, since all the other maps factor through $0$. But a generator for this top cohomology is the $m-1$ fold cup product of the class in $H^1$, so this must be the zero map (since $S^{m-1}$ is simply connected). Thus, no map can exist in this case either.