Continuous maps $f:S^n \to \mathbb{C}P^m$ with $f(x)\perp f(-x) $ 
Question 1: What is  a  complete classification of  all positive integers $m,n$  with the  following  property:
There is  a  continuous  map $f:S^n \to \mathbb{C}P^m$ such that $f$ maps  antipodal points to orthogonal lines. Namely  for every $x\in S^n$ we have $\;f(x) \perp f(-x)$. Here the  later perpendicularity is  meant as $f(x)$ is  orthogonal to $f(-x)$  with respect to the  standard inner product  of  $\mathbb{C}^{m+1}$
In particular, is it true to say that such map does not exist if $n>2m$?

Motivation:
One  can prove the  three  dimensional Borsuk Ulam theorem without any explicit or  implicit  use of homology-cohomology as follows:
(An Equivalent  formulation of )Borsuk_Ulam  in dimension $3$:  There is  no  an  odd  continuous  function $f:S^3\to S^2$.
Proof:  We  identify $S^2$  with $\mathbb{C}P^1$. Then, as I learned from Sebastian Goette via  his  MO  comment,  the  antipodal points  of  $S^2$  corresponds to  orthogonal lines in $\mathbb{C}P^1$.  So we  have  to prove that there is  no  a  continuous  map  $f:S^3\to  \mathbb{C}P^1$ with the property that $f$ maps  antipodal points  to  orthogonal lines. For  the  contrary assume that such $f$ exist. Every  map $f:S^3 \to \mathbb{C}P^1$  determines  a  complex line  bundle  $\ell$ over $S^3$ where  $\ell$ is  athe pull back of the  tautological line  bundle  over $\mathbb{C}P^1$  so is a
subbundle  of the  trivial  bundles $S^3 \times \mathbb{C}^2$. Obviousely every  line  bundle over  $S^3$  is a  trivial  bundle  because the  corresponding  clutching function $K:S^2 \to GL(1,\mathbb{C})$ is  null homotp because any such $K$  has a  logarithm by  the  lifting lemma  in the  covering space theory. We  take  a non vanishing section $S:S^3 \to  \mathbb{C}^2$ for  the  line  bundle  $\ell$. Put  $\omega= dx\wedge dy$, the  natural  determinant $2\_$ form on $\mathbb{C}^2$. Then $\omega(S(x), S(-x)):S^3 \to  \mathbb{C} \setminus \{0\}$ is  an odd  continuous  function, a  contradiction by the  2  dimensional BU  where the  later  has an elementary  proof. Because every  continuous  function $S^2\to S^1$  has  a  logarithm so obviously  it  can not  be  an odd map $\;\blacksquare$
Remark 1: Instead of  identification $S^2$  with $\mathbb{C}P^1$, one  can identify $S^2$  with the space  of projections  of  $M_2(\mathbb{C})$. In this  case antipodal maps  correspond to  orthogonal projections. The  precise identification is the following: $(x,y,z)\mapsto 1/2\begin{pmatrix} 1-z&x+yi\\x-yi&1+z \end{pmatrix}$
Remark 2 The  above  proof,  which is  independent of  homology or  cohomology, and  involves the  orthogonality of  lines  in the  projective space or projections  of the  matrix algebra, not  only motivates  the  question $1$ above  but  also motivates the  following question:

Question 2: Can one  generalize the  above  proof  to  find  new  proof  of the  higher dimensional BU, without involving  Homology-Cohomology?

 A: Such a map $S^n\to \mathbb CP^m$ exists if and only if either $n<2m$ or $n=2m=2$.
To see this, first note that such a map is the same as a $\mathbb Z/2$-equivariant map from $S^n$ to a certain subspace of $\mathbb CP^m\times \mathbb CP^m$, namely the space of pairs $(L,M)$ such that $L\perp M$. Now note that the latter subspace is an equivariant deformation retract of the space of pairs $(L,M)$ such that $L\neq M$. So the question is, for $X=\mathbb CP^m$, is there a map $f:S^n\to X$ such that $f(-x)\neq f(x)$ for all $x$?
If $n<2m$ then the answer is yes because you can embed
$$S^n\subset\mathbb R^{n+1}\subseteq\mathbb R^{2m}=\mathbb C^m\subset \mathbb CP^m.$$
To get a negative result when $n=2m$, note that for any map $f:S^n\to N$ to a smooth $n$-manifold there is a mod $2$ obstruction to having $f(x)\neq f(-x)$ for all $x$. It can be defined homologically, or it can be defined more geometrically by first putting $f$ in general position and then counting how many unordered pairs $(x,-x)$ satisfy $f(x)=f(-x)$. This element of $\mathbb Z/2$ depends only on the homotopy class of $f$. If $f$ is homotopic to a constant then you can work out (replacing $N$ by $\mathbb R^n$ if you like) that the obstruction is nontrivial. (This is a way of thinking about Borsuk-Ulam.) When $N=\mathbb CP^m$ then every map $S^{2m}\to X$ is homotopic to a constant unless $m=1$. But there are more homotopy classes of maps $S^2\to \mathbb CP^1$, and in fact for the maps of odd degree that obstruction is trivial.
The negative result for $n>2m$ follows from the negative result for $n=2m$ (when $m>1$). The negative result for $S^3\to\mathbb CP^1$ is a case of Borsuk-Ulam, as you observed.
Edit: The geometric idea for defining the mod $2$ invariant is this. An equivariant map $F:S^n\to N\times N$ (such as $x\mapsto (f(x),f(-x)$) is always homotopic through equivariant maps to a smooth map such that $F$ is transverse to the diagonal $\Delta_N\subset N\times N$. Then $F^{-1}(\Delta_N)$ is a finite set (zero-dimensional compact manifold) in $S^n$ invariant under $x\mapsto -x$, giving a finite set in $\mathbb RP^n$. The cardinality of this, reduced mod $2$, is the invariant. It is independent of the choice of $F$ within the homotopy class because if $F_0$ and $F_1$, both smooth and transverse to $\Delta_N$, are equivariantly homotopic then the homotopy $H:S^n\times I\to N\times N$ may be chosen to be transverse to $\Delta_N$, and then $H^{-1}(\Delta_N)$ is a cobordism in $S^n\times I$ giving a cobordism in $\mathbb RP^n\times I$, and a compact one-dimensional manifold must have an even number of boundary points.
