If $m$ is 2, the direction constraint simplifies to a dot product which can be represented as a single linear expression. On the other hand, if $m$ is 3, the wedge product can be replaced with a cross product, but I don't see how that could be represented with fewer than 3 linear equations. I'd guess, then, that for a given $m$, the number of constraints implied by $T\vec{x}\wedge\vec{y}=0$ is $^mc_2$. However, that's more than should be necessary to reduce the feasible region to a point for $m\ge4$, but by that logic, the number of linear constraints implied by $T\vec{x}\wedge\vec{y}=0$ should be no more than $m-1$. How many linear constraints does the direction constraint ($T\vec{x}\wedge\vec{y}=0$) imply in terms of $m$?

Say $\vec y = \mathbf e_1$ in a set of orthogonal basis vectors $\{\mathbf e_i\}_{i=1}^m$. While there are $m \choose 2$ basis bivectors, the only ones we care about are of the form $\mathbf e_1 \wedge \mathbf e_i$ for $i \neq 1$ (as $\mathbf e_1 \wedge \mathbf e_1 = 0$), so there are $m-1$ constraints. This also makes sense because knowing the value of the dot product with $\vec y$ (assuming $\vec y \neq \mathbf 0$) would only be one linear equation, but the combination of the wedge product and the dot product would completely determine the result (because we can define the geometric product in terms of these and the geometric product is invertible for [non-zero] vectors).