What is the dimension of the space of planes in $\Bbb R^3$ and how do we reach the answer?
Clarification: What I am searching for is what is the least number of parameters that I need. For example, the space of all directions in $\Bbb R^3$ would have dimension $2$, the space of all lines - dimension $4$.
Since we already know that the space of directions in $\Bbb R^3$ is $2$ (indeed, it can be identified with the unit sphere $\Bbb S^2 \subset \Bbb R^3$), we can see that the space $\Bbb P^2$ of lines (through the origin) in $\Bbb R^3$ also has dimension $2$ (the map $\Bbb S^2 \to \Bbb P^2$ that sends a direction to the line through the origin parallel to that direction is a $2:1$ covering).
On the other hand, if we pick any inner product on $\Bbb R^3$, we get for free a bijection from $\Bbb P^2$ to the space $(\Bbb P^2)^*$ of planes through the origin in $\Bbb R^2$, which identifies a line $\ell \in \Bbb P^2$ with the subspace consisting of all vectors orthogonal to $\ell$. Thus, in the topology induced by this bijection, the space of all planes through the origin in $\Bbb R^3$ also has dimension $2$.
Alternatively, the group $GL(3, \Bbb R)$ acts on $\Bbb R^3$ by matrix multiplication and maps lines through the origin to lines through the origin, so it acts on the space $(\Bbb P^2)^*$, and it's easy to see that this action is transitive. On the other hand, the stabilizer in $GL(3, \Bbb R)$ of the $xy$ plane, $$\left\{\begin{pmatrix}\ast\\ \ast\\ 0\end{pmatrix}\right\},$$ is $$H := \left\{\begin{pmatrix}\ast & \ast & \ast\\ \ast & \ast & \ast\\ 0 & 0 & \ast \end{pmatrix}\right\},$$ so we can identify $(\Bbb P^2)^*$ with the homogeneous space $$GL(3, \Bbb R) / H,$$ which (as a topological manifold) has dimension $$\dim (\Bbb P^2)^* = \dim (GL(3, \Bbb R) / H) = \dim GL(3, \Bbb R) - \dim H = 9 - 7 = 2.$$
By an analogous argument, the (real) Grassmannian $Gr(k, n)$, that is, the space of $k$-planes through the origin in $\Bbb R^n$, has a natural topological manifold structure with dimension $k (n - k)$.