I am seeking the geometric intuition behind the following two cases using matrix $A_{m \times n}$ with rank $r$.
When $m>n>r$, $A\textbf{x} = \textbf{b}$ has a solution if $\textbf{b}$ is in $C(A)$, otherwise no solution.
When $n>m>r$, $A\textbf{x} = \textbf{b}$ has infinitely many solutions if $\textbf{b}$ is in $C(A)$, otherwise no solution.
I can understand these two results from the number of equations vs the number of unknowns perspective. But, I have hard time interpreting them geometrically. In other words, I am seeking an interpretation using geometrical terms such as "subspace", "projection", etc.
Update:
Based on the answer provided by @TSF, I have drawn a sketch to better reflect these two cases. 
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First of all, in order for $A\textbf{x} = \textbf{b}$ to have a solution, we need $\textbf{b}$ to be in $C(A)$.
When $m>n>r$, we can see $\textbf{x}$ lives in a lower dimensional subspace than $\textbf{b}$. Thus, $\textbf{x}$ has no "free components", hence the unique solution.
When $n>m>r$, we can see $\textbf{x}$ lives in a higher dimensional subspace than $\textbf{b}$. Thus, $\textbf{x}$ has "free components" that we can assign any values to, hence infinitely many solutions.
First $b$ must be in $C(A)$ for obvious reasons otherwise no solution can exist. You can see this geometrically if you are in $\mathbb{R}^3$ and $A$ maps to the $xy$-plane, if $b$ is not in the $xy$-plane there cannot be a solution. So let's assume now that $b\in C(A)$. Now, if $m$ is bigger than $n$ than we are mapping to a "bigger" space, there will be only one solution. But if $m$ is smaller than $n$ than we are losing some subspaces, shrinking them the way the projection on the $xy$-plane in $\mathbb{R}^3$ shrinks the subspace generated by the $z$-axis. In this case, it wouldn't matter what your $z$ coordinate is and so you would have infinitely many solutions, since all $(x,y,z)$ always maps to $(x,y)$ regardless of your $z$ coordinate.