Why doesn't voltage drop across this resistor when transistor is off?
Think of the transistor as having a very large resistance when OFF. So you essentially have this:







simulate this circuit &ndash; Schematic created using CircuitLab

The current is then:

$$ I=\frac{V_{DD}}{R_D+R_{off}}$$

And the voltage across \$R_D\$ is:

$$V=IR_D $$

What happens if \$R_D\$ is very large (say 10M\$\Omega\$)? The current is essentially zero then, and the voltage across the resistor is 0V. 

In the ideal case, \$R_{off}\$ is 'infinite' which would mean you'd have an open circuit and no current flows through an open loop. But even for practical values, \$R_{off}\$ forces the current to be negligible. 

If there were some other circuitry connected to \$V_o\$ in your schematic, then we'd need to know the equivalent resistance looking into that other circuit because the current may no longer be negligible, but in your case, it is unloaded at \$V_o\$.