An ideal Zener diode with breakdown voltage of -3 V is reverse...

Given that the Zener diode is reverse biased with an input voltage of magnitude $|V_i| = 5$ V.

The breakdown voltage of the ideal Zener diode is $|V_Z| = 3$ V.

Since the applied voltage is greater than the Zener breakdown voltage ($5$ V $> 3$ V), the Zener diode operates in the breakdown region.

In the breakdown region, the voltage drop across the ideal Zener diode remains constant at its breakdown voltage. Therefore, the potential difference between points $C$ and $B$ is $V_{CB} = 3$ V.

The total voltage is applied across the series combination of the Zener diode and the resistor $R$. By Kirchhoff's Voltage Law, the sum of the voltage drops across the Zener diode and the resistor equals the total applied voltage.

Let $V_{BA}$ be the voltage difference between points $B$ and $A$, which is the voltage across the resistor $R$.

$|V_i| = V_{CB} + V_{BA}$

$5 = 3 + V_{BA}$

$V_{BA} = 2$ V.

The magnitude of the voltage difference between points $B$ and $A$ is $2$ V.

Answer: $2$ V