To analyze this problem, we need to understand each step involved in using moments to prove the properties related to the action line of the force vector $\vec{F} = 3\hat{i} - 4\hat{j}$ acting at point $A = (0, 2)$.

Step 1: Verify if the Action Line of $\vec{F}$ Passes Through Point $B$

To determine if the line of action of $\vec{F}$ passes through $B = (3, -2)$, we use the concept of moments.

1. Calculate the Moment of $\vec{F}$ about Point $B$: The moment $M_B$ of a force $\vec{F}$ about a point $B$ is given by the cross product of the position vector $\vec{r}_{AB}$ (from point $A$ to $B$) and the force vector $\vec{F}$: $M_B = \vec{r}_{AB} \times \vec{F}$

2. Find $\vec{r}_{AB}$: The position vector $\vec{r}_{AB}$ from $A = (0, 2)$ to $B = (3, -2)$ is: $\vec{r}_{AB} = (3 - 0)\hat{i} + (-2 - 2)\hat{j} = 3\hat{i} - 4\hat{j}$

3. Compute the Cross Product $\vec{r}_{AB} \times \vec{F}$: Since $\vec{r}_{AB} = 3\hat{i} - 4\hat{j}$ and $\vec{F} = 3\hat{i} - 4\hat{j}$, we find that $\vec{r}_{AB}$ is parallel to $\vec{F}$.

   When the position vector is parallel to the force vector, the cross product $\vec{r}_{AB} \times \vec{F}$ equals zero, meaning there is no moment about point $B$. This implies that the action line of $\vec{F}$ passes through $B$.

This completes the first part of the proof.

Would you like further details or clarification on this explanation?

Here are some additional questions to consider:

1. What would change if the vector $\vec{F}$ acted at a different point?
2. How would you find the action line of $\vec{F}$ algebraically?
3. Can moments be used to prove parallelism of lines?
4. Why does a zero moment about a point imply that the line of action passes through it?
5. How does the cross product of two vectors represent the moment?

Tip: When dealing with vectors and points, it's often helpful to write down the components and interpret them geometrically.