To solve the problem of finding the vector moment of the force $\vec{F} = 4\hat{i} - 3\hat{j}$ acting at point $A(2, 3)$ with respect to point $B(-1, 2)$, we need to use the concept of torque (moment of a force), which is given by:

$\vec{M} = \vec{r} \times \vec{F}$

Where:

• $\vec{M}$ is the moment vector,
• $\vec{r}$ is the position vector from point $B$ to point $A$,
• $\vec{F}$ is the force vector,
• $\times$ denotes the cross product.

Step 1: Find the position vector $\vec{r}$

The position vector $\vec{r}$ from point $B(-1, 2)$ to point $A(2, 3)$ is:

$\vec{r} = \vec{A} - \vec{B} = (2 - (-1))\hat{i} + (3 - 2)\hat{j} = 3\hat{i} + 1\hat{j}$

Step 2: Set up the cross product

Now we compute the cross product $\vec{r} \times \vec{F}$, where:

$\vec{r} = 3\hat{i} + 1\hat{j}, \quad \vec{F} = 4\hat{i} - 3\hat{j}$

The cross product in two dimensions gives a vector in the $\hat{k}$-direction: