A force `vec(F) = 3 hat(i) + 2hat(j) - 4hat(k)` is applied at the point `(1, -1, 2)`. What is the moment of the force about the point `(2 , -1, 3)`?

Options

A. `hat(i)+ 4hat(j) + 4hat(k)`
B. `2hat(i) + hat(j) + 2hat(k)`
C. `2hat(i) - 7hat(j) -2hat(k)`
D. `2hat(i) + 4hat(j) + hat(k)`

Answer

To find the moment of the force about a point, we can use the formula for torque (moment), which is given by:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

where:
- \(\vec{\tau}\) is the torque,
- \(\vec{r}\) is the position vector from the point about which we are calculating the moment to the point where the force is applied,
- \(\vec{F}\) is the force vector.

### Step 1: Identify the vectors
Given:
- Force vector: \(\vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k}\)
- Point A (where the force is applied): \(A(1, -1, 2)\)
- Point B (about which we are calculating the moment): \(B(2, -1, 3)\)

### Step 2: Calculate the position vector \(\vec{r}\)
The position vector \(\vec{r}\) is calculated as:

\[ \vec{r} = \vec{B} - \vec{A} \]

Substituting the coordinates of points A and B:

\[ \vec{B} = 2\hat{i} - 1\hat{j} + 3\hat{k} \]
\[ \vec{A} = 1\hat{i} - 1\hat{j} + 2\hat{k} \]

Thus,

\[ \vec{r} = (2 - 1)\hat{i} + (-1 + 1)\hat{j} + (3 - 2)\hat{k} = 1\hat{i} + 0\hat{j} + 1\hat{k} = \hat{i} + \hat{k} \]

### Step 3: Calculate the cross product \(\vec{r} \times \vec{F}\)
Now we need to calculate the cross product \(\vec{r} \times \vec{F}\):

\[ \vec{r} = \hat{i} + 0\hat{j} + \hat{k} \]
\[ \vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k} \]

Setting up the determinant for the cross product:

\[ \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 3 & 2 & -4 \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant:

\[ \vec{r} \times \vec{F} = \hat{i} \begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 3 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 0 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} = (0 \cdot -4) - (1 \cdot 2) = -2\)

2. \(\begin{vmatrix} 1 & 1 \\ 3 & -4 \end{vmatrix} = (1 \cdot -4) - (1 \cdot 3) = -4 - 3 = -7\)

3. \(\begin{vmatrix} 1 & 0 \\ 3 & 2 \end{vmatrix} = (1 \cdot 2) - (0 \cdot 3) = 2\)

Putting it all together:

\[ \vec{r} \times \vec{F} = -2\hat{i} + 7\hat{j} + 2\hat{k} \]

### Step 5: Final result for the moment
Thus, the moment of the force about the point \(B\) is:

\[ \vec{\tau} = -2\hat{i} + 7\hat{j} + 2\hat{k} \]