A force of `(hat(i) + 2hat(j) + 3hat(k))N` is acting on a body having position vector `(3 hat(i) + hat(j) + 2hat(k))` in the same frame of reference
The moment of the force about the origin is

Options

A. `- hat(i) - 7hat(j) + 5hat(k)`
B. `-7 hat(i) - 11 hat(j) + 5 hat(k)`
C. `-hat(i) + 11 hat(j) - 7 hat(k)`
D. `7 hat(i) + 10 hat(j) + 7 hat(k)`

Answer

To find the moment of the force about the origin, we can use the formula for torque, which is given by the cross product of the position vector \( \mathbf{r} \) and the force vector \( \mathbf{F} \):

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

### Step 1: Identify the vectors
Given:
- Position vector \( \mathbf{r} = 3\hat{i} + \hat{j} + 2\hat{k} \)
- Force vector \( \mathbf{F} = \hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Set up the cross product
To calculate the cross product \( \mathbf{r} \times \mathbf{F} \), we can use the determinant method. We will set up a 3x3 determinant:

\[ \mathbf{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can expand this determinant as follows:

\[ \mathbf{\tau} = \hat{i} \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} = (3)(3) - (2)(1) = 9 - 2 = 7 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5 \]

### Step 4: Combine the results
Now substituting back into the equation for \( \mathbf{\tau} \):

\[ \mathbf{\tau} = -1\hat{i} - 7\hat{j} + 5\hat{k} \]

Thus, we can write:

\[ \mathbf{\tau} = -\hat{i} - 7\hat{j} + 5\hat{k} \]

### Final Answer
The moment of the force about the origin is:

\[ \mathbf{\tau} = -\hat{i} - 7\hat{j} + 5\hat{k} \, \text{N m} \]

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