The torque of force `F =(2hati-3hatj+4hatk)` newton acting at the point `r=(3hati+2hatj+3hatk)` metre about origin is (in N-m)

Options

A. `6hati-6 hatj+12 hatk`
B. `17 hati-6hatj-13hatk`
C. `-6hati+6hatj-12hatk`
D. `-17hati+6hatj+13hatk`

Explanation

To find the torque of the force \( \mathbf{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \) newtons acting at the point \( \mathbf{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \) meters about the origin, we will use the formula for torque:

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

### Step 1: Write the vectors in matrix form

We will set up the cross product using the determinant of a matrix:

\[ \mathbf{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} \]

### Step 2: Calculate the determinant

To calculate the determinant, we can expand it as follows:

\[ \mathbf{\tau} = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

1. For \( \hat{i} \):

\[ \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} = (2)(4) - (3)(-3) = 8 + 9 = 17 \]

2. For \( \hat{j} \):

\[ \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} = (3)(4) - (3)(2) = 12 - 6 = 6 \]

3. For \( \hat{k} \):

\[ \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (2)(2) = -9 - 4 = -13 \]

### Step 4: Substitute back into the torque equation

Now substituting these values back into the torque equation:

\[ \mathbf{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \]

### Final Result

Thus, the torque \( \mathbf{\tau} \) is:

\[ \mathbf{\tau} = (17\hat{i} - 6\hat{j} - 13\hat{k}) \, \text{N-m} \]