The torque of the force `vecF=(2hati-3hatj+4hatk)N` acting at the point `vecr=(3hati+2hatj+3hatk)m` about the origin be

Options

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

Answer

To find the torque \(\vec{\tau}\) of the force \(\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \, \text{N}\) acting at the point \(\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \, \text{m}\) about the origin, we can use the formula for torque:

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

### Step 1: Write down the vectors
We have:
\[ \vec{F} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
\[ \vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k} \]

### Step 2: Set up the cross product
The cross product \(\vec{r} \times \vec{F}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
To compute the determinant, we can expand it as follows:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} \]

Calculating each of these 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: Combine the results
Now substituting back into the torque equation:
\[ \vec{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \]

### Step 5: Write the final answer
Thus, the torque \(\vec{\tau}\) is:
\[ \vec{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \, \text{N m} \]