Find the torque of a force `F=-3hat(i)+2hat(j)+hat(k)` acting at the point `r=8hat(i)+2hat(j)+3hat(k),(iftau=rxxF)`

Options

A. `14hat(i)-38hat(j)+16hat(k)`
B. `4hat(i)+4hat(j)+6hat(k)`
C. `-14hat(i)+38hat(j)-16hat(k)`
D. `-4hat(i)-17hat(j)+22hat(k)`

Explanation

To find the torque \(\tau\) of the force \(\mathbf{F} = -3\hat{i} + 2\hat{j} + \hat{k}\) acting at the point \(\mathbf{r} = 8\hat{i} + 2\hat{j} + 3\hat{k}\), we will use the formula for torque given by:

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

### Step 1: Write down the vectors
We have:
- \(\mathbf{F} = -3\hat{i} + 2\hat{j} + \hat{k}\)
- \(\mathbf{r} = 8\hat{i} + 2\hat{j} + 3\hat{k}\)

### Step 2: Set up the cross product
To find the cross product \(\mathbf{r} \times \mathbf{F}\), we can use the determinant of a matrix:

\[ \tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & 2 & 3 \\ -3 & 2 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
We will expand this determinant using the first row:

\[ \tau = \hat{i} \begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 8 & 3 \\ -3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 8 & 2 \\ -3 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} = (2 \cdot 1) - (3 \cdot 2) = 2 - 6 = -4 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 8 & 3 \\ -3 & 1 \end{vmatrix} = (8 \cdot 1) - (3 \cdot -3) = 8 + 9 = 17 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 8 & 2 \\ -3 & 2 \end{vmatrix} = (8 \cdot 2) - (2 \cdot -3) = 16 + 6 = 22 \]

### Step 4: Combine the results
Putting it all together, we have:

\[ \tau = -4\hat{i} - 17\hat{j} + 22\hat{k} \]

Thus, the torque \(\tau\) is:

\[ \tau = -4\hat{i} - 17\hat{j} + 22\hat{k} \]

### Final Answer
The torque of the force is:

\[ \tau = -4\hat{i} - 17\hat{j} + 22\hat{k} \]

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