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

Answer

To find the torque 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:

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

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

### Step 2: Set up the cross product using the determinant
To compute the cross product \( \mathbf{r} \times \mathbf{F} \), we can use the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{r} \) and \( \mathbf{F} \):

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

### Step 3: Calculate the determinant
Using the determinant formula, we expand it as follows:

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

Calculating each of the 2x2 determinants:

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

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

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

### Step 4: Combine the results
Now substituting back into the torque equation:

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

### Final Result
Thus, the torque vector is:

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