Find the torque of a force `vecF=-3hati+2hatj+hatk` acting at the point `vecr=8hati+2hatj+3hatk` about origin

Options

A. `14hati--38hatj+16hatk`
B. `4hati+4hatj+6hatk`
C. `-14hati+38hatj-16hatk`
D. `-4hati+17hatj-22hatk`

Explanation

To find the torque \( \vec{\tau} \) of a force \( \vec{F} \) acting at a point \( \vec{r} \) about the origin, we can use the formula:

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

where \( \times \) denotes the cross product.

Given:
- \( \vec{F} = -3 \hat{i} + 2 \hat{j} + 1 \hat{k} \)
- \( \vec{r} = 8 \hat{i} + 2 \hat{j} + 3 \hat{k} \)

### Step 1: Set up the cross product

We will set up the cross product using the determinant form:

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

### Step 2: Calculate the determinant

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

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

### Step 3: Calculate the individual 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: Substitute back into the torque equation

Now substituting back into the torque equation:

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

### Step 5: Write the final answer

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

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