The torque of a force `F = -2 hat(i) +2 hat(j) +3 hat(k)` acting on a point `r = hat(i) - 2 hat(j)+hat(k)` about origin will be

Options

A. `8 hat(i)+5 hat(j)+2 hat(k)`
B. `-8 hat(i)-5hat(j)-2 hat(k)`
C. `8 hat(i)-5 hat(j)+2 hat(k)`
D. `-8 hat(i)+5 hat(j)-2 hat(k)`

Explanation

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

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

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

### Step 1: Identify the vectors
Given:
- Force \( \mathbf{F} = -2 \hat{i} + 2 \hat{j} + 3 \hat{k} \)
- Position vector \( \mathbf{r} = \hat{i} - 2 \hat{j} + \hat{k} \)

### Step 2: Set up the cross product
We can express the cross product in determinant form:

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

### Step 3: Calculate the determinant
To calculate the determinant, we expand it as follows:

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

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -2 & 1 \\ 2 & 3 \end{vmatrix} = (-2)(3) - (1)(2) = -6 - 2 = -8 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix} = (1)(3) - (1)(-2) = 3 + 2 = 5 \]

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

### Step 4: Combine the results
Now substituting back into the expression for \( \tau \):

\[ \tau = -8 \hat{i} - 5 \hat{j} - 2 \hat{k} \]

### Final Result
Thus, the torque \( \tau \) is:

\[ \tau = -8 \hat{i} - 5 \hat{j} - 2 \hat{k} \]

### Conclusion
The correct answer is option 2: \( -8 \hat{i} - 5 \hat{j} - 2 \hat{k} \).

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