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

Options

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

Answer

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

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

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

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

### Step 2: Set up the determinant for the cross product
We will use the determinant method to calculate the cross product. The formula for the cross product in determinant form is:

\[ \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -2 & 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we will calculate the determinant:

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

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 equation for \( \vec{\tau} \):

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

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

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