The torpue of force `vecF=-2hati+2hatj+3hatk` acting on a point `vecr=hati-2hatj+hatk` about origin will be :

Options

A. 288 W
B. 280 W
C. 290 W
D. None

Answer

To find the torque of the force \(\vec{F} = -2\hat{i} + 2\hat{j} + 3\hat{k}\) acting on a point \(\vec{r} = \hat{i} - 2\hat{j} + \hat{k}\) about the origin, we can use the formula for torque:

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

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

### Step 2: Set up the determinant for the cross product
The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

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

### Step 3: Calculate the determinant
The determinant can be expanded as follows:

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

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 \(-\hat{j}\):
\[ \begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix} = (1)(3) - (1)(-2) = 3 + 2 = 5 \]
So, we have \(-\hat{j} \cdot 5 = -5\hat{j}\).

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

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

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

### Step 5: Write the final answer
Thus, the torque \(\vec{\tau}\) about the origin is:

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

### Step 6: Calculate the magnitude of the torque
The magnitude of the torque can be calculated as follows:

\[ |\vec{\tau}| = \sqrt{(-8)^2 + (-5)^2 + (-2)^2} = \sqrt{64 + 25 + 4} = \sqrt{93} \]

### Final Answer
The torque of the force acting on the point about the origin is:

\[ \sqrt{93} \text{ Newton meter} \]