A force F=`2hati+3hatj-hatk` acts at a point (2,-3,1). Then magnitude of torque about point (0,0,2) will be

Options

A. 6 units
B. `3sqrt5` units
C. `6sqrt5` units
D. None of these

Answer

To find the magnitude of torque about a point (0,0,2) due to a force \( \mathbf{F} = 2\hat{i} + 3\hat{j} - \hat{k} \) acting at the point (2,-3,1), we will follow these steps:

### Step 1: Determine the position vector \( \mathbf{R} \)
The position vector \( \mathbf{R} \) from the point (0,0,2) to the point (2,-3,1) can be calculated as:

\[ \mathbf{R} = (2 - 0)\hat{i} + (-3 - 0)\hat{j} + (1 - 2)\hat{k} = 2\hat{i} - 3\hat{j} - 1\hat{k} \]

### Step 2: Write the force vector \( \mathbf{F} \)
The force vector is given as:

\[ \mathbf{F} = 2\hat{i} + 3\hat{j} - \hat{k} \]

### Step 3: Calculate the torque \( \mathbf{\tau} \)
The torque \( \mathbf{\tau} \) is given by the cross product of the force vector \( \mathbf{F} \) and the position vector \( \mathbf{R} \):

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

Using the determinant method:

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

### Step 4: Calculate the determinant
Calculating the determinant:

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

Calculating each of the 2x2 determinants:

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

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & -1 \\ 2 & -1 \end{vmatrix} = (2)(-1) - (-1)(2) = -2 + 2 = 0 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & -3 \\ 2 & 3 \end{vmatrix} = (2)(3) - (-3)(2) = 6 + 6 = 12 \]

Putting it all together:

\[ \mathbf{\tau} = 6\hat{i} - 0\hat{j} + 12\hat{k} = 6\hat{i} + 12\hat{k} \]

### Step 5: Calculate the magnitude of the torque
The magnitude of the torque \( |\mathbf{\tau}| \) is given by:

\[ |\mathbf{\tau}| = \sqrt{(6)^2 + (0)^2 + (12)^2} = \sqrt{36 + 0 + 144} = \sqrt{180} = 6\sqrt{5} \]

### Final Answer
Thus, the magnitude of the torque about the point (0,0,2) is:

\[ \boxed{6\sqrt{5}} \]