If force `vecF=3hati+4hatj-2hatk` acts on a particle having position vector `2hati+hatj+2hatk` then, the torque about the origin will be :

Options

A. `3hati+4hatj-2hatk`
B. `-10hati+10hatj+5hatk`
C. `10hati+5hatj-10hatk`
D. `10hati+hatj-5hatk`

Explanation

To find the torque \(\vec{\tau}\) about the origin due to the force \(\vec{F}\) acting on a particle at position \(\vec{r}\), we can use the formula:

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

### Step 1: Identify the position vector and force vector
The position vector \(\vec{r}\) is given as:
\[ \vec{r} = 2\hat{i} + \hat{j} + 2\hat{k} \]

The force vector \(\vec{F}\) is given as:
\[ \vec{F} = 3\hat{i} + 4\hat{j} - 2\hat{k} \]

### Step 2: Set up the cross product
To find the torque, we need to compute the cross product \(\vec{r} \times \vec{F}\). We can represent this using the determinant of a matrix:

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding this determinant, we have:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} 1 & 2 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & 2 \\ 4 & -2 \end{vmatrix} = (1)(-2) - (2)(4) = -2 - 8 = -10 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (2)(3) = -4 - 6 = -10 \quad \Rightarrow \quad -(-10) = 10 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix} = (2)(4) - (1)(3) = 8 - 3 = 5 \]

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

\[ \vec{\tau} = -10\hat{i} + 10\hat{j} + 5\hat{k} \]

### Final Result
Thus, the torque about the origin is:

\[ \vec{\tau} = -10\hat{i} + 10\hat{j} + 5\hat{k} \]

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