If for some particle its position vector and linear momentum are respectively be `2hati+hatj+hatk` and `2hati-3hatj+hatk`. Its angular momentum will be:

Options

A. `2hati-4hatk`
B. `4hati-8hatk`
C. `4hati-4hatj+2hatk`
D. `4hati-8hatj`

Answer

To find the angular momentum of a particle given its position vector and linear momentum, we can use the formula for angular momentum \( \mathbf{L} \):

\[ \mathbf{L} = \mathbf{r} \times \mathbf{p} \]

where \( \mathbf{r} \) is the position vector and \( \mathbf{p} \) is the linear momentum vector.

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

### Step 2: Set up the cross product
The cross product \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

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

### Step 3: Calculate the determinant
To calculate the determinant, we will expand it using the first row:

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

Calculating each of the 2x2 determinants:

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

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

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

### Step 4: Combine the results
Now substituting back into the equation for \( \mathbf{L} \):

\[ \mathbf{L} = 4\hat{i} - 0\hat{j} - 8\hat{k} \]

Thus, we can simplify this to:

\[ \mathbf{L} = 4\hat{i} - 8\hat{k} \]

### Final Result
The angular momentum \( \mathbf{L} \) is:

\[ \mathbf{L} = 4\hat{i} - 8\hat{k} \]