Given `r=4hatj`and`p=2hati+3hatj+hatk`. The angular momentum is

Options

A. `4hati-8hatk`
B. `8hati-8hatk`
C. `8hatj`
D. `9hatk`

Answer

To find the angular momentum \( L \) given the position vector \( \mathbf{r} \) and momentum vector \( \mathbf{p} \), we use the formula:

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

### Step 1: Identify the vectors
Given:
- \( \mathbf{r} = 4\hat{j} \)
- \( \mathbf{p} = 2\hat{i} + 3\hat{j} + 1\hat{k} \)

### Step 2: Write the vectors in component form
We can express the vectors in component form:
- \( \mathbf{r} = 0\hat{i} + 4\hat{j} + 0\hat{k} \)
- \( \mathbf{p} = 2\hat{i} + 3\hat{j} + 1\hat{k} \)

### Step 3: Set up the determinant for the cross product
To find the cross product \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \), we can use the determinant of a matrix:

\[ \mathbf{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 0 \\ 2 & 3 & 1 \end{vmatrix} \]

### Step 4: Calculate the determinant
Now we calculate the determinant:

\[ \mathbf{L} = \hat{i} \begin{vmatrix} 4 & 0 \\ 3 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 4 \\ 2 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 4 & 0 \\ 3 & 1 \end{vmatrix} = (4)(1) - (0)(3) = 4 \]

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

3. For \( \hat{k} \):
\[ \begin{vmatrix} 0 & 4 \\ 2 & 3 \end{vmatrix} = (0)(3) - (4)(2) = -8 \]

### Step 5: Combine the results
Putting it all together, we have:

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

Thus, the angular momentum vector is:

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

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

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