The position of a particle is given by `vec r = hat i + 2hat j - hat k` and its momentum is `vec p = 3 hat i + 4 hat j - 2 hat k`. The angular momentum is perpendicular to

Options

A. `x`-axis
B. `y`-axis
C. `z-`axis
D. line at equal angles to all the axes

Explanation

To find the direction of the angular momentum vector, we will use the formula for angular momentum \(\vec{L}\), which is given by:

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

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

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

### Step 2: Set up the cross product
The cross product \(\vec{L} = \vec{r} \times \vec{p}\) can be calculated using the determinant of a matrix:

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

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

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

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 2 & -1 \\ 4 & -2 \end{vmatrix} = (2)(-2) - (-1)(4) = -4 + 4 = 0 \]

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

3. For \(\hat{k}\):
\[ \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = (1)(4) - (2)(3) = 4 - 6 = -2 \]

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

\[ \vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k} \]

### Step 5: Determine the direction
The angular momentum vector \(\vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k}\) indicates that the angular momentum is in the direction of the negative \(y\) and negative \(z\) axes. Since the coefficient of \(\hat{i}\) is zero, this means that \(\vec{L}\) is perpendicular to the \(x\)-axis.

### Conclusion
Thus, the angular momentum is perpendicular to the \(x\)-axis.

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