The position of a particle is given by `vecr=(hati+2hatj-hatk)` and momentum `vecp=(3hati+4hatj-2hatk)`. The angular momentum is perpendicular to the

Options

A. x-axis
B. y-axis
C. z-axis
D. yz-plane

Explanation

To determine the direction of the angular momentum vector given the position vector \(\vec{r}\) and momentum vector \(\vec{p}\), we will follow these steps:

### Step 1: Identify the vectors
The position vector \(\vec{r}\) and momentum vector \(\vec{p}\) are given as:
\[ \vec{r} = \hat{i} + 2\hat{j} - \hat{k} \]
\[ \vec{p} = 3\hat{i} + 4\hat{j} - 2\hat{k} \]

### Step 2: Calculate the angular momentum vector \(\vec{L}\)
The angular momentum \(\vec{L}\) is calculated using the cross product of the position vector \(\vec{r}\) and the momentum vector \(\vec{p}\):
\[ \vec{L} = \vec{r} \times \vec{p} \]

Using the determinant form for the cross product:
\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & 4 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \vec{L} = \hat{i} \begin{vmatrix} 2 & -1 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \]

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

Putting it all together:
\[ \vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k} \]

### Step 4: Analyze the components of \(\vec{L}\)
The angular momentum vector \(\vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k}\) has:
- No component in the \(x\)-direction (0 along \(\hat{i}\))
- A negative component in the \(y\)-direction (-1 along \(\hat{j}\))
- A negative component in the \(z\)-direction (-2 along \(\hat{k}\))

### Step 5: Determine the plane of \(\vec{L}\)
Since \(\vec{L}\) has no component in the \(x\)-direction, it lies in the \(yz\)-plane. Therefore, it is perpendicular to the \(x\)-axis.

### Conclusion
The angular momentum vector \(\vec{L}\) is perpendicular to the \(x\)-axis.