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(A) The angular momentum $\vec{L}$ is given by the cross product of the position vector $\vec{r}$ and the momentum vector $\vec{p}$.$\vec{L} = \vec{r}

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$X$-axis

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(A) The angular momentum $\vec{L}$ is given by the cross product of the position vector $\vec{r}$ and the momentum vector $\vec{p}$.$\vec{L} = \vec{r} 	imes \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 3 & 4 & -2 \end{vmatrix}$$\vec{L} = \hat{i}(2(-2) - (-1)(4)) - \hat{j}(1(-2) - (-1)(3)) + \hat{k}(1(4) - 2(3))$$\vec{L} = \hat{i}(-4 + 4) - \hat{j}(-2 + 3) + \hat{k}(4 - 6)$$\vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k}$.For a vector to be perpendicular to an axis,their dot product must be zero.The $X$-axis is represented by the unit vector $\hat{i} = (1\hat{i} + 0\hat{j} + 0\hat{k})$.Dot product of $\vec{L}$ and $\hat{i}$ is: $(0)(1) + (-1)(0) + (-2)(0) = 0$.Since the dot product is zero,the angular momentum is perpendicular to the $X$-axis.

The position vector of a particle is $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k})$ and its momentum is $\vec{p} = (3\hat{i} + 4\hat{j} - 2\hat{k})$. The angular momentum of this particle is perpendicular to:

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