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(A) The angular momentum $\overrightarrow {L}$ is given by the cross product of position vector $\overrightarrow {r}$ and momentum vector $\overrighta

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$X$-axis

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(A) The angular momentum $\overrightarrow {L}$ is given by the cross product of position vector $\overrightarrow {r}$ and momentum vector $\overrightarrow {P}$:$\overrightarrow {L} = \overrightarrow {r} 	imes \overrightarrow {P} = \begin{vmatrix} \hat i & \hat j & \hat k \ 1 & 2 & -1 \ 3 & 4 & -2 \end{vmatrix}$$= \hat i(2(-2) - (-1)(4)) - \hat j(1(-2) - (-1)(3)) + \hat k(1(4) - 2(3))$$= \hat i(-4 + 4) - \hat j(-2 + 3) + \hat k(4 - 6)$$= 0\hat i - 1\hat j - 2\hat k = -\hat j - 2\hat k$To check if $\overrightarrow {L}$ is perpendicular to the $X$-axis,we take the dot product of $\overrightarrow {L}$ with the unit vector along the $X$-axis,which is $\hat i$:$\overrightarrow {L} \cdot \hat i = (0\hat i - 1\hat j - 2\hat k) \cdot (1\hat i + 0\hat j + 0\hat k) = 0(1) + (-1)(0) + (-2)(0) = 0$Since the dot product is $0$,the angular momentum is perpendicular to the $X$-axis.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

The position of a particle is given by $\overrightarrow {r} = (\hat i + 2\hat j - \hat k)$ and momentum $\overrightarrow {P} = (3\hat i + 4\hat j - 2\hat k)$. The angular momentum is perpendicular to:

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