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

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$x-axis$

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(A) The angular momentum $\vec L$ is given by the cross product of position vector $\vec r$ and momentum vector $\vec p$: $\vec L = \vec r 	imes \vec p = \begin{vmatrix} \hat i & \hat j & \hat k \ 1 & 2 & -1 \ 3 & 4 & -2 \end{vmatrix}$ $= \hat i(2(-2) - (-1)(4)) - \hat j(1(-2) - (-1)(3)) + \hat k(1(4) - 2(3))$ $= \hat i(-4 + 4) - \hat j(-2 + 3) + \hat k(4 - 6)$ $= 0\hat i - 1\hat j - 2\hat k$ $= -\hat j - 2\hat k$ Since the angular momentum vector $\vec L = 0\hat i - 1\hat j - 2\hat k$ has no component along the $x$-axis (the coefficient of $\hat i$ is $0$),the vector $\vec L$ lies in the $yz$-plane. Therefore,the angular momentum is perpendicular to the $x$-axis.

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

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