Two particles,each of mass $m$ and speed $v$,travel in opposite directions along parallel lines separated by a distance $d$. Show that the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken.

(N/A) Let at a certain instant two particles be at points $P$ and $Q$,as shown in the figure.
Angular momentum of the system about point $P$:
$\vec{L}_{P} = m v \times 0 + m v \times d = m v d$ ... $(i)$
Angular momentum of the system about point $Q$:
$\vec{L}_{Q} = m v \times d + m v \times 0 = m v d$ ... $(ii)$
Consider a point $R$,which is at a distance $y$ from point $Q$,i.e.,$QR = y$ and $PR = d - y$.
Angular momentum of the system about point $R$:
$\vec{L}_{R} = m v \times (d - y) + m v \times y$
$= m v d - m v y + m v y$
$= m v d$ ... $(iii)$
Comparing equations $(i)$,$(ii)$,and $(iii)$,we get:
$\vec{L}_{P} = \vec{L}_{Q} = \vec{L}_{R}$ ... $(iv)$
We infer from equation $(iv)$ that the angular momentum of a system of two particles moving in opposite directions along parallel lines does not depend on the point about which it is calculated.