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Find the components along the x, y, z-axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z- component.

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Linear momentum of particle, →p=px^i+py^j+pz^k Position vector of the particle, →r=x^i+y^j+z^k Angular momentum, →l=→r×→p ⟹lx^i+ly^j+lz^k=^i(ypz−zpy)−^j(xpz−zpx)+^k(xpy−ypx) Therefore on comparison of coefficients, lx=ypz−zpy ly=zpx−xpz lz=xpy−ypx The...
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Linear momentum of particle, p=px^i+py^j+pz^k

Position vector of the particle, r=x^i+y^j+z^k
Angular momentum, l=r×p
lx^i+ly^j+lz^k=^i(ypzzpy)^j(xpzzpx)+^k(xpyypx)
Therefore on comparison of coefficients,
lx=ypzzpy
ly=zpxxpz
lz=xpyypx
The particle moves in the x-y plane. Hence the z component of the position vector and linear momentum vector becomes zero.
z=pz=0
Thus lx=0
ly=0
lz=xpyypx
Thus when particle is confined to move in the x-y plane, the angular momentum of particle is along the z-direction.
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