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Find the components along the x,y,z axes...

Find the components along the `x,y,z` axes of the angular momentum `overset rarr(L)` of a particle, whose position vector is `overset rarr(r )` with components `x,y,z` and momentum is `overset rarr(p)` with components `p_(x), p_(y) and p_(z)`. Show that if the particle moves only in the `x-y` plane, the angular momentum has only a z-component.

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Verified by Experts

We know that angular momentum `vec(l)` of a particle having position vector `vec (r )` and momentum `vec(p)` is given by
`vec(l)=vec(r )xx vec(p)`
But `vec(r )=[xhat(i)+yhat(j)+zhat(k)]` where x, y, z are the components of
`vec(r ) " and " vec(p)=[p_(x)vec(i)+p_(y)hat(j)+p_(z)hat(k)]`
`therefore " " vec (l)=vec(r )xxvec(p)=[xhat(i)+yhat(j)+zhat(k)]xx[p_(x)hat(i)+p_(y)hat(j)+p_(z)hat(k)]`
or ` (l_(x)hat(i)+l_(y)hat(j)+l_(z)hat(k))= |{:(hat(i), hat(j), hat(k)),(x,y,z),(p_(x),p_(y),p_(z)):}|`
`=(yp_(z)-zp_(y))hat(i)+(zp_(x)-xp_(z))hat(j)+(xp_(y)-yp_(x))hat(k)`
From this releation, we conclude that
`l_(x)=yp_(z)-zp_(y) l_(y)=zp_(x)-xp_(z) " and " l_(z)=xp_(y)-yp_(x)`
If the given particle moves only in the x-y plane, then z=0 and `p_(z)=0` and hence,
`vec(l)=(xp_(y)-yp_(z))hat(k)`, which is only the z-comonent of `vec(l)`.
It means that for a particle moving only in the x-y plane, the angular momentum has only the z-component.
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