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 `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.

Explanation

If position vector of a particle is `vecr` and linear momentum `vecp` than angular momentum
`vecl=vecrxxvecp`
but `vecr=xhati+yhatj+zhatk` and `vecp=p_(x)hati+p_(y)hatj+p_(z)hatk` are the respectively components of `vecr and vecp`
`vecl=l_(x)hati+l_(y)hatj+l_(z)hatk`
`:.vecl=[xhati+yhatj+zhatk]xx[p_(x)hati+p_(y)hatj+p_(z)hatk]`
`l_(x)hati+l_(y)hatj+l_(z)hatk=|(hati,hatj,hatk),(x,y,z),(p_(x),p_(y),p_(z))|`
`=hati(yp_(z)-zp_(y))+j(zp_(x)-xp_(z))+hatk(xp_(y)-yp_(x))`
From this
`l_(x)=yp_(z)-zp_(y)`
`l_(y)=zp_(x)-xp_(z)` and
`l_(z)=xp_(y)-yp_(x)`
The particle moves in the XY plane hence `Z=0 and p_(z)=0`
`:.vecl=0hati+0hatj+(xp_(y)-yp_(x))hatk`
`:.vecl=(xp_(y)-yp_(x))hatk`
Therefore, when the particle is confined to move in the XY-plane, the direction of angular momentum `(vecl)` is along the Z-direction.