Let p be the momentum of a particle which is defined as,
p= p x i ^ + p y j ^ + p z k ^ (1)
Here, p x is the angular momentum in x-axis, p y is the angular momentum in y-axis and p z is the angular momentum in z-axis.
Let r be the position vector of the particle which is defined as,
r=x i ^ +y j ^ +z k ^ (2)
Here, x is the position vector along x-axis, y is the position vector along y-axis and z is the position vector along z-axis.
Let l be the angular momentum of the particle which is defined as,
l= l x i ^ + l y j ^ + l z k ^
Here, l x is the angular momentum along x-axis, l y is the angular momentum along y-axis and l z is the angular momentum along z-axis.
The relation of l with p and r is,
l=r×p
From equation (1) and (2),
l=( x i ^ +y j ^ +z k ^ )×( p x i ^ + p y j ^ + p z k ^ )
The above vector product can be solved as,
l=| i ^ j ^ k ^ x y z p x p y p z | l x i ^ + l y j ^ + l z k ^ = i ^ ( y p z −z p y )− j ^ ( x p z −z p x )+ k ^ ( x p y −y p x )
Hence, from the above expressions,
l x =( y p z −z p y ) l y =−( x p z −z p x ) l z =( x p y −y p x ) (3)
As the particle moves only in x-y plane, hence z component of both the positions and linear momentum vectors will be 0.
Substitute z=0 and p z =0 in equation (3),
l x =0 l y =0 l z =( x p y −y p x )
Thus, the above expression shows that if the particle moves in x-y plane, then the angular momentum has only z component.