A particle of mass M is moving in a straight line with uniform speed v parallel to x-axis in X- Y plane at a height ‘h’ from the X-axis. Its angular momentum about the origin is

Angular momentum is defined as the product of the moment of inertia and angular velocity and is a vector quantity, it can be calculated as the cross product of the position vector r and linear momentum vector p = mv.

In this case, the position vector of the particle is$\overrightarrow{ \mathrm{r} }= \mathrm{x}\widehat{\mathrm{i} }+ \mathrm{h}\widehat{\mathrm{j} }+ 0\widehat{\mathrm{k} }$ (i,j,k are the unit vectors along the x,y and z-axis respectively) and linear momentum vector is p = mvi.

So, the angular momentum of the particle is

 $\therefore {\overrightarrow{\mathrm{L}}}_0=\overrightarrow{\mathrm{r}} \mathrm{X} \mathrm{m}\overrightarrow{\mathrm{v}}=\mathrm{m}\left[\right(\mathrm{x}\hat{\mathrm{i}}+\mathrm{h}\hat{\mathrm{j}}) \times \mathrm{v}\hat{\mathrm{i}}]$ 

$=\mathrm{mvh}(\hat{\mathrm{j}} \mathrm{X} \hat{\mathrm{i}})=-\left(\mathrm{mvh}\right)\hat{\mathrm{K}}$    (k is the unit vector along the z-axis)

So, the angular momentum of the particle about the origin is L= $\left(mvh\right)\hat{K}$ which is a vector quantity and its magnitude is L = mvh and direction is along negative z-axis.