Find the angular momentum of the particle about `O` in the following situations.
(`a`) A particle of mass `m` moving with velocity `v` parallel to the x-axis at distance `d` from it.
(`b`) A particle of mass `m` moving with velocity `v` on a line `y=x+3`.

Explanation

(`a`)
Angular momentum of particle about `O`
`=` linear momentum of particle `xx bot^(ar)` distance from `O`
`=mvd`
In the vector form
`vecl=vecrxxvecp=vecrxxmvecv`
`=(dhatj)xx(mvhati)`
`= -mvdhatk`
The direction of angular momentum is inside the plane of paper, i.e. along the ` -z` -direction.
(`b`) `y=x+3`, comparing it with straight line equation `y=mx+c m=tantheta=1` , a line makes an angle `theta=45^(@)` with the `+x-`axis and intercepts on the `y`-axis is `3`
Drop a perpendicular from `O` to the line `y=x+3`
`OA=d=3sin45^(@)=(3)/(sqrt2)`
Angular momentum about `O`
`=` linear momentum `xxbot^(ar)` distance from `O`
`=mvxx(3)/(sqrt2)=(3mv)/(sqrt2)`