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The z component of the angular momentum of a particle whose position vector is

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Correct Answer - a
Here, `vecr=xhati+yhatj+zhatk`
`vecp=p_(x)hati+p_(y)hatj+p_(z)hatk`
`vecp=p_(x)hati+p_(y)hatj+p_(z)hatk`
Let `vecL=L_(x)hati+L_(y)hatj+L_(Z)hatk`……………(i)
As `vecL=vecr xx vecp = |{:[hati,hatj,hatk],[x,y,x],[p_(x),p_(y),p_(z)]:}|`
`=hati(yp_(z)-zp_(y))+hatj(zp_(x)-xp_(z))+hatk(xp_(y)-yp_(x))`............(ii)
Comparing the coefficients of `hati, hatj` and `hatk` in Eqs. (i) and (ii), we get
`L_(x) = yp_(z)-zp_(y)`
`L_(y)=zp_(x)-xp_(z)`
`L_(z)=xp_(y)-yp_(x)`

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