If the centre of mass of three particles of masses of 1kg, 2 kg, 3 kg is at (2, 2, 2), then where should a fourth particle of mass 4 kg be placed so that the combined centre of mass may be at (0, 0, 0).

Explanation

Let `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(2))` and `(x_(3),y_(3),z_(3))` be the positions of masses `1kg`, `2kg`, `3kg` and let the co-ordinates of centre of mass of the three particle system is `(x_(cm),y_(cm),z_(cm))` respectively.
`x_(cm)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
`implies2=(1xx x_(1)+2xx x_(2)+3xx x_(3))/(1+2+3)`
(or) `x_(1)+2x_(2)+3x_(3)=12`.............`(1)`
Suppose the fourth particle of mass `4kg` is placed at `(x_(1),y_(1),z_(1))` so that centre of mass of new system shifts to `(0,0,0)`. For X-co-ordinate of new centre of mass we have
`0=(1xx x_(1)+2xx x_(2)+3xx x_(3)+4xx x_(4))/(1+2+3+4)`
`impliesx_(1)+2x_(2)+3x_(3)+4x_(4)=0`...........`(2)`
From equations `(1)` and `(2)`
`12+4x_(4)=0impliesx_(4)=-3`
Similarly `y_(4)=-3` and `x_(4)=-3`
Therefore 4kg should be placed at `(-3,-3,-3)`.