If the centre of mass of three particles of masses of `1kg, 2kg, 3kg` is at `(2,2,2)`, then where should a fourth particle of mass `4kg` 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 1 kg , 3 kg , 3 kg and let the co - ordinate 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)) rArr 2 = (1 xx x_(1) +2 xx x_(3) + 3 xx x_(2))/(1+2+3)`
or `x_(1)+2x_(2) + 3x_(3) = 12 ..... (1) `
Suppose the fourth particle of mass 4kg is placed at `(x_(4),y_(4),z_(4))` so that centre of mass of new system shifts to (0,0,0) . For a coordinate of new centre of mass we have
` 0 = (1 xx x_(1) +2xx x_(2) + 3 xx x_(3) +4 xx x_(4))/ (1+2+3+4)`
`rArr x_(1) + 2x_(2) + 3x_(3) + 4x_(4) = 0 .....(2)`
from equations (1) and (2)
`12 + 4x_(4) = 0 rArr x_(4) = - 3 `
similarly , `y_(4) = - 3 and z_(4) = - 3 `
Therefore 4kg should be placed at `(-3,-3,-3)` m