The centre of mass of three particles of masses `1 kg, 2kg and 3 kg` lies at the point `(3m, 3m, 3m)` where should a fourth particles of mass `4 kg` be positioned so that centre of mass of the four particle system at `(1m, 1m, 1m)`?

Explanation

We know that centre of mass of a solid body is given by
`X _(CM) = ( m _(1) x _(1) + m _(2) x _(2) + m _(3) x _(3))/( m _(1) + m _(2) + m _(3))`
`Y _(CM) = ( m _(1) y _(1) + m _(2) y _(2) + m _(3) y _(3))/(m _(1) + m _(2) + m _(3)) `
`Z _(CM ) = (m _(1) z _(1) + m _(2) z _(2) + m _(3) z _(3) )/( m _(1) + m _(2) + m _(23)`
coordinates of centre of mass,
`(X _(CM) , Y _(CM), Z _(CM)) = (3,3,3)`
THen
`1 xx x _(1) + 2 xx x _(2) + 3 xx x _(3) = (1 + 2 + 3) 3...(i)`
When cenre of mass shifts to `(1,1,1)`
`X _(CM ) = Y _(CM ) = Z _(CM ) =1 `(given)
`(1 (1 + 2 + 3 + 4) = 1 + 2x _(2) + 3 x _(1) + 4 x _(4)...(ii)`
SOlving Eq. (1) and (ii), we get
`4x _(4) = 10 - 18`
`implies x _(4) =- 2`
Similarly, `y _(4) =- 2, z _(4) =- 2 `
The fourth particle must be placed at the point `(1.1)`