# The center of mass of three particles of masses 1 kg,2 kg and 3 kg is at (2,2,2). The position of the fourth mass of 4 kg to be placed in the system as that the new center of mass is at (0,0,0) is :

$\begin {array} {1 1} (a)\;(-3,-3,-3) & \quad (b)\;(-3,3,-3) \\ (c)\;(2,3,-3) & \quad (d)\;(2,-2,3) \end {array}$

(A) (-3,-3,-3)
Hence A is the correct answer.

As the c.m. of three particles is at (2,2,2) ∴The total mass = 1+2+3 = 6kg Now consider the 4kg mass at the position (x,y,z) Now centre of mass of total system at (0,0,0) ∴ 624 0 10 × + x = ⇒ 12 = -4x ⇒ x = -3 Similarly 624 0 12 × + y = ⇒ y = -3 624 0 12 × + z = ⇒ z = -3 From the above we can conclude that x = -3, y = -3, z = -3