Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1, 2, 3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1, 3, -2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system?

According to the definition of centre of mass, we can imagine one particle of mass (1 + 2 + 3) kg at (1,2, 3); another particle of mass (2 + 3) kg at (-1,3,-2).

Let the third particle of mass 5 kg put at (${\mathrm{x}}_3, {\mathrm{y}}_3, {\mathrm{z}}_3$) i.e.

 ${\mathrm{m}}_1 = 6 \mathrm{kg}, ({\mathrm{x}}_1, {\mathrm{y}}_1, {\mathrm{z}}_1) = (1, 2, 3)$

${\mathrm{m}}_2 = 5 \mathrm{kg}, ({\mathrm{x}}_2, {\mathrm{y}}_2, {\mathrm{z}}_2) = (-1, 3,-2)$

${\mathrm{m}}_3=5 \mathrm{kg},$ let its position is $\left(x_{3,}{y}_{3.}{z}_3\right)$

Given, $({\mathrm{X}}_{\mathrm{CM}}, {\mathrm{Y}}_{\mathrm{CM}}, {\mathrm{Z}}_{\mathrm{CM}}) = (1, 2, 3)$

1 = $\frac{6\times 1+5\times (-1)+5{\mathrm{x}}_3}{6+5+5}$

$5{\mathrm{x}}_3 = 16-1 = 15 \mathrm{or} {\mathrm{x}}_3 = 3$

Similarly, $y_3 = 1 and z_3 = 8$