The position vectors of three particles of mass `m_(1)= 1kg, m_(2)=2 kg` and `m_(3)=4 kg` are `vec(r )_(1)=(hat(i)+4hat(j)+hat(k))m, " " vec(r )_(2)=(hat(i)+hat(j)+hat(k))m`, and `vec(r )_(3)=(2hat(i)-hat(j)-2hat(k))m` respectively. Find the position vector of their centre of mass.

Explanation

The position vector of centre of mass of the three particles is given by
`vec(r )_(c )=(m_(1)vec(r )_(1)+m_(2)vec(r )_(2)+m_(3) vec(r )_(3))/(m_(1)+m_(2)+m_(3))`
`vec(r )_(c )=(1(hat(i)+4hat(j)+hat(k))+2(hat(i)+hat(j)+hat(k))+4(2hat(i)-hat(j)-2hat(k)))/(1+2+4)`
` = ((11hat(i)+2hat(j)-5hat(k)))/(7)=(1)/(7)(11hat(i)+2hat(j)-5hat(k))m`