The position vectors of three particles of mass `m_(1)=1kg, m_(2)=2kg` and `m_(3)=4` 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
`vecr_(c )=(m_(1)vecr_(1)+m_(2)vecr_(2)+m_(3)vecr_(3))/(m_(1)+m_(2)+m_(3))`
`vecr_(c )=(1(hati+4hatj+hatk)+2(hati+hatj+hatk)+4(2hati-hatj-2hatk))/(1+2+4)`
`=((11hati+2hatj-5hatk))/(7)=(1)/(7)(11i+2j-5k)m`