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

Explanation

The position ventor of CM of the three particles `(bar(r)_(CM))` will be given by
`bar(r)_(CM) = (m_(1)vec(r_(1)) + m_(2)vec(r_(2)) +m_(3)vec(r_(3)))/(m_(1)+m_(2)+m_(3))`
Substituting the values , we get
` bar(r)_(CM) = ((1)(hat(i)+4hat(j)+hat(k))+(2)(hat(i)+hat(j)+hat(k))+(3)(2hat(i)-hat(j)-2hat(k)))/(1+2+3) `
` = (9hat(i)+3hat(j)-3hat(k))/6 , bar(r)_(CM) = 1/2 (3hat(i)+hat(j)-hat(k)) m `