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The position vectors of three particles ...

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

Text Solution

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The correct Answer is:
`(1)/(2)(3i+j-k) m `
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Knowledge Check

  • Position vector of the centre of mass of a system of N particles of total mass M is

    A
    `(sum^(vecMr)p)/(M)`
    B
    `sum_((p=1)/(r))^(N)vecr_(p)`
    C
    `(sum_((p=1)/(r))^(N)m_(p)vecr_(p))/(sum_(p=1)^(N)m_(p))`
    D
    `(sum_((p=1)/(r))^(N)m_(p)vecr_(p))/(sum_(p=1)^(N)r_(p))`

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