Two bodies of mass $1 ext{ kg}$ and $3 ext{ kg}$ have position vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-3\

Question

Two bodies of mass $1 	ext{ kg}$ and $3 	ext{ kg}$ have position vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-3\hat{i}-2\hat{j}+\hat{k}$, respectively. The centre of mass of this system has a position vector:

Accepted Answer

The position vector of the centre of mass $\vec{R}_{CM}$ for a system of two particles is given by:
$\vec{R}_{CM} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$
Given $m_1 = 1 	ext{ kg}$, $\vec{r}_1 = \hat{i}+2\hat{j}+\hat{k}$ and $m_2 = 3 	ext{ kg}$, $\vec{r}_2 = -3\hat{i}-2\hat{j}+\hat{k}$.
Substituting the given values:
$\vec{R}_{CM} = \frac{1(\hat{i}+2\hat{j}+\hat{k}) + 3(-3\hat{i}-2\hat{j}+\hat{k})}{1 + 3}$
$\vec{R}_{CM} = \frac{\hat{i}+2\hat{j}+\hat{k} - 9\hat{i}-6\hat{j}+3\hat{k}}{4}$
$\vec{R}_{CM} = \frac{-8\hat{i}-4\hat{j}+4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k}$

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