Two bodies of mass 1 kg and 3 kg have posit | vedclass

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(A) Given:$m_1 = 1 \ kg$,$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$$m_2 = 3 \ kg$,$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$The position vector of t

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$-2\hat{i} - \hat{j} + \hat{k}$

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

Two bodies of mass $1 \ kg$ and $3 \ kg$ 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:

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