Two bodies of mass 1,kg and 3,kg have positio | vedclass

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(C) Given masses are $m_{1} = 1\,kg$ and $m_{2} = 3\,kg$.Position vectors are $\vec{r}_{1} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{r}_{2} = -3\hat{i

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

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(C) Given masses are $m_{1} = 1\,kg$ and $m_{2} = 3\,kg$.Position vectors are $\vec{r}_{1} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{r}_{2} = -3\hat{i} - 2\hat{j} + \hat{k}$.The position vector of the centre of mass $\vec{r}_{cm}$ is given by the formula $\vec{r}_{cm} = \frac{m_{1}\vec{r}_{1} + m_{2}\vec{r}_{2}}{m_{1} + m_{2}}$.Substituting the values: $\vec{r}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$.Simplifying the numerator: $\vec{r}_{cm} = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4}$.Combining like terms: $\vec{r}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$.Dividing by $4$: $\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$.

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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