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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
A
$-\hat{i} + \hat{j} + \hat{k}$
B
$-2\hat{i} + 2\hat{k}$
C
$-2\hat{i} - \hat{j} + \hat{k}$
D
$2\hat{i} - \hat{j} - 2\hat{k}$
Detailed Solution
$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$
$m_1\vec{r}_1 = 1(\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} + 2\hat{j} + \hat{k}$
$m_2\vec{r}_2 = 3(-3\hat{i} - 2\hat{j} + \hat{k}) = -9\hat{i} - 6\hat{j} + 3\hat{k}$
Sum $= -8\hat{i} - 4\hat{j} + 4\hat{k}$
$\vec{r}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$
$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$
$m_1\vec{r}_1 = 1(\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} + 2\hat{j} + \hat{k}$
$m_2\vec{r}_2 = 3(-3\hat{i} - 2\hat{j} + \hat{k}) = -9\hat{i} - 6\hat{j} + 3\hat{k}$
Sum $= -8\hat{i} - 4\hat{j} + 4\hat{k}$
$\vec{r}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$
$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$