Two particles whose masses are 30 kg and 10 kg and their position vectors are −ı^−ȷ^−k^ and i^+j^+k^ respectively would have the centre of mass at position :

∵ R→cm=m1r→1+m2r→2m1+m2=10(i^+j^+k^)+30(−i^−j^−k^)10+30=10i^+10j^+10k^−30i^−30j^−30k^40=−20i^−20j^−20k^40R→cm=−(i^+j^+k^)2

Two particles whose masses are 30 kg and 10 kg and their position vectors are −ı^−ȷ^−k^ and i^+j^+k^ respectively would have the centre of mass at position :

∵ R→cm=m1r→1+m2r→2m1+m2=10(i^+j^+k^)+30(−i^−j^−k^)10+30=10i^+10j^+10k^−30i^−30j^−30k^40=−20i^−20j^−20k^40R→cm=−(i^+j^+k^)2

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Two particles whose masses are 30 kg and 10 kg and their position vectors are $- \hat{ı} - \hat{ȷ} - \hat{k}$ and $\hat{i} + \hat{j} + \hat{k}$ respectively would have the centre of mass at position :

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$(\hat{ı} + \hat{ȷ} + \hat{k}) / 2$

$- (\hat{ı} + \hat{ȷ} + \hat{k}) / 4$

$\hat{ı} + \hat{ȷ} + \hat{k} / 2$

$- (\hat{ı} + \hat{ȷ} + \hat{k}) / 2$

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Detailed Solution

$∵ {\vec{R}}_{cm} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$

$= \frac{10 (\hat{i} + \hat{j} + \hat{k}) + 30 (- \hat{i} - \hat{j} - \hat{k})}{10 + 30}$

$= \frac{10 \hat{i} + 10 \hat{j} + 10 \hat{k} - 30 \hat{i} - 30 \hat{j} - 30 \hat{k}}{40}$

$= \frac{- 20 \hat{i} - 20 \hat{j} - 20 \hat{k}}{40}$

${\vec{R}}_{c m} = \frac{- (\hat{i} + \hat{j} + \hat{k})}{2}$

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Two particles whose masses are 30 kg and 10 kg and their position vectors are −ı^−ȷ^−k^ and i^+j^+k^ respectively would have the centre of mass at position :

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Two particles whose masses are 30 kg and 10 kg and their position vectors are $- \hat{ı} - \hat{ȷ} - \hat{k}$ and $\hat{i} + \hat{j} + \hat{k}$ respectively would have the centre of mass at position :