The coordinates of the positions of particles | vedclass

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Question

(C) Given masses are $m_1 = 7 \, g$,$m_2 = 4 \, g$,and $m_3 = 10 \, g$.Their position vectors are $\vec{r}_1 = (1\hat{i} + 5\hat{j} - 3\hat{k}) \, cm$

Vedclass · Accepted Answer

$\left( \frac{15}{7}, \frac{85}{21}, -\frac{1}{7} \right) \, cm$

Vedclass · Answer

(C) Given masses are $m_1 = 7 \, g$,$m_2 = 4 \, g$,and $m_3 = 10 \, g$.Their position vectors are $\vec{r}_1 = (1\hat{i} + 5\hat{j} - 3\hat{k}) \, cm$,$\vec{r}_2 = (2\hat{i} + 5\hat{j} + 7\hat{k}) \, cm$,and $\vec{r}_3 = (3\hat{i} + 3\hat{j} - 1\hat{k}) \, cm$.The position of the centre of mass $\vec{R}_{cm}$ is given by $\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$.Substituting the values:$\vec{R}_{cm} = \frac{7(1\hat{i} + 5\hat{j} - 3\hat{k}) + 4(2\hat{i} + 5\hat{j} + 7\hat{k}) + 10(3\hat{i} + 3\hat{j} - 1\hat{k})}{7 + 4 + 10}$$\vec{R}_{cm} = \frac{(7\hat{i} + 35\hat{j} - 21\hat{k}) + (8\hat{i} + 20\hat{j} + 28\hat{k}) + (30\hat{i} + 30\hat{j} - 10\hat{k})}{21}$$\vec{R}_{cm} = \frac{(7+8+30)\hat{i} + (35+20+30)\hat{j} + (-21+28-10)\hat{k}}{21}$$\vec{R}_{cm} = \frac{45\hat{i} + 85\hat{j} - 3\hat{k}}{21} = \frac{45}{21}\hat{i} + \frac{85}{21}\hat{j} - \frac{3}{21}\hat{k}$$\vec{R}_{cm} = \frac{15}{7}\hat{i} + \frac{85}{21}\hat{j} - \frac{1}{7}\hat{k}$.Thus,the coordinates are $\left( \frac{15}{7}, \frac{85}{21}, -\frac{1}{7} \right) \, cm$.

The coordinates of the positions of particles of mass $7 \, g$,$4 \, g$,and $10 \, g$ are $(1, 5, -3) \, cm$,$(2, 5, 7) \, cm$,and $(3, 3, -1) \, cm$ respectively. The position of the centre of mass of the system would be:

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