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Centre of mass of two particles with masses are $3 k g$ and $4 k g$ located at $(1, 0, 1)$ and $(2, 2, 0)$ has the co-ordinates of

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$\frac{11}{7}, \frac{8}{7}, \frac{3}{7}$

$\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$

$\frac{2}{3}, \frac{2}{3}, \frac{4}{3}$

$\frac{4}{3}, \frac{2}{3}, \frac{2}{3}$

answer is B.

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Detailed Solution

$x c m = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} = \frac{3 (1) + 4 (2)}{7}$

$= \frac{3 + 8}{7} = 11 / 7$

$y c m = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} = \frac{3 (0) + 4 (2)}{7}$

$= \frac{8}{7}$

$z c m = \frac{m_{1} z_{1} + m_{2} z_{2}}{m_{1} + z_{2}} = \frac{3 (1) + 4 (0)}{7}$ $= \frac{3}{7}$

$(x c m, y c m, z c m) = (\frac{11}{7}, \frac{8}{7}, \frac{3}{7})$

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