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Particles of masses 1kg and 3kg are at $(2 i + 5 j + 13 k) m$ and $(- 6 i + 4 j - 2 k) m$ then instantaneous position of their centre of mass is

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$\frac{1}{4} (- 16 i + 17 j + 7 k) m$

$\frac{1}{4} (- 8 i + 17 j + 7 k) m$

$\frac{1}{4} (- 6 i + 17 j + 7 k) m$

$\frac{1}{4} (- 6 i + 17 j + 5 k) m$

answer is A.

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Detailed Solution

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

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