Particles of masses 2 Kg and 3 Kg are at (2i^+3j^+4k^) and (4i^+2j^+3k^) respectively. Then position of centre of mass of those two is

m1=2kg,m2=3kg r1¯=2i^+3j^+4k^, r2¯=4i^+2j^+3k^ rcm¯=m1r1+m2r2m1+m2=2(2i^+3j^+4k^)+3(4i^+2j^+3k^)2+3 =4i^+6j^+8k^+12i^+6j^+9k^5=16i^+12j^+17k^5

Particles of masses 2 Kg and 3 Kg are at (2i^+3j^+4k^) and (4i^+2j^+3k^) respectively. Then position of centre of mass of those two is

m1=2kg,m2=3kg r1¯=2i^+3j^+4k^, r2¯=4i^+2j^+3k^ rcm¯=m1r1+m2r2m1+m2=2(2i^+3j^+4k^)+3(4i^+2j^+3k^)2+3 =4i^+6j^+8k^+12i^+6j^+9k^5=16i^+12j^+17k^5

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Particles of masses $2 K g$ and $3 K g$ are at $(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ and $(4 \hat{i} + 2 \hat{j} + 3 \hat{k})$ respectively. Then position of centre of mass of those two is

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$\frac{17 \hat{i} + 12 \hat{j} + 16 \hat{k}}{5}$

$\frac{16 \hat{i} + 17 \hat{j} + 12 \hat{k}}{5}$

$\frac{16 \hat{i} + 12 \hat{j} + 17 \hat{k}}{5}$

$\frac{12 \hat{i} + 16 \hat{j} + 17 \hat{k}}{5}$

answer is C.

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Detailed Solution

$m_{1} = 2 k g, m_{2} = 3 k g$
$\bar{r_{1}} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \bar{r_{2}} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\bar{r_{c m}} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}} = \frac{2 (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + 3 (4 \hat{i} + 2 \hat{j} + 3 \hat{k})}{2 + 3}$
$= \frac{4 \hat{i} + 6 \hat{j} + 8 \hat{k} + 12 \hat{i} + 6 \hat{j} + 9 \hat{k}}{5} = \frac{16 \hat{i} + 12 \hat{j} + 17 \hat{k}}{5}$