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Find the position vector of the centre of mass of three particles of masses 1 kg, 2 kg and 3 kg located at r1 = (4i + 2j - 3k) m, r2 = (i - 4j + 2k) m and r3 = (2i - 2j + k) m.

1. 2i - 2j + 2/3 k
2. i - j + 2/3 k
3. 2i + 2j - 2/3 k
4. 4i - 4j + 4/3 k

Correct answer: 2i - 2j + 2/3 k

Solution

Weighted average gives x = 12/6 = 2, y = -12/6 = -2, z = 4/6 = 2/3, so R_cm = 2i - 2j + (2/3)k.

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