Two particles of mass $1 k g$ and $3 k g$ have position vectors $2 ¯ i + 3 ¯ j + 4 ¯ ¯ ¯ k$ and $- 2 ¯ i + 3 ¯ j - 4 ¯ ¯ ¯ k$ respectively. The position vector of centre of mass of the system is:

- ¯ i + 3 ¯ j - 2 ¯ ¯ ¯ k

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- ¯ i + 3 ¯ j + 2 ¯ ¯ ¯ k

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- ¯ i - 3 ¯ j - 2 ¯ ¯ ¯ k

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¯ i + 3 ¯ j - 2 ¯ ¯ ¯ k

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Solution

The correct option is A $- ¯ i + 3 ¯ j - 2 ¯ ¯ ¯ k$
We know that the position of center of mass is given by the formula,

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$ where, $x_{1}$ and $x_{2}$ are position vectors of the two masses.

Thus, in this case $x_{c m} = \frac{1 (2 ¯ i + 3 ¯ j + 4 ¯ ¯ ¯ k) + 3 (- 2 ¯ i + 3 ¯ j - 4 ¯ ¯ ¯ k)}{1 + 3} = - ¯ i + 3 ¯ j - 2 ¯ ¯ ¯ k$

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\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 1 and \vec{r} \cdot (- \hat{i} + \hat{j}) = 4

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