The centre of mass of a system of three particles of masses $1 g$ , $2 g$ and $3 g$ is taken as the origin of a coordinate system. The position vector of a fourth particle of mass $4 g$ such that the centre of mass of the four particle system lies at the point $(1, 2, 3)$ is $α (i^{\land} + 2 j^{\land} + k^{\land}),$ where $α$ is a constant. The value of $α$ is :-

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$\frac{5}{2}$

$\frac{2}{5}$

$\frac{10}{3}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

$\begin{aligned} {\vec{r}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2} + m_{3} {\vec{r}}_{3}}{m_{1} + m_{2} + m_{3}} \\ {\vec{r}}_{c m} = 0 \\ \Rightarrow m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2} + m_{3} {\vec{r}}_{3} = 0 \\ {\vec{r}}_{c m}^{'} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2} + m_{3} {\vec{r}}_{3} + m_{4} {\vec{r}}_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} \\ (\hat{i} + 2 \hat{j} + 3 \hat{k}) = \frac{0 + 4 α (\hat{i} + 2 \hat{j} + 3 \hat{k})}{1 + 2 + 3 + 4} \Rightarrow α = \frac{5}{2} \end{aligned}$