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 `alpha(hati+2hatj+3hatk)`, where `alpha` is a constant. The value of `alpha` is

Options

A. `(10)/(3)`
B. `(5)/(2)`
C. `(1)/(2)`
D. `(2)/(5)`

Answer

To find the value of \( \alpha \) in the position vector of the fourth particle, we will follow these steps:

### Step 1: Understand the System
We have three particles with masses:
- \( m_1 = 1 \, \text{g} \)
- \( m_2 = 2 \, \text{g} \)
- \( m_3 = 3 \, \text{g} \)

The center of mass (COM) of these three particles is taken as the origin of the coordinate system, which means their combined position vector is zero.

### Step 2: Calculate the Center of Mass of the First Three Particles
The formula for the center of mass \( \vec{R}_{\text{COM}} \) of a system of particles is given by:
\[ \vec{R}_{\text{COM}} = \frac{1}{M} \sum_{i=1}^{n} m_i \vec{r}_i \]
where \( M \) is the total mass and \( \vec{r}_i \) are the position vectors of the particles.

For our three particles, the total mass \( M \) is:
\[ M = m_1 + m_2 + m_3 = 1 + 2 + 3 = 6 \, \text{g} \]

Since the COM is at the origin, we have:
\[ \frac{1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3}{6} = 0 \]
This simplifies to:
\[ x_1 + 2x_2 + 3x_3 = 0 \quad (1) \]

### Step 3: Introduce the Fourth Particle
Now, we introduce a fourth particle with mass \( m_4 = 4 \, \text{g} \) and position vector \( \vec{r}_4 = \alpha (\hat{i} + 2\hat{j} + 3\hat{k}) \).

### Step 4: Calculate the New Center of Mass
The new center of mass for the four particles is given by:
\[ \vec{R}_{\text{COM}} = \frac{1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3 + 4 \cdot x_4}{M_{\text{total}}} \]
where \( M_{\text{total}} = 1 + 2 + 3 + 4 = 10 \, \text{g} \).

We want this new center of mass to be at the point \( (1, 2, 3) \):
\[ \frac{1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3 + 4 \cdot x_4}{10} = (1, 2, 3) \]

### Step 5: Set Up the Equations
From the x-coordinate:
\[ \frac{0 + 4 \cdot \alpha}{10} = 1 \]
This simplifies to:
\[ 4\alpha = 10 \quad \Rightarrow \quad \alpha = \frac{10}{4} = \frac{5}{2} \quad (2) \]

For the y-coordinate:
\[ \frac{0 + 4 \cdot 2\alpha}{10} = 2 \]
This simplifies to:
\[ 8\alpha = 20 \quad \Rightarrow \quad \alpha = \frac{20}{8} = \frac{5}{2} \]

For the z-coordinate:
\[ \frac{0 + 4 \cdot 3\alpha}{10} = 3 \]
This simplifies to:
\[ 12\alpha = 30 \quad \Rightarrow \quad \alpha = \frac{30}{12} = \frac{5}{2} \]

### Conclusion
In all three cases, we find that:
\[ \alpha = \frac{5}{2} \]

### Final Answer
Thus, the value of \( \alpha \) is \( \frac{5}{2} \).

---