The centre of mass of three particles of masses `1kg, 2 kg` and `3kg` lies at the point (3m, 3m, 3m). Where should a fourth particle of mass 4 kg be positioned so that the centre of mass of the four particle system lies at the point (1m, 1m, 1m)?

Answer

To find the position of the fourth particle of mass 4 kg such that the center of mass of the four-particle system lies at the point (1m, 1m, 1m), we can follow these steps:

### Step 1: Understand the formula for the center of mass
The center of mass (CM) of a system of particles is given by the formula:
\[ \text{CM} = \frac{\sum m_i \cdot r_i}{\sum m_i} \]
where \( m_i \) is the mass of each particle and \( r_i \) is the position vector of each particle.

### Step 2: Identify the known values
We have three particles with the following masses and positions:
- Particle 1: Mass \( m_1 = 1 \, \text{kg} \), Position \( (3, 3, 3) \)
- Particle 2: Mass \( m_2 = 2 \, \text{kg} \), Position \( (3, 3, 3) \)
- Particle 3: Mass \( m_3 = 3 \, \text{kg} \), Position \( (3, 3, 3) \)

We want to find the position \( (x, y, z) \) of the fourth particle \( m_4 = 4 \, \text{kg} \) such that the center of mass of the system is at \( (1, 1, 1) \).

### Step 3: Set up the equations for the center of mass
The total mass of the system is:
\[ M = m_1 + m_2 + m_3 + m_4 = 1 + 2 + 3 + 4 = 10 \, \text{kg} \]

The x-coordinate of the center of mass can be expressed as:
\[ x_{CM} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2 + m_3 \cdot x_3 + m_4 \cdot x_4}{M} \]
Substituting the known values:
\[ 1 = \frac{1 \cdot 3 + 2 \cdot 3 + 3 \cdot 3 + 4 \cdot x}{10} \]

### Step 4: Solve for \( x \)
Calculating the left side:
\[ 1 = \frac{3 + 6 + 9 + 4x}{10} \]
\[ 1 = \frac{18 + 4x}{10} \]
Multiplying both sides by 10:
\[ 10 = 18 + 4x \]
Rearranging gives:
\[ 4x = 10 - 18 \]
\[ 4x = -8 \]
\[ x = -2 \]

### Step 5: Repeat for y and z coordinates
Since the positions of the first three particles are the same for y and z, the calculations will be identical:
\[ y_{CM} = \frac{m_1 \cdot y_1 + m_2 \cdot y_2 + m_3 \cdot y_3 + m_4 \cdot y_4}{M} \]
\[ 1 = \frac{1 \cdot 3 + 2 \cdot 3 + 3 \cdot 3 + 4 \cdot y}{10} \]
Following the same steps as above, we find:
\[ y = -2 \]
And similarly for z:
\[ z = -2 \]

### Final Position of the Fourth Particle
Thus, the position of the fourth particle should be:
\[ (x, y, z) = (-2, -2, -2) \]