If the centre of mass of three particles of masses of `1kg, 2kg, 3kg` is at `(2,2,2)`, then where should a fourth particle of mass `4kg` be placed so that the combined centre of mass may be at `(0,0,0).`

Options

A. (-3,-3,-3)
B. (3,3,3)
C. (-1,-2,-3)
D. NONE OF THESE

Explanation

To find the position where the fourth particle of mass `4 kg` should be placed so that the combined center of mass of all four particles is at `(0,0,0)`, we can follow these steps:

### Step 1: Understand the center of mass formula
The center of mass (CM) of a system of particles is given by the formula:

\[ \text{CM} = \frac{\sum m_i \vec{r_i}}{\sum m_i} \]

where \( m_i \) is the mass of each particle and \( \vec{r_i} \) is the position vector of each particle.

### Step 2: Calculate the total mass
We have three particles with masses:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 2 \, \text{kg} \)
- \( m_3 = 3 \, \text{kg} \)

The total mass of these three particles is:

\[ M = m_1 + m_2 + m_3 = 1 + 2 + 3 = 6 \, \text{kg} \]

### Step 3: Find the position vector of the center of mass of the first three particles
The center of mass of the first three particles is given as \( (2, 2, 2) \). Therefore, we can express the position vector of the center of mass as:

\[ \vec{R}_{\text{CM}} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k} \]

### Step 4: Set up the equation for the combined center of mass
Let the position vector of the fourth particle (mass \( 4 \, \text{kg} \)) be \( \vec{R} = (x, y, z) \). The total mass of the system with the fourth particle is:

\[ M_{\text{total}} = 6 + 4 = 10 \, \text{kg} \]

We want the combined center of mass to be at the origin \( (0, 0, 0) \). Therefore, we can set up the equation:

\[ \frac{(6 \cdot (2 \hat{i} + 2 \hat{j} + 2 \hat{k}) + 4 \cdot (x \hat{i} + y \hat{j} + z \hat{k}))}{10} = 0 \]

### Step 5: Simplify the equation
Multiplying both sides by \( 10 \) gives:

\[ 6 \cdot (2 \hat{i} + 2 \hat{j} + 2 \hat{k}) + 4 \cdot (x \hat{i} + y \hat{j} + z \hat{k}) = 0 \]

Calculating the left-hand side:

\[ 12 \hat{i} + 12 \hat{j} + 12 \hat{k} + 4x \hat{i} + 4y \hat{j} + 4z \hat{k} = 0 \]

### Step 6: Set up equations for each component
This gives us three equations:

1. \( 12 + 4x = 0 \)
2. \( 12 + 4y = 0 \)
3. \( 12 + 4z = 0 \)

### Step 7: Solve for \( x, y, z \)
From the first equation:

\[ 4x = -12 \implies x = -3 \]

From the second equation:

\[ 4y = -12 \implies y = -3 \]

From the third equation:

\[ 4z = -12 \implies z = -3 \]

### Step 8: Conclusion
The position of the fourth particle should be at:

\[ \vec{R} = (-3, -3, -3) \]

Thus, the fourth particle of mass `4 kg` should be placed at the coordinates \( (-3, -3, -3) \).

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