Centre of mass of three particles of masses `1 kg, 2 kg and 3 kg` lies at the point `(1,2,3)` and centre of mass of another system of particles `3 kg and 2 kg` lies at the point `(-1, 3, -2)`. Where should we put a particle of mass `5 kg` so that the centre of mass of entire system lies at the centre of mass of first system ?

Options

A. (0, 0, 0)
B. (1, 3, 2)
C. (-1, 2, 3)
D. (3, 1, 8)

Explanation

To find the position where we should place a particle of mass \(5 \, \text{kg}\) so that the center of mass of the entire system lies at the center of mass of the first system, we can follow these steps:

### Step 1: Understand the Center of Mass
The center of mass (CM) of a system of particles can be calculated using 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: Define the Given Data
1. For the first system of particles (masses \(1 \, \text{kg}\), \(2 \, \text{kg}\), and \(3 \, \text{kg}\)):
- Center of mass is at \((1, 2, 3)\).
- Total mass \(M_1 = 1 + 2 + 3 = 6 \, \text{kg}\).

2. For the second system of particles (masses \(3 \, \text{kg}\) and \(2 \, \text{kg}\)):
- Center of mass is at \((-1, 3, -2)\).
- Total mass \(M_2 = 3 + 2 = 5 \, \text{kg}\).

### Step 3: Set Up the Equation for the New Center of Mass
Let the position of the \(5 \, \text{kg}\) particle be \((x, y, z)\). The total mass of the entire system will be:
\[ M = 6 + 5 = 11 \, \text{kg} \]

We want the center of mass of the entire system to be at \((1, 2, 3)\). Therefore, we can set up the equations for the center of mass in each coordinate direction.

### Step 4: Calculate the Center of Mass in the X-direction
Using the formula for the center of mass in the x-direction:
\[ \frac{(6 \cdot 1) + (5 \cdot (-1)) + (5 \cdot x)}{11} = 1 \]
This simplifies to:
\[ \frac{6 - 5 + 5x}{11} = 1 \]
\[ 6 - 5 + 5x = 11 \]
\[ 5x = 10 \implies x = 2 \]

### Step 5: Calculate the Center of Mass in the Y-direction
Using the formula for the center of mass in the y-direction:
\[ \frac{(6 \cdot 2) + (5 \cdot 3) + (5 \cdot y)}{11} = 2 \]
This simplifies to:
\[ \frac{12 + 15 + 5y}{11} = 2 \]
\[ 12 + 15 + 5y = 22 \]
\[ 5y = -5 \implies y = -1 \]

### Step 6: Calculate the Center of Mass in the Z-direction
Using the formula for the center of mass in the z-direction:
\[ \frac{(6 \cdot 3) + (5 \cdot (-2)) + (5 \cdot z)}{11} = 3 \]
This simplifies to:
\[ \frac{18 - 10 + 5z}{11} = 3 \]
\[ 18 - 10 + 5z = 33 \]
\[ 5z = 25 \implies z = 5 \]

### Final Result
The position where we should place the \(5 \, \text{kg}\) particle is:
\[ (x, y, z) = (2, -1, 5) \]

### Summary
The coordinates where the \(5 \, \text{kg}\) particle should be placed are \((2, -1, 5)\).

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