Two particles of masses 1 kg and 3 kg have position vectors `2hati+3hatj+4hatk and-2hati+3hatj-4hatk` respectively. The centre of mass has a position vector

Options

A. `hati-3hatj-2hatk`
B. `-hati-3hatj-2hatk`
C. `-hati+3hatj+2hatk`
D. `-hati+3hatj-2hatk`

Answer

To find the position vector of the center of mass of two particles, we can use the formula:

\[ \vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \]

where:
- \( m_1 \) and \( m_2 \) are the masses of the two particles,
- \( \vec{r}_1 \) and \( \vec{r}_2 \) are their respective position vectors.

### Given:
- Mass of particle 1, \( m_1 = 1 \, \text{kg} \)
- Position vector of particle 1, \( \vec{r}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
- Mass of particle 2, \( m_2 = 3 \, \text{kg} \)
- Position vector of particle 2, \( \vec{r}_2 = -2\hat{i} + 3\hat{j} - 4\hat{k} \)

### Step 1: Calculate \( m_1 \vec{r}_1 \) and \( m_2 \vec{r}_2 \)

\[ m_1 \vec{r}_1 = 1 \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k} \]

\[ m_2 \vec{r}_2 = 3 \cdot (-2\hat{i} + 3\hat{j} - 4\hat{k}) = -6\hat{i} + 9\hat{j} - 12\hat{k} \]

### Step 2: Add \( m_1 \vec{r}_1 \) and \( m_2 \vec{r}_2 \)

\[ m_1 \vec{r}_1 + m_2 \vec{r}_2 = (2\hat{i} + 3\hat{j} + 4\hat{k}) + (-6\hat{i} + 9\hat{j} - 12\hat{k}) \]

Combine the components:

\[ = (2 - 6)\hat{i} + (3 + 9)\hat{j} + (4 - 12)\hat{k} \]

\[ = -4\hat{i} + 12\hat{j} - 8\hat{k} \]

### Step 3: Divide by the total mass \( m_1 + m_2 \)

\[ m_1 + m_2 = 1 + 3 = 4 \, \text{kg} \]

Now, we find the position vector of the center of mass:

\[ \vec{R}_{cm} = \frac{-4\hat{i} + 12\hat{j} - 8\hat{k}}{4} \]

### Step 4: Simplify the expression

\[ \vec{R}_{cm} = -\hat{i} + 3\hat{j} - 2\hat{k} \]

### Final Result

The position vector of the center of mass is:

\[ \vec{R}_{cm} = -\hat{i} + 3\hat{j} - 2\hat{k} \]