Two bodies of mass `1 kg` and `3 kg` have position vectors `hat i+ 2 hat j + hat k` and `- 3 hat i- 2 hat j+ hat k`, respectively. The centre of mass of this system has a position vector.

Options

A. `-2hat(i)-hat(j)+hat(k)`
B. `2hat(i)-hat(j)-2hat(k)`
C. `-hat(i)+hat(j)+hat(k)`
D. `-2hat(i)+2hat(k)`

Answer

To find the center of mass of the two bodies, we will use the formula for the center of mass (R_cm) of a system of particles:

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

### Step-by-Step Solution:

1. **Identify the masses and position vectors**:
- Mass \( m_1 = 1 \, \text{kg} \) with position vector \( \mathbf{r}_1 = \hat{i} + 2\hat{j} + \hat{k} \)
- Mass \( m_2 = 3 \, \text{kg} \) with position vector \( \mathbf{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k} \)

2. **Substitute the values into the center of mass formula**:
\[ \mathbf{R}_{cm} = \frac{1 \cdot (\hat{i} + 2\hat{j} + \hat{k}) + 3 \cdot (-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3} \]

3. **Calculate the numerator**:
- First, calculate \( 1 \cdot (\hat{i} + 2\hat{j} + \hat{k}) \):
\[ = \hat{i} + 2\hat{j} + \hat{k} \]
- Then calculate \( 3 \cdot (-3\hat{i} - 2\hat{j} + \hat{k}) \):
\[ = -9\hat{i} - 6\hat{j} + 3\hat{k} \]
- Now combine these results:
\[ \hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k} = (-8\hat{i} - 4\hat{j} + 4\hat{k}) \]

4. **Calculate the denominator**:
\[ m_1 + m_2 = 1 + 3 = 4 \]

5. **Combine the results**:
\[ \mathbf{R}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} \]

6. **Simplify the expression**:
\[ \mathbf{R}_{cm} = -2\hat{i} - \hat{j} + \hat{k} \]

### Final Answer:
The position vector of the center of mass is:
\[ \mathbf{R}_{cm} = -2\hat{i} - \hat{j} + \hat{k} \]