The position vectors of three particles of mass `m_(1) = 1kg, m_(2) = 2kg and m_(3) = 3kg` are `r_(1) = (hat(i) + 4hat(j) +hat(k))`m, `r_(2) = (hat(i)+hat(j)+hat(k))`m and `r_(3) = (2hat(i) - hat(j) -hat2k)`m, respectively. Find the position vector of their center of mass.

Options

A. `r_(cm) =1/2(3i + j -2k)m`
B. `r_(cm) =1/2(3i + 2j -k)m`
C. `r_(cm) =1/2(3i + 3j -k)m`
D. `r_(cm) =1/2(3i + j -k)m`

Explanation

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

\[ \mathbf{r}_{cm} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_1 + m_2 + m_3} \]

### Step 1: Identify the given values
- Masses:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 2 \, \text{kg} \)
- \( m_3 = 3 \, \text{kg} \)

- Position vectors:
- \( \mathbf{r}_1 = \hat{i} + 4\hat{j} + \hat{k} \)
- \( \mathbf{r}_2 = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{r}_3 = 2\hat{i} - \hat{j} - 2\hat{k} \)

### Step 2: Calculate the total mass
\[ m_{total} = m_1 + m_2 + m_3 = 1 + 2 + 3 = 6 \, \text{kg} \]

### Step 3: Calculate the weighted sum of the position vectors
\[ m_1 \mathbf{r}_1 = 1 \cdot (\hat{i} + 4\hat{j} + \hat{k}) = \hat{i} + 4\hat{j} + \hat{k} \]
\[ m_2 \mathbf{r}_2 = 2 \cdot (\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + 2\hat{k} \]
\[ m_3 \mathbf{r}_3 = 3 \cdot (2\hat{i} - \hat{j} - 2\hat{k}) = 6\hat{i} - 3\hat{j} - 6\hat{k} \]

### Step 4: Sum the weighted position vectors
\[ m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3 = (\hat{i} + 4\hat{j} + \hat{k}) + (2\hat{i} + 2\hat{j} + 2\hat{k}) + (6\hat{i} - 3\hat{j} - 6\hat{k}) \]

Combine the components:
- \( \hat{i} \) components: \( 1 + 2 + 6 = 9 \)
- \( \hat{j} \) components: \( 4 + 2 - 3 = 3 \)
- \( \hat{k} \) components: \( 1 + 2 - 6 = -3 \)

So, the total weighted sum is:
\[ 9\hat{i} + 3\hat{j} - 3\hat{k} \]

### Step 5: Calculate the position vector of the center of mass
\[ \mathbf{r}_{cm} = \frac{9\hat{i} + 3\hat{j} - 3\hat{k}}{6} \]

Breaking it down:
\[ \mathbf{r}_{cm} = \frac{9}{6}\hat{i} + \frac{3}{6}\hat{j} - \frac{3}{6}\hat{k} = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k} \]

### Final Answer
The position vector of the center of mass is:
\[ \mathbf{r}_{cm} = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k} \]

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