Particles of masses 100 and 300 gram have position vectors `(2hat(i)+5hat(j)+13hat(k))` and `(-6hat(i)+4hat(j)+2hat(k))`. Position vector of their centre of mass is

Options

A. `-(16)/(4)hati+(17)/(4)hatj+(7)/(4)hatk`
B. `(20)/(4)hati+(17)/(4)hatj+(7)/(4)hatk`
C. `-(16)/(4)hati+(17)/(4)hatj+(19)/(4)hatk`
D. `-(16)/(4)hati+(13)/(4)hatj+(19)/(4)hatk`

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.

### Step 1: Identify the given values
- Mass of the first particle, \( m_1 = 100 \) grams
- Position vector of the first particle, \( \vec{r}_1 = 2\hat{i} + 5\hat{j} + 13\hat{k} \)
- Mass of the second particle, \( m_2 = 300 \) grams
- Position vector of the second particle, \( \vec{r}_2 = -6\hat{i} + 4\hat{j} + 2\hat{k} \)

### Step 2: Calculate the total mass
\[ m_{total} = m_1 + m_2 = 100 \, \text{g} + 300 \, \text{g} = 400 \, \text{g} \]

### Step 3: Calculate the numerator of the center of mass formula
\[ m_1 \vec{r}_1 + m_2 \vec{r}_2 = 100 \, \text{g} \cdot (2\hat{i} + 5\hat{j} + 13\hat{k}) + 300 \, \text{g} \cdot (-6\hat{i} + 4\hat{j} + 2\hat{k}) \]

Calculating each term:
- For the first particle:
\[ 100 \cdot (2\hat{i} + 5\hat{j} + 13\hat{k}) = 200\hat{i} + 500\hat{j} + 1300\hat{k} \]

- For the second particle:
\[ 300 \cdot (-6\hat{i} + 4\hat{j} + 2\hat{k}) = -1800\hat{i} + 1200\hat{j} + 600\hat{k} \]

### Step 4: Add the two results together
\[ m_1 \vec{r}_1 + m_2 \vec{r}_2 = (200\hat{i} - 1800\hat{i}) + (500\hat{j} + 1200\hat{j}) + (1300\hat{k} + 600\hat{k}) \]
\[ = -1600\hat{i} + 1700\hat{j} + 1900\hat{k} \]

### Step 5: Divide by the total mass to find the center of mass
\[ \vec{R}_{cm} = \frac{-1600\hat{i} + 1700\hat{j} + 1900\hat{k}}{400} \]
\[ = -4\hat{i} + 4.25\hat{j} + 4.75\hat{k} \]

### Final Result
The position vector of the center of mass is:
\[ \vec{R}_{cm} = -4\hat{i} + 4.25\hat{j} + 4.75\hat{k} \]