Two particles whose masses are 10 kg and 30 kg and their position vectors are `hati+hatj+hatk and -hati-hatj-hatk` respectively would have the centre of mass at position :-

Options

A. `-((hati+hatj+hatk)/2)`
B. `((hati+hatj+hatk)/2)`
C. `-((hati+hatj+hatk)/4)`
D. `((hati+hatj+hatk)/4)`

Answer

To find the center of mass of two particles with given masses and position vectors, we can use the formula for the center of mass (COM):

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

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

### Step-by-Step Solution:

1. **Identify the masses and position vectors:**
- Mass of particle 1, \(m_1 = 10 \, \text{kg}\)
- Position vector of particle 1, \(\vec{r}_1 = \hat{i} + \hat{j} + \hat{k}\)
- Mass of particle 2, \(m_2 = 30 \, \text{kg}\)
- Position vector of particle 2, \(\vec{r}_2 = -\hat{i} - \hat{j} - \hat{k}\)

2. **Substitute the values into the COM formula:**
\[ \vec{R}_{cm} = \frac{10(\hat{i} + \hat{j} + \hat{k}) + 30(-\hat{i} - \hat{j} - \hat{k})}{10 + 30} \]

3. **Calculate the numerator:**
- Calculate \(10(\hat{i} + \hat{j} + \hat{k}) = 10\hat{i} + 10\hat{j} + 10\hat{k}\)
- Calculate \(30(-\hat{i} - \hat{j} - \hat{k}) = -30\hat{i} - 30\hat{j} - 30\hat{k}\)
- Now combine these:
\[ 10\hat{i} + 10\hat{j} + 10\hat{k} - 30\hat{i} - 30\hat{j} - 30\hat{k} = (10 - 30)\hat{i} + (10 - 30)\hat{j} + (10 - 30)\hat{k} = -20\hat{i} - 20\hat{j} - 20\hat{k} \]

4. **Calculate the denominator:**
\[ m_1 + m_2 = 10 + 30 = 40 \]

5. **Combine the results:**
\[ \vec{R}_{cm} = \frac{-20\hat{i} - 20\hat{j} - 20\hat{k}}{40} \]
\[ \vec{R}_{cm} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k} \]

6. **Final result:**
\[ \vec{R}_{cm} = -\frac{1}{2}(\hat{i} + \hat{j} + \hat{k}) \]

### Answer:
The center of mass position vector is:
\[ \vec{R}_{cm} = -\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} - \frac{1}{2} \hat{k} \]