Particles of masses `1kg` and `3kg` are at `2i + 5j + 13k)m` and `(-6i + 4j - 2k)m` then instantaneous position of their centre of mass is

Options

A. `1/4(-16i+17j+7k)m`
B. `1/4(-8i+17j+7k)m`
C. `1/4(-6i+17j+7k)m`
D. `1/4(-6i+17j+5k)m`

Explanation

To find the instantaneous position of the center of mass of two particles, we can use the formula for the center of mass (R_cm) given by:

\[ R_{cm} = \frac{m_1 \cdot r_1 + m_2 \cdot r_2}{m_1 + m_2} \]

where:
- \( m_1 \) and \( m_2 \) are the masses of the particles,
- \( r_1 \) and \( r_2 \) are their respective position vectors.

### Step 1: Identify the given values
- Mass of the first particle, \( m_1 = 1 \, \text{kg} \)
- Position vector of the first particle, \( r_1 = 2i + 5j + 13k \)
- Mass of the second particle, \( m_2 = 3 \, \text{kg} \)
- Position vector of the second particle, \( r_2 = -6i + 4j - 2k \)

### Step 2: Substitute the values into the center of mass formula
Substituting the known values into the formula:

\[ R_{cm} = \frac{1 \cdot (2i + 5j + 13k) + 3 \cdot (-6i + 4j - 2k)}{1 + 3} \]

### Step 3: Calculate the numerator
Calculating each term in the numerator:

1. For the first particle:
\[ 1 \cdot (2i + 5j + 13k) = 2i + 5j + 13k \]

2. For the second particle:
\[ 3 \cdot (-6i + 4j - 2k) = -18i + 12j - 6k \]

Now, combine these results:
\[ R_{cm} = \frac{(2i + 5j + 13k) + (-18i + 12j - 6k)}{4} \]

### Step 4: Simplify the numerator
Combine the vectors:
\[ = \frac{(2 - 18)i + (5 + 12)j + (13 - 6)k}{4} \]
\[ = \frac{-16i + 17j + 7k}{4} \]

### Step 5: Final expression for the center of mass
Now, divide each component by 4:
\[ R_{cm} = -4i + \frac{17}{4}j + \frac{7}{4}k \]

### Conclusion
Thus, the instantaneous position of the center of mass is:
\[ R_{cm} = \frac{1}{4}(-16i + 17j + 7k) \]

### Final Answer
The correct answer is option A:
\[ \frac{1}{4}(-16i + 17j + 7k) \]