Two blocks of mass `1kg` and `3 kg` have position v ectors ` hat(i) + 2 hat(j) + hat(k)` and `3 hat(i) - 2 hat(j) + hat(k)` , respectively . The center of mass of this system has a position vector.

Options

A. `- 2 hat(i) + 2 hat(k)`
B. `-2 hat(i) - hat(j) + hat(k)`
C. `2.5 hat(i) - hat(j) - hat(k)`
D. `- hat(i) + hat(j) + hat(k)`

Explanation

To find the center of mass of the two blocks with given masses and position vectors, we can follow these steps:

### Step 1: Identify the masses and position vectors
- Mass of block 1, \( m_1 = 1 \, \text{kg} \)
- Position vector of block 1, \( \vec{r_1} = \hat{i} + 2\hat{j} + \hat{k} \)
- Mass of block 2, \( m_2 = 3 \, \text{kg} \)
- Position vector of block 2, \( \vec{r_2} = 3\hat{i} - 2\hat{j} + \hat{k} \)

### Step 2: Use the formula for the center of mass
The formula for the center of mass \( \vec{R} \) of a system of particles is given by:

\[ \vec{R} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} \]

### Step 3: Substitute the values into the formula
Substituting the values we have:

\[ \vec{R} = \frac{1 \cdot (\hat{i} + 2\hat{j} + \hat{k}) + 3 \cdot (3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3} \]

### Step 4: Calculate the numerator
Calculating \( m_1 \vec{r_1} + m_2 \vec{r_2} \):

\[ = 1 \cdot (\hat{i} + 2\hat{j} + \hat{k}) + 3 \cdot (3\hat{i} - 2\hat{j} + \hat{k}) \]

Calculating each term:

- For \( m_1 \vec{r_1} \):
\[ = \hat{i} + 2\hat{j} + \hat{k} \]

- For \( m_2 \vec{r_2} \):
\[ = 3 \cdot 3\hat{i} + 3 \cdot (-2\hat{j}) + 3 \cdot \hat{k} = 9\hat{i} - 6\hat{j} + 3\hat{k} \]

Now, adding these vectors together:

\[ \hat{i} + 2\hat{j} + \hat{k} + 9\hat{i} - 6\hat{j} + 3\hat{k} = (1 + 9)\hat{i} + (2 - 6)\hat{j} + (1 + 3)\hat{k} \]

This simplifies to:

\[ 10\hat{i} - 4\hat{j} + 4\hat{k} \]

### Step 5: Calculate the total mass
The total mass \( m_1 + m_2 \) is:

\[ 1 + 3 = 4 \, \text{kg} \]

### Step 6: Divide the result by the total mass
Now, we can find the center of mass:

\[ \vec{R} = \frac{10\hat{i} - 4\hat{j} + 4\hat{k}}{4} \]

This simplifies to:

\[ \vec{R} = \frac{10}{4}\hat{i} - \frac{4}{4}\hat{j} + \frac{4}{4}\hat{k} = 2.5\hat{i} - 1\hat{j} + 1\hat{k} \]

### Final Result
Thus, the position vector of the center of mass is:

\[ \vec{R} = 2.5\hat{i} - 1\hat{j} + 1\hat{k} \]

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