For any vectors `veca and vecb, (veca xx hati) + (vecb xx hati) + ( veca xx hatj) . (vecb xx hatj) + (veca xx hatk ) .(vecb xx hatk)` is always equal to

Options

A. `veca. vecb`
B. ` 2 veca. Vecb`
C. zero
D. none of these

Explanation

To solve the expression \((\vec{a} \times \hat{i}) + (\vec{b} \times \hat{i}) + (\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j}) + (\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k})\), we will follow these steps:

### Step 1: Calculate \((\vec{a} \times \hat{i}) \cdot (\vec{b} \times \hat{i})\)
Using the vector triple product identity:
\[ \vec{a} \times \hat{i} \cdot \vec{b} \times \hat{i} = (\vec{a} \cdot \vec{b})(\hat{i} \cdot \hat{i}) - (\vec{a} \cdot \hat{i})(\vec{b} \cdot \hat{i}) \]
Since \(\hat{i} \cdot \hat{i} = 1\), this simplifies to:
\[ \vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{i})(\vec{b} \cdot \hat{i}) \]

### Step 2: Calculate \((\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j})\)
Using the same identity:
\[ \vec{a} \times \hat{j} \cdot \vec{b} \times \hat{j} = (\vec{a} \cdot \vec{b})(\hat{j} \cdot \hat{j}) - (\vec{a} \cdot \hat{j})(\vec{b} \cdot \hat{j}) \]
This simplifies to:
\[ \vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{j})(\vec{b} \cdot \hat{j}) \]

### Step 3: Calculate \((\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k})\)
Using the same identity:
\[ \vec{a} \times \hat{k} \cdot \vec{b} \times \hat{k} = (\vec{a} \cdot \vec{b})(\hat{k} \cdot \hat{k}) - (\vec{a} \cdot \hat{k})(\vec{b} \cdot \hat{k}) \]
This simplifies to:
\[ \vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{k})(\vec{b} \cdot \hat{k}) \]

### Step 4: Combine all parts
Now, we add the results from Steps 1, 2, and 3:
\[ (\vec{a} \times \hat{i}) \cdot (\vec{b} \times \hat{i}) + (\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j}) + (\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k}) \]
This leads to:
\[ \left(\vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{i})(\vec{b} \cdot \hat{i})\right) + \left(\vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{j})(\vec{b} \cdot \hat{j})\right) + \left(\vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{k})(\vec{b} \cdot \hat{k})\right) \]
Combining these gives:
\[ 3(\vec{a} \cdot \vec{b}) - \left((\vec{a} \cdot \hat{i})(\vec{b} \cdot \hat{i}) + (\vec{a} \cdot \hat{j})(\vec{b} \cdot \hat{j}) + (\vec{a} \cdot \hat{k})(\vec{b} \cdot \hat{k})\right) \]

### Step 5: Substitute vector components
Let:
\[ \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}, \quad \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \]
Then:
\[ \vec{a} \cdot \hat{i} = a_1, \quad \vec{b} \cdot \hat{i} = b_1 \]
\[ \vec{a} \cdot \hat{j} = a_2, \quad \vec{b} \cdot \hat{j} = b_2 \]
\[ \vec{a} \cdot \hat{k} = a_3, \quad \vec{b} \cdot \hat{k} = b_3 \]
Substituting these into our expression gives:
\[ 3(\vec{a} \cdot \vec{b}) - (a_1 b_1 + a_2 b_2 + a_3 b_3) = 2(\vec{a} \cdot \vec{b}) \]

### Final Result
Thus, we conclude that:
\[ (\vec{a} \times \hat{i}) + (\vec{b} \times \hat{i}) + (\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j}) + (\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k}) = 2(\vec{a} \cdot \vec{b}) \]