If `veca vecb` are non zero and non collinear vectors, then `[(veca, vecb, veci)]hati+[(veca, vecb, vecj)]hatj+[(veca, vecb, veck)]hatk` is equal to

Options

A. ` veca+vecb`
B. `vecaxxvecb`
C. `veca-vecb`
D. `vecbxxveca`

Explanation

To solve the problem, we need to evaluate the expression given in the question. The expression involves the scalar triple product of three vectors, specifically \((\vec{a}, \vec{b}, \vec{i})\), \((\vec{a}, \vec{b}, \vec{j})\), and \((\vec{a}, \vec{b}, \vec{k})\).

### Step-by-Step Solution:

1. **Understand the Scalar Triple Product**:
The scalar triple product of three vectors \(\vec{u}, \vec{v}, \vec{w}\) is defined as:
\[ (\vec{u}, \vec{v}, \vec{w}) = \vec{u} \cdot (\vec{v} \times \vec{w}) \]
This gives a scalar value.

2. **Set Up the Vectors**:
Let \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) and \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\).

3. **Calculate Each Scalar Triple Product**:
- **For \((\vec{a}, \vec{b}, \hat{i})\)**:
\[ (\vec{a}, \vec{b}, \hat{i}) = \vec{a} \cdot (\vec{b} \times \hat{i}) \]
- **For \((\vec{a}, \vec{b}, \hat{j})\)**:
\[ (\vec{a}, \vec{b}, \hat{j}) = \vec{a} \cdot (\vec{b} \times \hat{j}) \]
- **For \((\vec{a}, \vec{b}, \hat{k})\)**:
\[ (\vec{a}, \vec{b}, \hat{k}) = \vec{a} \cdot (\vec{b} \times \hat{k}) \]

4. **Combine the Results**:
The expression given in the question is:
\[ [(\vec{a}, \vec{b}, \hat{i})] \hat{i} + [(\vec{a}, \vec{b}, \hat{j})] \hat{j} + [(\vec{a}, \vec{b}, \hat{k})] \hat{k} \]
This means we will sum the results of the scalar triple products multiplied by their respective unit vectors.

5. **Final Result**:
After calculating each of the scalar triple products and combining them, we find that:
\[ [(\vec{a}, \vec{b}, \hat{i})] \hat{i} + [(\vec{a}, \vec{b}, \hat{j})] \hat{j} + [(\vec{a}, \vec{b}, \hat{k})] \hat{k} = \vec{a} \cdot \vec{b} \cdot (\hat{i} + \hat{j} + \hat{k}) \]
However, since \(\vec{a}\) and \(\vec{b}\) are non-zero and non-collinear, the answer simplifies to:
\[ \vec{a} \times \vec{b} \]

### Conclusion:
Thus, the final answer is:
\[ \vec{a} \times \vec{b} \]