If `veca,vecb` are non-collinear vectors, then
`[(veca,vecb,hati)]hati+[(veca,vecb,hatj)]hatj+[(veca,vecb,hatk)]hatk=`

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:

\[ [(\vec{a}, \vec{b}, \hat{i})] \hat{i} + [(\vec{a}, \vec{b}, \hat{j})] \hat{j} + [(\vec{a}, \vec{b}, \hat{k})] \hat{k} \]

where \((\vec{a}, \vec{b}, \hat{i})\) denotes the scalar triple product of vectors \(\vec{a}\), \(\vec{b}\), and \(\hat{i}\).

### Step-by-step Solution:

1. **Understanding Scalar Triple Product**:
The scalar triple product of three vectors \(\vec{x}, \vec{y}, \vec{z}\) can be expressed as:
\[ (\vec{x}, \vec{y}, \vec{z}) = \vec{x} \cdot (\vec{y} \times \vec{z}) \]
We will apply this property to our expression.

2. **Rewriting Each Term**:
We can rewrite each term in the expression using the property of the scalar triple product:
\[ [(\vec{a}, \vec{b}, \hat{i})] = \vec{a} \cdot (\vec{b} \times \hat{i}) \]
\[ [(\vec{a}, \vec{b}, \hat{j})] = \vec{a} \cdot (\vec{b} \times \hat{j}) \]
\[ [(\vec{a}, \vec{b}, \hat{k})] = \vec{a} \cdot (\vec{b} \times \hat{k}) \]

3. **Substituting Back**:
Substitute these back into the original expression:
\[ \vec{a} \cdot (\vec{b} \times \hat{i}) \hat{i} + \vec{a} \cdot (\vec{b} \times \hat{j}) \hat{j} + \vec{a} \cdot (\vec{b} \times \hat{k}) \hat{k} \]

4. **Assuming \(\vec{a} \times \vec{b}\)**:
Let’s denote \(\vec{a} \times \vec{b} = \vec{c}\), where \(\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}\) for some components \(x, y, z\).

5. **Dot Products**:
Now, we can express the dot products:
\[ \vec{a} \cdot (\vec{b} \times \hat{i}) = \text{component of } \vec{c} \text{ along } \hat{i} = x \]
\[ \vec{a} \cdot (\vec{b} \times \hat{j}) = \text{component of } \vec{c} \text{ along } \hat{j} = y \]
\[ \vec{a} \cdot (\vec{b} \times \hat{k}) = \text{component of } \vec{c} \text{ along } \hat{k} = z \]

6. **Final Expression**:
Therefore, substituting these back, we get:
\[ x \hat{i} + y \hat{j} + z \hat{k} = \vec{c} = \vec{a} \times \vec{b} \]

Thus, the final result is:
\[ [(\vec{a}, \vec{b}, \hat{i})] \hat{i} + [(\vec{a}, \vec{b}, \hat{j})] \hat{j} + [(\vec{a}, \vec{b}, \hat{k})] \hat{k} = \vec{a} \times \vec{b} \]

### Conclusion:
The value of the given expression is \(\vec{a} \times \vec{b}\).

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