For any vector `vec(alpha)`, what is `(vec(alpha). hat( i)) hat(i)+(vec(alpha). hat(j)) hat(j)+(vec(alpha). hat(k)) hat(k)` equal to ?

Options

A. `vec(alpha)`
B. `3 vec(alpha)`
C. `-vec(alpha)`
D. `vec(0)`

Explanation

To solve the problem, we need to evaluate the expression:

\[ (\vec{\alpha} \cdot \hat{i}) \hat{i} + (\vec{\alpha} \cdot \hat{j}) \hat{j} + (\vec{\alpha} \cdot \hat{k}) \hat{k} \]

where \(\vec{\alpha}\) is a vector expressed in terms of its components along the basis vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\).

### Step-by-Step Solution:

1. **Express the vector \(\vec{\alpha}\)**:
Let \(\vec{\alpha} = a \hat{i} + b \hat{j} + c \hat{k}\), where \(a\), \(b\), and \(c\) are the components of the vector along the respective axes.

2. **Calculate \(\vec{\alpha} \cdot \hat{i}\)**:
\[ \vec{\alpha} \cdot \hat{i} = (a \hat{i} + b \hat{j} + c \hat{k}) \cdot \hat{i} = a \]
This is because the dot product of \(\hat{i}\) with \(\hat{j}\) and \(\hat{k}\) is zero.

3. **Calculate \(\vec{\alpha} \cdot \hat{j}\)**:
\[ \vec{\alpha} \cdot \hat{j} = (a \hat{i} + b \hat{j} + c \hat{k}) \cdot \hat{j} = b \]

4. **Calculate \(\vec{\alpha} \cdot \hat{k}\)**:
\[ \vec{\alpha} \cdot \hat{k} = (a \hat{i} + b \hat{j} + c \hat{k}) \cdot \hat{k} = c \]

5. **Substitute these results back into the original expression**:
\[ (\vec{\alpha} \cdot \hat{i}) \hat{i} + (\vec{\alpha} \cdot \hat{j}) \hat{j} + (\vec{\alpha} \cdot \hat{k}) \hat{k} = a \hat{i} + b \hat{j} + c \hat{k} \]

6. **Recognize that this is simply the vector \(\vec{\alpha}\)**:
\[ a \hat{i} + b \hat{j} + c \hat{k} = \vec{\alpha} \]

### Conclusion:
Thus, the expression evaluates to:

\[ (\vec{\alpha} \cdot \hat{i}) \hat{i} + (\vec{\alpha} \cdot \hat{j}) \hat{j} + (\vec{\alpha} \cdot \hat{k}) \hat{k} = \vec{\alpha} \]