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

Options

A. `alpha`
B. `3alpha`
C. `-alpha`
D. 0

Answer

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

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

where \(\alpha\) is a vector represented as:

\[ \alpha = x \hat{i} + y \hat{j} + z \hat{k} \]

### Step 1: Calculate the dot products

1. **Calculate \(\alpha \cdot \hat{i}\)**:
\[ \alpha \cdot \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{i} = x (\hat{i} \cdot \hat{i}) + y (\hat{j} \cdot \hat{i}) + z (\hat{k} \cdot \hat{i}) = x \cdot 1 + 0 + 0 = x \]

2. **Calculate \(\alpha \cdot \hat{j}\)**:
\[ \alpha \cdot \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{j} = x (\hat{i} \cdot \hat{j}) + y (\hat{j} \cdot \hat{j}) + z (\hat{k} \cdot \hat{j}) = 0 + y \cdot 1 + 0 = y \]

3. **Calculate \(\alpha \cdot \hat{k}\)**:
\[ \alpha \cdot \hat{k} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{k} = x (\hat{i} \cdot \hat{k}) + y (\hat{j} \cdot \hat{k}) + z (\hat{k} \cdot \hat{k}) = 0 + 0 + z \cdot 1 = z \]

### Step 2: Substitute the dot products back into the expression

Now substituting the results back into the original expression:

\[ (\alpha \cdot \hat{i}) \hat{i} + (\alpha \cdot \hat{j}) \hat{j} + (\alpha \cdot \hat{k}) \hat{k} = x \hat{i} + y \hat{j} + z \hat{k} \]

### Step 3: Conclusion

Thus, we have:

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

So the final answer is:

\[ \alpha \]