`(a*hati)hati+(a*hatj)hatj+(a*hatk)hatk` is equal to

Options

A. `a`
B. 2a
C. 3a
D. 0

Answer

To solve the expression \((a \hat{i}) \hat{i} + (a \hat{j}) \hat{j} + (a \hat{k}) \hat{k}\), we will break it down step by step.

### Step 1: Understand the Components
Let \( \mathbf{a} = x \hat{i} + y \hat{j} + z \hat{k} \), where \(x\), \(y\), and \(z\) are the components of the vector \( \mathbf{a} \) along the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) directions respectively.

### Step 2: Substitute \( \mathbf{a} \) into the Expression
We can rewrite the expression as follows:
\[ (\mathbf{a} \cdot \hat{i}) \hat{i} + (\mathbf{a} \cdot \hat{j}) \hat{j} + (\mathbf{a} \cdot \hat{k}) \hat{k} \]
Now, we will calculate each dot product.

### Step 3: Calculate the Dot Products
1. **Calculate \( \mathbf{a} \cdot \hat{i} \)**:
\[ \mathbf{a} \cdot \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{i} = x (\hat{i} \cdot \hat{i}) + y (\hat{j} \cdot \hat{i}) + z (\hat{k} \cdot \hat{i}) = x \cdot 1 + 0 + 0 = x \]
Thus, \( (\mathbf{a} \cdot \hat{i}) \hat{i} = x \hat{i} \).

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

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

### Step 4: Combine the Results
Now, we can combine all the results:
\[ (\mathbf{a} \cdot \hat{i}) \hat{i} + (\mathbf{a} \cdot \hat{j}) \hat{j} + (\mathbf{a} \cdot \hat{k}) \hat{k} = x \hat{i} + y \hat{j} + z \hat{k} \]
This is simply \( \mathbf{a} \).

### Final Result
Thus, the expression \((a \hat{i}) \hat{i} + (a \hat{j}) \hat{j} + (a \hat{k}) \hat{k}\) is equal to:
\[ \mathbf{a} \]