`(r*hat(i))^(2)+(r*hat(j))^(2)+(r*hat(k))^(2)` is equal to

Options

A. `3r^(2)`
B. `r^(2)`
C. `0`
D. None of these

Answer

To solve the problem `(r*hat(i))^(2)+(r*hat(j))^(2)+(r*hat(k))^(2)`, we will follow these steps:

### Step 1: Define the vector r
Assume the vector \( \mathbf{r} \) can be expressed in terms of its components along the unit vectors \( \hat{i}, \hat{j}, \hat{k} \):
\[ \mathbf{r} = x \hat{i} + y \hat{j} + z \hat{k} \]

### Step 2: Calculate \( \mathbf{r} \cdot \hat{i} \)
To find \( \mathbf{r} \cdot \hat{i} \):
\[ \mathbf{r} \cdot \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{i} \]
Using the dot product properties, we have:
\[ \mathbf{r} \cdot \hat{i} = x (\hat{i} \cdot \hat{i}) + y (\hat{j} \cdot \hat{i}) + z (\hat{k} \cdot \hat{i}) = x \cdot 1 + y \cdot 0 + z \cdot 0 = x \]

### Step 3: Calculate \( \mathbf{r} \cdot \hat{j} \)
Similarly, for \( \mathbf{r} \cdot \hat{j} \):
\[ \mathbf{r} \cdot \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{j} \]
This gives us:
\[ \mathbf{r} \cdot \hat{j} = x (\hat{i} \cdot \hat{j}) + y (\hat{j} \cdot \hat{j}) + z (\hat{k} \cdot \hat{j}) = x \cdot 0 + y \cdot 1 + z \cdot 0 = y \]

### Step 4: Calculate \( \mathbf{r} \cdot \hat{k} \)
Now, for \( \mathbf{r} \cdot \hat{k} \):
\[ \mathbf{r} \cdot \hat{k} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{k} \]
This results in:
\[ \mathbf{r} \cdot \hat{k} = x (\hat{i} \cdot \hat{k}) + y (\hat{j} \cdot \hat{k}) + z (\hat{k} \cdot \hat{k}) = x \cdot 0 + y \cdot 0 + z \cdot 1 = z \]

### Step 5: Calculate the squares
Now we can find the squares of these dot products:
\[ (\mathbf{r} \cdot \hat{i})^2 = x^2 \]
\[ (\mathbf{r} \cdot \hat{j})^2 = y^2 \]
\[ (\mathbf{r} \cdot \hat{k})^2 = z^2 \]

### Step 6: Add the squares
Now, we add these results together:
\[ (\mathbf{r} \cdot \hat{i})^2 + (\mathbf{r} \cdot \hat{j})^2 + (\mathbf{r} \cdot \hat{k})^2 = x^2 + y^2 + z^2 \]

### Step 7: Relate to the magnitude of r
The expression \( x^2 + y^2 + z^2 \) is equal to the square of the magnitude of the vector \( \mathbf{r} \):
\[ x^2 + y^2 + z^2 = |\mathbf{r}|^2 \]

### Conclusion
Thus, we conclude that:
\[ (\mathbf{r} \cdot \hat{i})^2 + (\mathbf{r} \cdot \hat{j})^2 + (\mathbf{r} \cdot \hat{k})^2 = |\mathbf{r}|^2 \]