Which of the following is true ?

Options

A. `hat(i).hat(j)=hat(j).hat(k)=hat(k).hat(i)=0`
B. `hat(i).hat(i)=hat(j).hat(j)=hat(k).hat(k)=0`
C. `hat(i)^(2)+hat(j)^(2)+hat(k)^(2)=0`
D. None of these

Answer

To determine which of the given statements is true regarding the dot products of the unit vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \), let's analyze each statement step by step.

### Step-by-Step Solution:

1. **Understanding the Unit Vectors**:
- The unit vectors in three-dimensional space are defined as:
- \( \mathbf{i} \) along the x-axis,
- \( \mathbf{j} \) along the y-axis,
- \( \mathbf{k} \) along the z-axis.

2. **Dot Product of Perpendicular Vectors**:
- The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
- Where \( \theta \) is the angle between the two vectors.
- Since \( \mathbf{i} \) and \( \mathbf{j} \) are perpendicular, the angle \( \theta \) is \( 90^\circ \):
\[ \mathbf{i} \cdot \mathbf{j} = |\mathbf{i}| |\mathbf{j}| \cos(90^\circ) = 1 \cdot 1 \cdot 0 = 0 \]
- Similarly, for \( \mathbf{j} \cdot \mathbf{k} \) and \( \mathbf{k} \cdot \mathbf{i} \):
\[ \mathbf{j} \cdot \mathbf{k} = 0 \quad \text{and} \quad \mathbf{k} \cdot \mathbf{i} = 0 \]

3. **Dot Product of a Vector with Itself**:
- The dot product of a vector with itself gives the square of its magnitude:
\[ \mathbf{i} \cdot \mathbf{i} = |\mathbf{i}|^2 = 1^2 = 1 \]
- Similarly:
\[ \mathbf{j} \cdot \mathbf{j} = 1 \quad \text{and} \quad \mathbf{k} \cdot \mathbf{k} = 1 \]

4. **Summing the Dot Products**:
- Now, if we sum the dot products:
\[ \mathbf{i} \cdot \mathbf{i} + \mathbf{j} \cdot \mathbf{j} + \mathbf{k} \cdot \mathbf{k} = 1 + 1 + 1 = 3 \]

5. **Conclusion**:
- From the above calculations, we find:
- \( \mathbf{i} \cdot \mathbf{j} = 0 \)
- \( \mathbf{j} \cdot \mathbf{k} = 0 \)
- \( \mathbf{k} \cdot \mathbf{i} = 0 \)
- \( \mathbf{i} \cdot \mathbf{i} = 1 \)
- \( \mathbf{j} \cdot \mathbf{j} = 1 \)
- \( \mathbf{k} \cdot \mathbf{k} = 1 \)
- Therefore, the total \( \mathbf{i} \cdot \mathbf{i} + \mathbf{j} \cdot \mathbf{j} + \mathbf{k} \cdot \mathbf{k} = 3 \), which is not equal to zero.

### Final Answer:
The correct statement is that \( \mathbf{i} \cdot \mathbf{i} = 1 \), \( \mathbf{j} \cdot \mathbf{j} = 1 \), \( \mathbf{k} \cdot \mathbf{k} = 1 \), and their sum is \( 3 \). Thus, the first option is true.