If `hat(i), hat(j)` and `hat(k)` represent unit vectors along the x, y and z axes respectively, then the angle `theta` between the vectors `(hat(i) + hat(j) + hat(k))` and `(hat(i) + hat(j))` is equal to :

Options

A. `sin^(-1)((1)/(sqrt(3)))`
B. `sin^(-1)(sqrt((2)/(3)))`
C. `cos^(-1)((1)/(sqrt(3)))`
D. `90^(@)`

Answer

To find the angle \( \theta \) between the vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} \), we can follow these steps:

### Step 1: Define the Vectors
Let:
- Vector \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \)
- Vector \( \mathbf{B} = \hat{i} + \hat{j} \)

### Step 2: Calculate the Dot Product
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j}) \]
Calculating the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = \hat{i} \cdot \hat{i} + \hat{i} \cdot \hat{j} + \hat{j} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{i} + \hat{k} \cdot \hat{j} \]
Since the dot product of different unit vectors is zero, we have:
\[ \mathbf{A} \cdot \mathbf{B} = 1 + 0 + 0 + 1 + 0 + 0 = 2 \]

### Step 3: Calculate the Magnitudes of the Vectors
The magnitude of vector \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{(\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

The magnitude of vector \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{(\hat{i} + \hat{j}) \cdot (\hat{i} + \hat{j})} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 4: Use the Cosine Formula
The angle \( \theta \) between two vectors can be calculated using the formula:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]
Substituting the values we calculated:
\[ \cos \theta = \frac{2}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \]

### Step 5: Find the Sine of the Angle
To convert \( \cos \theta \) to \( \sin \theta \), we can use the identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Calculating \( \sin^2 \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{\sqrt{6}}{3}\right)^2 = 1 - \frac{6}{9} = \frac{3}{9} = \frac{1}{3} \]
Thus,
\[ \sin \theta = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]

### Step 6: Find the Angle \( \theta \)
Finally, we can find \( \theta \):
\[ \theta = \sin^{-1} \left(\frac{1}{\sqrt{3}}\right) \]

### Conclusion
The angle \( \theta \) between the vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} \) is:
\[ \theta = \sin^{-1} \left(\frac{1}{\sqrt{3}}\right) \]