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 use the dot product formula. Here’s a step-by-step solution:

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

### Step 2: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)
Using the dot product formula:
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j}) \]
Expanding this:
\[ \mathbf{A} \cdot \mathbf{B} = \hat{i} \cdot \hat{i} + \hat{i} \cdot \hat{j} + \hat{j} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{i} + \hat{k} \cdot \hat{j} \]
Since \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{j} \cdot \hat{j} = 1 \), and all other dot products are zero:
\[ \mathbf{A} \cdot \mathbf{B} = 1 + 0 + 0 + 1 + 0 + 0 = 2 \]

### Step 3: Calculate the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \)
Magnitude of \( \mathbf{A} \):
\[ |\mathbf{A}| = |\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

Magnitude of \( \mathbf{B} \):
\[ |\mathbf{B}| = |\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 4: Use the dot product to find \( \cos \theta \)
Using the formula:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
Substituting the values we found:
\[ 2 = \sqrt{3} \cdot \sqrt{2} \cdot \cos \theta \]
This simplifies to:
\[ 2 = \sqrt{6} \cos \theta \]
Thus:
\[ \cos \theta = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \]

### Step 5: Find \( \sin \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \left(\frac{2}{\sqrt{6}}\right)^2 = 1 - \frac{4}{6} = \frac{2}{6} = \frac{1}{3} \]
So:
\[ \sin \theta = \frac{1}{\sqrt{3}} \]

### Step 6: Determine \( \theta \)
Thus:
\[ \theta = \sin^{-1} \left(\frac{1}{\sqrt{3}}\right) \]

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