What is the angle between `hati+hatj+hatk and hatj`?

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. none of these

Explanation

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

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

### Step 2: Calculate the dot product of the vectors
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{j} = \hat{i} \cdot \hat{j} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{j} \]
Since \( \hat{i} \cdot \hat{j} = 0 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{k} \cdot \hat{j} = 0 \):
\[ \mathbf{A} \cdot \mathbf{B} = 0 + 1 + 0 = 1 \]

### Step 3: Calculate the magnitudes of the vectors
The magnitude of \( \mathbf{A} \) is given by:
\[ |\mathbf{A}| = \sqrt{(\hat{i})^2 + (\hat{j})^2 + (\hat{k})^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = |\hat{j}| = 1 \]

### Step 4: Use the dot product to find the angle
The formula for the dot product in terms of the angle \( \theta \) between the two vectors is:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
Substituting the values we found:
\[ 1 = \sqrt{3} \cdot 1 \cdot \cos \theta \]
This simplifies to:
\[ \cos \theta = \frac{1}{\sqrt{3}} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Conclusion
Since the options provided were \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \) and none of these, we can conclude that the angle \( \theta \) does not match any of the given options. Thus, the answer is "none of these".