The angle between the vector `2hati+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 vector \( \mathbf{A} = 2\hat{i} + \hat{j} + \hat{k} \) and the vector \( \mathbf{B} = \hat{j} \), we can use the dot product formula. The steps to solve the problem are as follows:

### Step 1: Write down the vectors
The vectors are:
\[ \mathbf{A} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{B} = \hat{j} \]

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

### Step 3: Calculate the magnitudes of the vectors
The magnitude of vector \( \mathbf{A} \) is calculated as:
\[ |\mathbf{A}| = \sqrt{(2^2) + (1^2) + (1^2)} = \sqrt{4 + 1 + 1} = \sqrt{6} \]
The magnitude of vector \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{(1^2)} = 1 \]

### Step 4: Use the dot product to find the angle
Using the formula for the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
Substituting the known values:
\[ 1 = (\sqrt{6})(1) \cos \theta \]
This simplifies to:
\[ \cos \theta = \frac{1}{\sqrt{6}} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \]

### Step 6: Determine if the angle matches the options
Now, we check the provided options: \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \) and \( \text{none of these} \). Since \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), and \( \cos \frac{\pi}{3} = \frac{1}{2} \), none of these match \( \frac{1}{\sqrt{6}} \).

Thus, the correct answer is **none of these**.