If `hat(i), hat(j) and hat(k)` are unit vectors along x,y and z-axis respectively, the angle `theta` between the vector `hat(i) + hat(j) + hat(k)" ""and vector" hat(j)` is given by

Options

A. `theta = cos^(-1) ((1)/(sqrt3))`
B. `theta = sin^(-1) ((1)/(sqrt3))`
C. `theta = cos^(-1) ((sqrt3)/(2))`
D. `theta = sin^(-1) ((sqrt3)/(2))`

Answer

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

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

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

The magnitude of vector \( \mathbf{B} \) is:
\[ |\mathbf{B}| = |\hat{j}| = 1 \]

### Step 3: 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 \( \hat{k} \cdot \hat{j} = 0 \), we have:
\[ \mathbf{A} \cdot \mathbf{B} = 0 + 1 + 0 = 1 \]

### Step 4: Use the Dot Product to Find the Angle
The formula for the dot product in terms of the angle \( \theta \) 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:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

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

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