The angle between the vectors `(hati+hatj)` and `(hatj+hatk)` is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle between the vectors \( \mathbf{v_1} = \hat{i} + \hat{j} \) and \( \mathbf{v_2} = \hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Identify the vectors
We have:
- \( \mathbf{v_1} = \hat{i} + \hat{j} \)
- \( \mathbf{v_2} = \hat{j} + \hat{k} \)

### Step 2: Calculate the dot product of the vectors
The dot product \( \mathbf{v_1} \cdot \mathbf{v_2} \) is calculated as follows:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = (\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) = \hat{i} \cdot \hat{j} + \hat{i} \cdot \hat{k} + \hat{j} \cdot \hat{j} + \hat{j} \cdot \hat{k} \]
Since \( \hat{i} \cdot \hat{j} = 0 \), \( \hat{i} \cdot \hat{k} = 0 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{j} \cdot \hat{k} = 0 \), we find:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = 0 + 0 + 1 + 0 = 1 \]

### Step 3: Calculate the magnitudes of the vectors
The magnitude of \( \mathbf{v_1} \) is:
\[ |\mathbf{v_1}| = \sqrt{(\hat{i}^2 + \hat{j}^2)} = \sqrt{1^2 + 1^2} = \sqrt{2} \]
The magnitude of \( \mathbf{v_2} \) is:
\[ |\mathbf{v_2}| = \sqrt{(\hat{j}^2 + \hat{k}^2)} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 4: Use the dot product to find the cosine of the angle
Using the formula for the cosine of the angle \( \theta \) between two vectors:
\[ \cos \theta = \frac{\mathbf{v_1} \cdot \mathbf{v_2}}{|\mathbf{v_1}| |\mathbf{v_2}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
This gives us:
\[ \theta = 60^\circ \]

### Final Answer
The angle between the vectors \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \) is \( 60^\circ \).

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