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

Options

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

Explanation

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

### Step 1: Define the Vectors
Let:
\[ \mathbf{a} = \hat{i} + \hat{j} \]
\[ \mathbf{b} = \hat{j} + \hat{k} \]

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

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

The magnitude of vector \(\mathbf{b}\) is:
\[ |\mathbf{b}| = \sqrt{(\hat{j})^2 + (\hat{k})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 4: Use the Cosine Formula
The cosine of the angle \(\theta\) between the two vectors is given by:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \]

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

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

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