The angle between the Vector `(hat(i)+hat(j))` and `(hat(j)+hat(k))` 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 will follow these steps:

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

### Step 2: Calculate the Dot Product
The dot product \( \vec{A} \cdot \vec{B} \) is calculated as follows:
\[ \vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) \]
Using the distributive property of dot products, we have:
\[ \vec{A} \cdot \vec{B} = \hat{i} \cdot \hat{j} + \hat{i} \cdot \hat{k} + \hat{j} \cdot \hat{j} + \hat{j} \cdot \hat{k} \]
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 find:
\[ \vec{A} \cdot \vec{B} = 0 + 0 + 1 + 0 = 1 \]

### Step 3: Calculate the Magnitudes of the Vectors
Next, we calculate the magnitudes of \( \vec{A} \) and \( \vec{B} \):
\[ |\vec{A}| = \sqrt{(\hat{i})^2 + (\hat{j})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]
\[ |\vec{B}| = \sqrt{(\hat{j})^2 + (\hat{k})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 4: Use the Dot Product to Find the Angle
The formula relating the dot product to the angle \( \theta \) between the two vectors is:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Substituting the values we have:
\[ 1 = (\sqrt{2})(\sqrt{2}) \cos \theta \]
\[ 1 = 2 \cos \theta \]
\[ \cos \theta = \frac{1}{2} \]

### Step 5: Determine the Angle
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

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

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