Assertion: The angle between the two vectors `(hati+hatj)` and `(hatj+hatk)` is `(pi)/(3)` radian.
Reason: Angle between two vectors `vecA` and `vecB` is given by `theta=cos^(-1)((vecA.vecB)/(AB))`

Options

A. Weber`//`ampere
B. Volt-second`//`ampere
C. Joule`//("ampee")^(2)`
D. ohm-second

Explanation

To solve the problem, we need to determine the angle between the two vectors \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \) using the formula for the angle between two vectors.

### Step-by-Step Solution:

1. **Identify the Vectors**:
Let \( \vec{A} = \hat{i} + \hat{j} \) and \( \vec{B} = \hat{j} + \hat{k} \).

2. **Calculate the Dot Product**:
The dot product \( \vec{A} \cdot \vec{B} \) is calculated as follows:
\[ \vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) = \hat{i} \cdot \hat{j} + \hat{i} \cdot \hat{k} + \hat{j} \cdot \hat{j} + \hat{j} \cdot \hat{k} \]
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:
\[ \vec{A} \cdot \vec{B} = 0 + 0 + 1 + 0 = 1 \]

3. **Calculate the Magnitudes of the Vectors**:
The magnitude of \( \vec{A} \) is:
\[ |\vec{A}| = \sqrt{(\hat{i}^2 + \hat{j}^2)} = \sqrt{1^2 + 1^2} = \sqrt{2} \]
The magnitude of \( \vec{B} \) is:
\[ |\vec{B}| = \sqrt{(\hat{j}^2 + \hat{k}^2)} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

4. **Use the Formula for the Angle**:
The angle \( \theta \) between the two vectors is given by:
\[ \theta = \cos^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}\right) \]
Substituting the values we calculated:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2} \cdot \sqrt{2}}\right) = \cos^{-1}\left(\frac{1}{2}\right) \]

5. **Find the Angle**:
We know that \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \) radians.

### Conclusion:
Thus, the angle between the two vectors \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \) is indeed \( \frac{\pi}{3} \) radians, confirming that both the assertion and the reason provided in the question are true.