If `theta` be the angle between the vectors `i+jandj+k`, then `theta` is

Options

A. 0
B. `pi//4`
C. `pi//2`
D. `pi//3`

Answer

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

### Step 1: Identify the vectors
Let:
\[ \mathbf{A} = \mathbf{i} + \mathbf{j} \]
\[ \mathbf{B} = \mathbf{j} + \mathbf{k} \]

### Step 2: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \]
For our vectors:
- \( \mathbf{A} = (1, 1, 0) \) (coefficients of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \))
- \( \mathbf{B} = (0, 1, 1) \)

Now, calculate the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = (1)(0) + (1)(1) + (0)(1) = 0 + 1 + 0 = 1 \]

### Step 3: Calculate the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \)
The magnitude of a vector \( \mathbf{V} = (x, y, z) \) is given by:
\[ |\mathbf{V}| = \sqrt{x^2 + y^2 + z^2} \]

For \( \mathbf{A} \):
\[ |\mathbf{A}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]

For \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 4: Use the cosine formula to find \( \theta \)
The angle \( \theta \) between two vectors can be found using the formula:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

Substituting the values we calculated:
\[ \cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \]

### Step 5: Find \( \theta \)
To find \( \theta \), take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
This gives:
\[ \theta = 60^\circ \quad \text{or} \quad \theta = \frac{\pi}{3} \text{ radians} \]

### Final Answer:
Thus, the angle \( \theta \) between the vectors \( \mathbf{i} + \mathbf{j} \) and \( \mathbf{j} + \mathbf{k} \) is:
\[ \theta = 60^\circ \text{ or } \frac{\pi}{3} \]