If `theta` be the angle between the vectors `4(i-k)` and `i+j+k`, then `theta` is

Options

A. `pi//4`
B. `pi//3`
C. `pi//2`
D. `cos^(-1)(1//sqrt(3))`

Answer

To find the angle \( \theta \) between the vectors \( \mathbf{A} = 4(\mathbf{i} - \mathbf{k}) \) and \( \mathbf{B} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), we can use the formula for the cosine of the angle between two vectors:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

### Step 1: Define the vectors
The vectors are given as:
\[ \mathbf{A} = 4(\mathbf{i} - \mathbf{k}) = 4\mathbf{i} - 4\mathbf{k} \]
\[ \mathbf{B} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]

### Step 2: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)
The dot product is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (4\mathbf{i} - 4\mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) \]
Using the distributive property of the dot product:
\[ = 4\mathbf{i} \cdot \mathbf{i} + 4\mathbf{i} \cdot \mathbf{j} - 4\mathbf{k} \cdot \mathbf{i} - 4\mathbf{k} \cdot \mathbf{j} - 4\mathbf{k} \cdot \mathbf{k} \]
Since \( \mathbf{i} \cdot \mathbf{i} = 1 \), \( \mathbf{j} \cdot \mathbf{j} = 1 \), \( \mathbf{k} \cdot \mathbf{k} = 1 \), and the dot products between different unit vectors are zero:
\[ = 4(1) + 0 + 0 - 0 - 4(1) = 4 - 4 = 0 \]

### Step 3: Calculate the magnitudes \( |\mathbf{A}| \) and \( |\mathbf{B}| \)
The magnitude of \( \mathbf{A} \):
\[ |\mathbf{A}| = \sqrt{(4)^2 + (0)^2 + (-4)^2} = \sqrt{16 + 0 + 16} = \sqrt{32} = 4\sqrt{2} \]

The magnitude of \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 4: Substitute into the cosine formula
Now substituting into the cosine formula:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{0}{(4\sqrt{2})(\sqrt{3})} = 0 \]

### Step 5: Find the angle \( \theta \)
Since \( \cos \theta = 0 \), we find:
\[ \theta = \frac{\pi}{2} \]

### Final Answer
Thus, the angle \( \theta \) between the vectors is:
\[ \theta = \frac{\pi}{2} \]