If `theta` is the angle between the vectors is `4( hat(i)- hat(k)) and hat(i)+hat(j)+hat(k)`, then what is `(sin theta + cos theta)` equal to ?

Options

A. 0
B. `1/2`
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( \sin \theta + \cos \theta \) where \( \theta \) is the angle between the vectors \( \mathbf{a} = 4\hat{i} - 4\hat{k} \) and \( \mathbf{b} = \hat{i} + \hat{j} + \hat{k} \).

### Step-by-Step Solution:

1. **Define the Vectors**:
\[ \mathbf{a} = 4\hat{i} - 4\hat{k} \]
\[ \mathbf{b} = \hat{i} + \hat{j} + \hat{k} \]

2. **Calculate the Dot Product**:
The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = (4\hat{i} - 4\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) \]
Using the properties of dot product:
\[ = 4(\hat{i} \cdot \hat{i}) + 4(0) - 4(\hat{k} \cdot \hat{k}) = 4(1) + 0 - 4(1) = 4 - 4 = 0 \]

3. **Calculate the Magnitudes of the Vectors**:
- Magnitude of \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{(4)^2 + (0)^2 + (-4)^2} = \sqrt{16 + 0 + 16} = \sqrt{32} = 4\sqrt{2} \]
- Magnitude of \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

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:
\[ \cos \theta = \frac{0}{(4\sqrt{2})(\sqrt{3})} = 0 \]

5. **Determine the Angle \( \theta \)**:
Since \( \cos \theta = 0 \), we have:
\[ \theta = \frac{\pi}{2} \text{ (90 degrees)} \]

6. **Calculate \( \sin \theta + \cos \theta \)**:
Now, we can find:
\[ \sin \theta + \cos \theta = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) \]
\[ = 1 + 0 = 1 \]

### Final Answer:
\[ \sin \theta + \cos \theta = 1 \]