Vector `vec(A)=hat(i)+hat(j)-2hat(k)` and `vec(B)=3hat(i)+3hat(j)-6hat(k)` are `:`

Options

A. Parallel
B. Antiparallel
C. Perpendicular
D. at acute angle with each other

Explanation

To determine the relationship between the vectors \(\vec{A} = \hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{B} = 3\hat{i} + 3\hat{j} - 6\hat{k}\), we will find the angle between them using the dot product formula.

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \vec{A} = \hat{i} + \hat{j} - 2\hat{k} \]
\[ \vec{B} = 3\hat{i} + 3\hat{j} - 6\hat{k} \]

2. **Calculate the dot product \(\vec{A} \cdot \vec{B}\)**:
\[ \vec{A} \cdot \vec{B} = (1)(3) + (1)(3) + (-2)(-6) \]
\[ = 3 + 3 + 12 = 18 \]

3. **Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)**:
- Magnitude of \(\vec{A}\):
\[ |\vec{A}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

- Magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{3^2 + 3^2 + (-6)^2} = \sqrt{9 + 9 + 36} = \sqrt{54} = 3\sqrt{6} \]

4. **Use the dot product to find \(\cos \theta\)**:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
\[ = \frac{18}{\sqrt{6} \cdot 3\sqrt{6}} = \frac{18}{3 \cdot 6} = \frac{18}{18} = 1 \]

5. **Determine the angle \(\theta\)**:
Since \(\cos \theta = 1\), we have:
\[ \theta = 0^\circ \]

6. **Conclusion**:
The angle between the vectors is \(0^\circ\), which means the vectors are parallel.

### Final Answer:
The vectors \(\vec{A}\) and \(\vec{B}\) are **parallel**.