Consider a tetrahedron with faces `F_(1),F_(2),F_(3),F_(4)`. Let `vec(V_(1)), vecV_(2),vecV_(3),vecV_(4)` be the vectors whose magnitude are respectively equal to areas of `F_(1).F_(2).F_(3).F_(4)` and whose directions are perpendicular to their faces in outward direction. Then `|vecV_(1) + vecV_(2) + vecV_(3) + vecV_(4)|`, equals:

Options

A. 1
B. 4
C. 0
D. None of these

Answer

To solve the problem, we need to find the magnitude of the vector sum of the vectors associated with the areas of the faces of a tetrahedron. Let's denote the vectors corresponding to the areas of the faces as \( \vec{V}_1, \vec{V}_2, \vec{V}_3, \vec{V}_4 \).

### Step-by-step Solution:

1. **Understanding the Faces and Vectors**:
- The tetrahedron has four faces: \( F_1, F_2, F_3, F_4 \).
- The vectors \( \vec{V}_1, \vec{V}_2, \vec{V}_3, \vec{V}_4 \) represent the areas of these faces and are directed outward, perpendicular to each face.

2. **Expressing the Area Vectors**:
- The area of face \( F_1 \) can be expressed as:
\[ \vec{V}_1 = \frac{1}{2} \vec{A} \times \vec{B} \]
- For face \( F_2 \):
\[ \vec{V}_2 = \frac{1}{2} \vec{B} \times \vec{C} \]
- For face \( F_3 \):
\[ \vec{V}_3 = \frac{1}{2} \vec{C} \times \vec{A} \]
- For face \( F_4 \):
\[ \vec{V}_4 = \frac{1}{2} (\vec{A} - \vec{B}) \times (\vec{C} - \vec{B}) \]

3. **Simplifying \( \vec{V}_4 \)**:
- We can expand \( \vec{V}_4 \):
\[ \vec{V}_4 = \frac{1}{2} \left( \vec{A} \times \vec{C} - \vec{A} \times \vec{B} - \vec{B} \times \vec{C} + \vec{B} \times \vec{B} \right) \]
- Since \( \vec{B} \times \vec{B} = 0 \), we have:
\[ \vec{V}_4 = \frac{1}{2} \left( \vec{A} \times \vec{C} - \vec{A} \times \vec{B} - \vec{B} \times \vec{C} \right) \]

4. **Summing the Vectors**:
- Now, we sum all the vectors:
\[ \vec{V}_1 + \vec{V}_2 + \vec{V}_3 + \vec{V}_4 = \frac{1}{2} \left( \vec{A} \times \vec{B} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} + \vec{A} \times \vec{C} - \vec{A} \times \vec{B} - \vec{B} \times \vec{C} \right) \]

5. **Canceling Terms**:
- Notice that \( \vec{A} \times \vec{B} \) and \( \vec{B} \times \vec{C} \) cancel out:
\[ \vec{V}_1 + \vec{V}_2 + \vec{V}_3 + \vec{V}_4 = \frac{1}{2} \left( \vec{C} \times \vec{A} - \vec{A} \times \vec{C} \right) = 0 \]

6. **Magnitude of the Sum**:
- The magnitude of the vector sum is:
\[ |\vec{V}_1 + \vec{V}_2 + \vec{V}_3 + \vec{V}_4| = |0| = 0 \]

### Final Answer:
\[ |\vec{V}_1 + \vec{V}_2 + \vec{V}_3 + \vec{V}_4| = 0 \]