The three concurrent edges of a parallelopiped represent the vectors `veca, vecb, vecc` such that `[(veca, vecb, vecc)]=V`. Then the volume of the parallelopiped whose three concurrent edges are the three diagonals of three faces of the given parallelopiped is

Options

A. `2V`
B. `3V`
C. `V`
D. `6V`

Explanation

To find the volume of the parallelepiped whose concurrent edges are the diagonals of the faces of a given parallelepiped defined by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that the volume \(V\) of the original parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) is given by the scalar triple product:
\[ V = [\vec{a}, \vec{b}, \vec{c}] \]

2. **Identifying the Diagonals**:
The diagonals of the three faces of the parallelepiped can be expressed as:
- Diagonal of the face formed by \(\vec{a}\) and \(\vec{b}\): \(\vec{d_1} = \vec{a} + \vec{b}\)
- Diagonal of the face formed by \(\vec{b}\) and \(\vec{c}\): \(\vec{d_2} = \vec{b} + \vec{c}\)
- Diagonal of the face formed by \(\vec{c}\) and \(\vec{a}\): \(\vec{d_3} = \vec{c} + \vec{a}\)

3. **Setting Up the Volume of the New Parallelepiped**:
The volume \(V'\) of the new parallelepiped formed by the diagonals \(\vec{d_1}\), \(\vec{d_2}\), and \(\vec{d_3}\) can be calculated using the scalar triple product:
\[ V' = [\vec{d_1}, \vec{d_2}, \vec{d_3}] \]

4. **Substituting the Diagonal Vectors**:
Substitute the expressions for the diagonals:
\[ V' = [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] \]

5. **Expanding the Scalar Triple Product**:
Using the properties of the scalar triple product, we can expand this:
\[ V' = [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = [\vec{a}, \vec{b}, \vec{c}] + [\vec{a}, \vec{b}, \vec{c}] + [\vec{b}, \vec{c}, \vec{a}] + [\vec{c}, \vec{a}, \vec{b}] \]
Each of the last three terms is equal to \(V\) because of the cyclic property of the scalar triple product.

6. **Calculating the Result**:
Therefore, we have:
\[ V' = V + V + V = 2V \]

7. **Final Answer**:
The volume of the parallelepiped whose concurrent edges are the three diagonals of the faces of the given parallelepiped is:
\[ \boxed{2V} \]