If `[vec(a) vec(b) vec(c)]=4` then `[vec(a)times vec(b) vec(b)times vec(c)vec(c) times vec(a)]`=

Options

A. 4
B. 2
C. 8
D. 16

Answer

To solve the problem, we need to find the value of the expression \([ \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} ]\) given that \([ \vec{a}, \vec{b}, \vec{c} ] = 4\).

### Step-by-step Solution:

1. **Understanding the Scalar Triple Product**:
The scalar triple product \([ \vec{a}, \vec{b}, \vec{c} ]\) is defined as the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). It can also be expressed in terms of the cross product:
\[ [ \vec{a}, \vec{b}, \vec{c} ] = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

2. **Using the Identity**:
There is a known identity that relates the scalar triple product and the cross products:
\[ [ \vec{a}, \vec{b}, \vec{c} ]^2 = [ \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} ] \]
This means that the scalar triple product squared is equal to the determinant of the matrix formed by the cross products of the vectors.

3. **Substituting the Given Value**:
From the problem, we know that:
\[ [ \vec{a}, \vec{b}, \vec{c} ] = 4 \]
Therefore, we can substitute this value into the identity:
\[ [ \vec{a}, \vec{b}, \vec{c} ]^2 = 4^2 = 16 \]

4. **Conclusion**:
Thus, we find that:
\[ [ \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} ] = 16 \]

### Final Answer:
\[ \boxed{16} \]