If `bar(A_(1))bar(B)` and `bar( C )` are three non-coplanar voctors, then `(vec(B).(vec(A)xx vec( C )))/((vec( C )xx vec(A)).vec(B))+(vec(B).(vec(A)xx vec( C )))/(vec( C ).(vec(A)xx vec(B)))` is equal to

Options

A. 2
B. `-2`
C. 0
D. none of these

Answer

To solve the given problem, we need to evaluate the expression:

\[ \frac{\vec{B} \cdot (\vec{A} \times \vec{C})}{(\vec{C} \times \vec{A}) \cdot \vec{B}} + \frac{\vec{B} \cdot (\vec{A} \times \vec{C})}{\vec{C} \cdot (\vec{A} \times \vec{B})} \]

### Step-by-Step Solution:

1. **Identify the Scalar Triple Product**:
The expression \(\vec{B} \cdot (\vec{A} \times \vec{C})\) represents the scalar triple product, which can be denoted as \([ \vec{A}, \vec{B}, \vec{C} ]\). This can also be represented as the determinant of a matrix formed by the vectors \(\vec{A}, \vec{B}, \vec{C}\):
\[ [ \vec{A}, \vec{B}, \vec{C} ] = \begin{vmatrix} \vec{A} \\ \vec{B} \\ \vec{C} \end{vmatrix} \]

2. **Rewrite the Expression**:
Using the notation for the scalar triple product, we can rewrite the expression as:
\[ \frac{[\vec{B}, \vec{A}, \vec{C}]}{[\vec{C}, \vec{A}, \vec{B}]} + \frac{[\vec{B}, \vec{A}, \vec{C}]}{[\vec{C}, \vec{A}, \vec{B}]} \]
This simplifies to:
\[ 2 \cdot \frac{[\vec{B}, \vec{A}, \vec{C}]}{[\vec{C}, \vec{A}, \vec{B}]} \]

3. **Use Properties of Determinants**:
We know that interchanging two rows of a determinant changes its sign. Therefore, if we interchange \(\vec{B}\) and \(\vec{C}\) in the determinant \([\vec{C}, \vec{A}, \vec{B}]\), we get:
\[ [\vec{C}, \vec{A}, \vec{B}] = -[\vec{B}, \vec{A}, \vec{C}] \]

4. **Substitute in the Expression**:
Substituting this result into our expression gives:
\[ 2 \cdot \frac{[\vec{B}, \vec{A}, \vec{C}]}{-[\vec{B}, \vec{A}, \vec{C}]} = 2 \cdot -1 = -2 \]

5. **Final Result**:
Therefore, the value of the given expression is:
\[ \boxed{-2} \]