If `vecx_(1),vecx_(2),vecx_(3)` are two sets of non-coplanar vectors such that `vecx_(r ).vecy_(s)={{:(0","" if "rnes),(2""" if "r=s):}` where r,s =1,2,3
what is the value of `[vecx_(1)vecx_(2)vecx_(3)][vecy_(1)vecy_(2)vecy_(3)]` ?

Answer

To solve the problem, we need to find the value of the scalar triple product \([ \vec{x}_1, \vec{x}_2, \vec{x}_3 ] [ \vec{y}_1, \vec{y}_2, \vec{y}_3 ]\) given the conditions on the dot products of the vectors.

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We are given that:
\[ \vec{x}_r \cdot \vec{y}_s = \begin{cases} 0 & \text{if } r \neq s \\ 2 & \text{if } r = s \end{cases} \]
This means that the dot product between different vectors from the two sets is zero, indicating orthogonality, while the dot product of the same indexed vectors is 2.

2. **Setting Up the Scalar Triple Product**:
We need to calculate the scalar triple product, which can be expressed in determinant form:
\[ [ \vec{x}_1, \vec{x}_2, \vec{x}_3 ] [ \vec{y}_1, \vec{y}_2, \vec{y}_3 ] = \begin{vmatrix} \vec{x}_1 \cdot \vec{y}_1 & \vec{x}_1 \cdot \vec{y}_2 & \vec{x}_1 \cdot \vec{y}_3 \\ \vec{x}_2 \cdot \vec{y}_1 & \vec{x}_2 \cdot \vec{y}_2 & \vec{x}_2 \cdot \vec{y}_3 \\ \vec{x}_3 \cdot \vec{y}_1 & \vec{x}_3 \cdot \vec{y}_2 & \vec{x}_3 \cdot \vec{y}_3 \end{vmatrix} \]

3. **Filling in the Determinant**:
Using the conditions provided:
- \(\vec{x}_1 \cdot \vec{y}_1 = 2\)
- \(\vec{x}_2 \cdot \vec{y}_2 = 2\)
- \(\vec{x}_3 \cdot \vec{y}_3 = 2\)
- All other dot products are \(0\).

This gives us the determinant:
\[ \begin{vmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix} \]

4. **Calculating the Determinant**:
The determinant of a diagonal matrix is the product of its diagonal elements:
\[ 2 \times 2 \times 2 = 8 \]

5. **Conclusion**:
Therefore, the value of the scalar triple product is:
\[ [ \vec{x}_1, \vec{x}_2, \vec{x}_3 ] [ \vec{y}_1, \vec{y}_2, \vec{y}_3 ] = 8 \]