Scalar Triple Product Calculation

The question asks us to determine how the scalar triple product \(\left[\vec a \vec b \vec c\right]\) depends on the variables $x$ and $y$. We are given three vectors:

• \(\vec a = \hat i - \hat k\)
• \(\vec b = x\hat i + \hat j + (1 - x)\hat k\)
• \(\vec c = y\hat i + x\hat j + (1 + x - y)\hat k\)

The scalar triple product \(\left[\vec a \vec b \vec c\right]\) can be found by calculating the determinant of the matrix formed by the components of these vectors.

Vector Components Identification

First, let's list the components of each vector:

• Vector \(\vec a\) has components \((1, 0, -1)\).
• Vector \(\vec b\) has components \((x, 1, 1-x)\).
• Vector \(\vec c\) has components \((y, x, 1+x-y)\).

Determinant Evaluation Steps

The scalar triple product \(\left[\vec a \vec b \vec c\right]\) is calculated as the determinant of the matrix formed by these components:

$$ \left[\vec a \vec b \vec c\right] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix} $$

We can expand this determinant along the first row. The formula for a 3x3 determinant expansion along the first row ($a_{11}, a_{12}, a_{13}$) is $a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$, where $C_{ij}$ is the cofactor.

$$ \left[\vec a \vec b \vec c\right] = 1 \cdot \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} - 0 \cdot \begin{vmatrix} x & 1-x \\ y & 1+x-y \end{vmatrix} + (-1) \cdot \begin{vmatrix} x & 1 \\ y & x \end{vmatrix} $$

Now, we calculate the individual 2x2 determinants:

1. The first 2x2 determinant: $$ \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} = (1)(1+x-y) - (x)(1-x) $$ $$ = 1 + x - y - x + x^2 $$ $$ = 1 - y + x^2 $$
2. The second 2x2 determinant (which is multiplied by 0) is not needed for the final calculation.
3. The third 2x2 determinant (multiplied by -1): $$ \begin{vmatrix} x & 1 \\ y & x \end{vmatrix} = (x)(x) - (1)(y) $$ $$ = x^2 - y $$

Substitute these results back into the expansion:

$$ \left[\vec a \vec b \vec c\right] = 1 \cdot (1 - y + x^2) - 0 + (-1) \cdot (x^2 - y) $$

$$ \left[\vec a \vec b \vec c\right] = (1 - y + x^2) - (x^2 - y) $$

$$ \left[\vec a \vec b \vec c\right] = 1 - y + x^2 - x^2 + y $$

Simplifying the expression:

$$ \left[\vec a \vec b \vec c\right] = 1 $$

Variable Dependency Analysis

The calculated value of the scalar triple product \(\left[\vec a \vec b \vec c\right]\) is 1. This is a constant value, meaning it does not change regardless of the values assigned to $x$ and $y$. Therefore, the scalar triple product depends on neither $x$ nor $y$.