1. Key Concepts and Formulas

• Scalar Triple Product (STP): For three vectors $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$, $\overrightarrow b = b_1\widehat i + b_2\widehat j + b_3\widehat k$, and $\overrightarrow c = c_1\widehat i + c_2\widehat j + c_3\widehat k$, the scalar triple product is given by the determinant of the matrix formed by their components: $[\overrightarrow a, \overrightarrow b, \overrightarrow c] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$ The STP represents the signed volume of the parallelepiped formed by the three vectors.
• Determinant of a 3x3 Matrix: The determinant of a 3x3 matrix can be calculated using cofactor expansion. For a matrix with the first row elements $a_1, a_2, a_3$: $\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix}$ The determinant of a 2x2 matrix $\begin{vmatrix} p & q \\ r & s \end{vmatrix}$ is $ps - qr$.

2. Step-by-Step Solution

Step 1: Identify the components of the given vectors. We are given the vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ in terms of their $\widehat i$, $\widehat j$, and $\widehat k$ components. We need to list these components to set up the determinant for the scalar triple product.

• $\overrightarrow a = \widehat i - \widehat k = 1\widehat i + 0\widehat j - 1\widehat k \implies (a_1, a_2, a_3) = (1, 0, -1)$
• $\overrightarrow b = x\widehat i + \widehat j + (1 - x)\widehat k \implies (b_1, b_2, b_3) = (x, 1, 1-x)$
• $\overrightarrow c = y\widehat i + x\widehat j + (1 + x - y)\widehat k \implies (c_1, c_2, c_3) = (y, x, 1+x-y)$

Step 2: Set up the determinant for the scalar triple product $[\overrightarrow a, \overrightarrow b, \overrightarrow c]$. Using the components identified in Step 1, we form a 3x3 matrix and calculate its determinant. $[\overrightarrow a, \overrightarrow b, \overrightarrow c] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix}$

Step 3: Evaluate the determinant using cofactor expansion along the first row. We expand the determinant along the first row, which contains a zero term, simplifying the calculation. $\begin{aligned} [\overrightarrow a, \overrightarrow b, \overrightarrow c] &= 1 \cdot \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} - 0 \cdot \begin{vmatrix} x & 1-x \\ y & 1+x-y \end{vmatrix} + (-1) \cdot \begin{vmatrix} x & 1 \\ y & x \end{vmatrix} \\ &= 1 \cdot [1 \cdot (1+x-y) - (1-x) \cdot x] - 0 + (-1) \cdot [x \cdot x - 1 \cdot y] \end{aligned}$

Step 4: Simplify the expression obtained from the determinant. Now, we perform the algebraic simplifications. $\begin{aligned} &= [ (1+x-y) - (x - x^2) ] - [ x^2 - y ] \\ &= [ 1+x-y - x + x^2 ] - x^2 + y \\ &= 1 + x - y - x + x^2 - x^2 + y \end{aligned}$

Step 5: Combine like terms to find the final value of the scalar triple product. We cancel out terms that are opposites of each other.

• The $+x$ and $-x$ terms cancel out.
• The $-y$ and $+y$ terms cancel out.
• The $+x^2$ and $-x^2$ terms cancel out. $[\overrightarrow a, \overrightarrow b, \overrightarrow c] = 1$

Step 6: Analyze the result to determine the dependency of the scalar triple product. The scalar triple product $[\overrightarrow a, \overrightarrow b, \overrightarrow c]$ simplifies to the constant value $1$. This means the value of the scalar triple product does not change regardless of the values assigned to $x$ and $y$. Therefore, it depends on neither $x$ nor $y$.

3. Common Mistakes & Tips

• Sign Errors in Determinant Expansion: Be extremely careful with the alternating signs ($+,-,+$) when expanding the determinant. A single sign error can lead to a completely wrong answer.
• Algebraic Simplification: Double-check every step of the algebraic simplification. Distributing negative signs correctly and combining like terms accurately are crucial.
• Missing Components: Ensure that all components of the vectors are correctly identified, especially when a coefficient is zero (e.g., the $\widehat j$ component of $\overrightarrow a$).

4. Summary

The problem requires calculating the scalar triple product of three given vectors. By extracting the components of each vector and setting up the corresponding 3x3 determinant, we evaluated the determinant using cofactor expansion. The simplification of the determinant resulted in a constant value of 1, which is independent of the variables $x$ and $y$. Thus, the scalar triple product depends on neither $x$ nor $y$.

The final answer is $\boxed{\text{neither } x \text{ nor } y}$.