Let `vec a = hati - hat k, vec b =x hati + hatj+ (1-x) hat k and vec c = y hat I +x hat j +(1+x-y) hatk`. Then `[vec a, vec b, vec c]` depends on

Options

A. only y
B. only x
C. both x and y
D. neither x nor y

Answer

To determine the dependence of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we will calculate the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) using the determinant of a matrix formed by these vectors.

### Step-by-Step Solution:

1. **Define the Vectors**:
\[ \vec{a} = \hat{i} - \hat{k} \]
\[ \vec{b} = x\hat{i} + \hat{j} + (1-x)\hat{k} \]
\[ \vec{c} = y\hat{i} + x\hat{j} + (1+x-y)\hat{k} \]

2. **Write the Coefficients**:
The coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) for each vector are:
- For \(\vec{a}\): \(1, 0, -1\)
- For \(\vec{b}\): \(x, 1, 1-x\)
- For \(\vec{c}\): \(y, x, 1+x-y\)

3. **Set Up the Determinant**:
The scalar triple product can be calculated using the determinant:
\[ [\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix} \]

4. **Calculate the Determinant**:
Using the determinant formula, we expand it as follows:
\[ = 1 \cdot \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} - 0 + (-1) \cdot \begin{vmatrix} x & 1 \\ y & x \end{vmatrix} \]

Now, calculating the first determinant:
\[ = 1 \cdot (1(1+x-y) - (1-x)x) = 1 + x - y - x + x^2 = x^2 - y + 1 \]

For the second determinant:
\[ = -1 \cdot (x \cdot x - 1 \cdot y) = - (x^2 - y) \]

Combining these results:
\[ [\vec{a}, \vec{b}, \vec{c}] = (x^2 - y + 1) - (x^2 - y) = 1 \]

5. **Conclusion**:
The scalar triple product \([\vec{a}, \vec{b}, \vec{c}] = 1\), which is a constant and does not depend on \(x\) or \(y\).

### Final Answer:
The vectors \([\vec{a}, \vec{b}, \vec{c}]\) depend on neither \(x\) nor \(y\).