Let `veca=hati-hatk, vecb=xhati+hatj+(1-x)hatk` and `vecc=yhati+xhatj+(1+x-y)hatk`, then `[veca vecb vecc]` depends on

Options

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

Explanation

To solve the problem, we need to find the determinant of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) given as:

\[ \vec{a} = \hat{i} - \hat{k} \]
\[ \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k} \]
\[ \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \]

### Step 1: Write the vectors in matrix form

We will represent the vectors as a matrix where each row corresponds to the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):

\[ \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1 - x \\ y & x & 1 + x - y \end{vmatrix} \]

### Step 2: Simplify the determinant

We can simplify the determinant by performing row operations. We will add the first row to the third row:

\[ \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1 - x \\ y + 1 & x & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant

Now, we can calculate the determinant using the formula for a \(3 \times 3\) matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

\[ D = 1 \cdot (1 \cdot 0 - (1 - x) \cdot x) - 0 + (-1) \cdot (x \cdot (y + 1) - 1 \cdot y) \]

Calculating the terms:

1. The first term simplifies to:
\[ 1 \cdot (0 - (1 - x)x) = - (1 - x)x = -x + x^2 \]

2. The second term is zero since it is multiplied by zero.

3. The third term simplifies to:
\[ -1 \cdot (xy + x - y) = -xy - x + y \]

Putting it all together:

\[ D = -x + x^2 - xy - x + y \]
\[ D = x^2 - xy - 2x + y \]

### Step 4: Analyze the dependency

Now, we need to determine if the determinant \(D\) depends on \(x\) and \(y\).

The expression \(D = x^2 - xy - 2x + y\) is a polynomial in \(x\) and \(y\). Since it contains terms involving both \(x\) and \(y\), it indicates that the determinant does depend on these variables.

### Conclusion

The determinant \([ \vec{a} \, \vec{b} \, \vec{c} ]\) depends on \(x\) and \(y\).