Let `veca =veci -veck, vecb = xveci+ vecj + (1-x)veck and vecc =y veci +xvecj + (1+x -y)veck`. Then `veca, vecb and vecc` are non-coplanar for

Options

A. some values of x
B. some values of y
C. no values of x and y
D. for all values of x and y

Explanation

To determine the conditions under which the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are non-coplanar, we will follow these steps:

### Step 1: Define the Vectors
We start with the given vectors:
- \(\vec{a} = \vec{i} - \vec{k}\)
- \(\vec{b} = x\vec{i} + \vec{j} + (1-x)\vec{k}\)
- \(\vec{c} = y\vec{i} + x\vec{j} + (1 + x - y)\vec{k}\)

### Step 2: Write the Coefficients
Next, we extract the coefficients of \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\) from each vector:
- For \(\vec{a}\): Coefficients are \(1\) (for \(\vec{i}\)), \(0\) (for \(\vec{j}\)), and \(-1\) (for \(\vec{k}\)).
- For \(\vec{b}\): Coefficients are \(x\) (for \(\vec{i}\)), \(1\) (for \(\vec{j}\)), and \(1-x\) (for \(\vec{k}\)).
- For \(\vec{c}\): Coefficients are \(y\) (for \(\vec{i}\)), \(x\) (for \(\vec{j}\)), and \(1+x-y\) (for \(\vec{k}\)).

### Step 3: Form the Determinant
To check for non-coplanarity, we calculate the determinant of the matrix formed by these coefficients:
\[ \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix} \]

### Step 4: Calculate the Determinant
We compute the determinant using the formula for a \(3 \times 3\) matrix:
\[ D = 1 \cdot \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} - 0 + (-1) \cdot \begin{vmatrix} x & 1-x \\ y & x \end{vmatrix} \]

Calculating the first \(2 \times 2\) determinant:
\[ \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} = 1(1+x-y) - (1-x)x = 1 + x - y - x + x^2 = x^2 + 1 - y \]

Calculating the second \(2 \times 2\) determinant:
\[ \begin{vmatrix} x & 1-x \\ y & x \end{vmatrix} = x \cdot x - (1-x)y = x^2 - y + xy \]

Putting it all together:
\[ D = (x^2 + 1 - y) - (x^2 - y + xy) = 1 - xy \]

### Step 5: Determine Non-Coplanarity Condition
The vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are non-coplanar if the determinant \(D \neq 0\):
\[ 1 - xy \neq 0 \implies xy \neq 1 \]

### Conclusion
Thus, the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are non-coplanar for the condition:
\[ xy \neq 1 \]