If `vec(a)=hat(i)-hat(k), vec(b)=xhat(i)+hat(j)+(1-x)hat(k)`
`vec(c)=yhat(i)+xhat(j)+(1+x-y)hat(k)`.
then `vec(a).(vec(b) xx vec (c))` depends on

Options

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

Answer

To solve the problem, we need to find the value of \(\vec{a} \cdot (\vec{b} \times \vec{c})\) where:

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

### Step 1: Calculate \(\vec{b} \times \vec{c}\)

To calculate the cross product \(\vec{b} \times \vec{c}\), we can use the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{b}\) and \(\vec{c}\):

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix} \]

### Step 2: Calculate the determinant

Expanding the determinant, we have:

\[ \vec{b} \times \vec{c} = \hat{i} \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} - \hat{j} \begin{vmatrix} x & 1-x \\ y & 1+x-y \end{vmatrix} + \hat{k} \begin{vmatrix} x & 1 \\ y & x \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} = 1(1+x-y) - (1-x)x = 1 + x - y - x + x^2 = x^2 + 1 - y \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} x & 1-x \\ y & 1+x-y \end{vmatrix} = x(1+x-y) - (1-x)y = x + x^2 - xy - y + xy = x + x^2 - y \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} x & 1 \\ y & x \end{vmatrix} = x^2 - y \]

Putting it all together, we have:

\[ \vec{b} \times \vec{c} = (x^2 + 1 - y)\hat{i} - (x + x^2 - y)\hat{j} + (x^2 - y)\hat{k} \]

### Step 3: Calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\)

Now we compute the dot product \(\vec{a} \cdot (\vec{b} \times \vec{c})\):

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (\hat{i} - \hat{k}) \cdot \left((x^2 + 1 - y)\hat{i} - (x + x^2 - y)\hat{j} + (x^2 - y)\hat{k}\right) \]

Calculating this gives:

\[ = (x^2 + 1 - y) - (x^2 - y) = 1 \]

### Final Result

Thus, the value of \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is:

\[ \boxed{1} \]