Let `vec(a) = hat(i) + hat(j) + hat(k),vec(b) = hat(i) - hat(j) + 2hat(k)` and `vec(c) = xhat(i) + (x-2)hat(j) - hat(k)`. If the vector `vec(c)` lies in the plane of `vec(a)` and `vec(b)` then x equals

Options

A. 0
B. 1
C. `-4`
D. `-2`

Answer

To solve the problem, we need to find the value of \( x \) such that the vector \( \vec{c} \) lies in the plane formed by the vectors \( \vec{a} \) and \( \vec{b} \). This can be determined using the scalar triple product, which states that if three vectors are coplanar, their scalar triple product is zero.

### Step-by-step Solution:

1. **Define the Vectors**:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \]
\[ \vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k} \]

2. **Set Up the Determinant**:
The condition for coplanarity is given by the determinant of the matrix formed by the vectors \( \vec{a}, \vec{b}, \vec{c} \) being equal to zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
We can expand the determinant using the first row:
\[ = 1 \cdot \begin{vmatrix} -1 & 2 \\ x-2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 2 \\ x & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ x & x-2 \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
- First determinant:
\[ = (-1)(-1) - (2)(x-2) = 1 - 2x + 4 = 5 - 2x \]
- Second determinant:
\[ = (1)(-1) - (2)(x) = -1 - 2x \]
- Third determinant:
\[ = (1)(x-2) - (-1)(x) = x - 2 + x = 2x - 2 \]

4. **Combine the Determinants**:
Now substituting back into the determinant:
\[ 1(5 - 2x) - 1(-1 - 2x) + 1(2x - 2) = 0 \]
Simplifying this gives:
\[ 5 - 2x + 1 + 2x + 2x - 2 = 0 \]
Combine like terms:
\[ 5 + 1 - 2 = 0 \implies 4 + 2x = 0 \]

5. **Solve for \( x \)**:
\[ 2x + 4 = 0 \implies 2x = -4 \implies x = -2 \]

### Final Answer:
\[ x = -2 \]