`vecA = (1, -1, 1), vecC = (-1,-1,0)` are given vectors then the vector B which satisfies `A xx B = C` and A.B = 1 is

Options

A. `(1, 0, 0)`
B. `(0, 0, 1)`
C. `(0, -1, 0)`
D. none

Answer

To solve the problem, we need to find the vector \( \vec{B} \) that satisfies the conditions \( \vec{A} \times \vec{B} = \vec{C} \) and \( \vec{A} \cdot \vec{B} = 1 \), given \( \vec{A} = (1, -1, 1) \) and \( \vec{C} = (-1, -1, 0) \).

### Step-by-Step Solution:

1. **Define the Vectors:**
\[ \vec{A} = (1, -1, 1) \quad \text{and} \quad \vec{C} = (-1, -1, 0) \]
Let \( \vec{B} = (x, y, z) \).

2. **Dot Product Condition:**
We have the condition \( \vec{A} \cdot \vec{B} = 1 \):
\[ \vec{A} \cdot \vec{B} = 1 \cdot x + (-1) \cdot y + 1 \cdot z = x - y + z = 1 \quad \text{(Equation 1)} \]

3. **Cross Product Condition:**
We also have the condition \( \vec{A} \times \vec{B} = \vec{C} \). The cross product can be calculated using the determinant:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ x & y & z \end{vmatrix} \]
Expanding this determinant:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -1 & 1 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ x & y \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i}((-1)z - 1y) - \hat{j}(1z - 1x) + \hat{k}(1y + 1x) \]
\[ = \hat{i}(-z - y) - \hat{j}(z - x) + \hat{k}(y + x) \]
Therefore,
\[ \vec{A} \times \vec{B} = (-z - y, -(z - x), y + x) \]
Setting this equal to \( \vec{C} = (-1, -1, 0) \):
\[ -z - y = -1 \quad \text{(Equation 2)} \]
\[ -(z - x) = -1 \quad \text{(Equation 3)} \]
\[ y + x = 0 \quad \text{(Equation 4)} \]

4. **Solving the Equations:**
From Equation 2:
\[ z + y = 1 \quad \Rightarrow \quad z = 1 - y \quad \text{(Equation 5)} \]
From Equation 3:
\[ z - x = 1 \quad \Rightarrow \quad z = x + 1 \quad \text{(Equation 6)} \]
Now we have two expressions for \( z \):
\[ 1 - y = x + 1 \]
Rearranging gives:
\[ x + y = 0 \quad \Rightarrow \quad y = -x \quad \text{(Equation 7)} \]
Substituting Equation 7 into Equation 1:
\[ x - (-x) + z = 1 \quad \Rightarrow \quad 2x + z = 1 \]
Using Equation 6:
\[ 2x + (x + 1) = 1 \]
\[ 3x + 1 = 1 \quad \Rightarrow \quad 3x = 0 \quad \Rightarrow \quad x = 0 \]
Substituting \( x = 0 \) into Equation 7:
\[ y = -0 = 0 \]
Finally, substituting \( y = 0 \) into Equation 5:
\[ z = 1 - 0 = 1 \]

5. **Final Result:**
Therefore, the vector \( \vec{B} \) is:
\[ \vec{B} = (0, 0, 1) \]

### Conclusion:
The vector \( \vec{B} \) that satisfies both conditions is:
\[ \vec{B} = (0, 0, 1) \]