If `a = 1, -1, 1, a. b =0, a xx b = c`, where `c = -2, -1, 1` then the vector b is

Options

A. `-1,0,1`
B. `0,1,1`
C. `-1,-1,0`
D. `1,0-1`

Answer

To find the vector \( \mathbf{b} \) given the conditions \( \mathbf{a} = (1, -1, 1) \), \( \mathbf{a} \cdot \mathbf{b} = 0 \), and \( \mathbf{a} \times \mathbf{b} = \mathbf{c} \) where \( \mathbf{c} = (-2, -1, 1) \), we can follow these steps:

### Step 1: Set Up the Vector \( \mathbf{b} \)
Let \( \mathbf{b} = (b_1, b_2, b_3) \). We know that \( \mathbf{a} \cdot \mathbf{b} = 0 \).

### Step 2: Calculate the Dot Product
The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = 1 \cdot b_1 + (-1) \cdot b_2 + 1 \cdot b_3 = b_1 - b_2 + b_3 \]
Setting this equal to zero gives us the equation:
\[ b_1 - b_2 + b_3 = 0 \quad \text{(Equation 1)} \]

### Step 3: Calculate the Cross Product
Next, we calculate the cross product \( \mathbf{a} \times \mathbf{b} \):
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
This determinant expands to:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{i}((-1) b_3 - 1 b_2) - \mathbf{j}(1 b_3 - 1 b_1) + \mathbf{k}(1 b_2 + 1 b_1) \]
Simplifying gives:
\[ \mathbf{a} \times \mathbf{b} = (-b_3 - b_2, -b_3 + b_1, b_2 + b_1) \]
Setting this equal to \( \mathbf{c} = (-2, -1, 1) \) gives us three equations:
1. \(-b_3 - b_2 = -2\) (Equation 2)
2. \(-b_3 + b_1 = -1\) (Equation 3)
3. \(b_2 + b_1 = 1\) (Equation 4)

### Step 4: Solve the System of Equations
Now we have four equations:
1. \(b_1 - b_2 + b_3 = 0\) (Equation 1)
2. \(-b_3 - b_2 = -2\) (Equation 2)
3. \(-b_3 + b_1 = -1\) (Equation 3)
4. \(b_2 + b_1 = 1\) (Equation 4)

From Equation 2:
\[ b_3 + b_2 = 2 \implies b_3 = 2 - b_2 \quad \text{(Substituting into Equation 1)} \]

Substituting \( b_3 \) into Equation 1:
\[ b_1 - b_2 + (2 - b_2) = 0 \implies b_1 - 2b_2 + 2 = 0 \implies b_1 = 2b_2 - 2 \quad \text{(Equation 5)} \]

Now substituting Equation 5 into Equation 4:
\[ b_2 + (2b_2 - 2) = 1 \implies 3b_2 - 2 = 1 \implies 3b_2 = 3 \implies b_2 = 1 \]

### Step 5: Find \( b_1 \) and \( b_3 \)
Using \( b_2 = 1 \) in Equation 5:
\[ b_1 = 2(1) - 2 = 0 \]
Using \( b_2 = 1 \) in the expression for \( b_3 \):
\[ b_3 = 2 - 1 = 1 \]

### Final Result
Thus, the vector \( \mathbf{b} \) is:
\[ \mathbf{b} = (0, 1, 1) \]