Let a = i- 2j + k and b = i-j+k be two vectors. If c is a vector such that b x c = b x a and c.a = 0, then c . b is equal to:

Options

A. `1//2`
B. `-3//2`
C. `-1//2`
D. `-1`

Answer

To solve the problem, we need to find the value of \( c \cdot b \) given the vectors \( a = \mathbf{i} - 2\mathbf{j} + \mathbf{k} \) and \( b = \mathbf{i} - \mathbf{j} + \mathbf{k} \) under the conditions that \( \mathbf{b} \times \mathbf{c} = \mathbf{b} \times \mathbf{a} \) and \( \mathbf{c} \cdot \mathbf{a} = 0 \).

### Step-by-Step Solution:

1. **Define the vector \( c \)**:
Let \( \mathbf{c} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \).

2. **Use the condition \( \mathbf{c} \cdot \mathbf{a} = 0 \)**:
\[ \mathbf{c} \cdot \mathbf{a} = (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \cdot (\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = x - 2y + z = 0 \]
This gives us our first equation:
\[ x - 2y + z = 0 \quad \text{(1)} \]

3. **Calculate \( \mathbf{b} \times \mathbf{c} \)**:
The cross product \( \mathbf{b} \times \mathbf{c} \) can be computed using the determinant:
\[ \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k} \quad \text{and} \quad \mathbf{c} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \]
\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ x & y & z \end{vmatrix} \]
Expanding this determinant:
\[ = \mathbf{i}((-1)z - 1y) - \mathbf{j}(1z - 1x) + \mathbf{k}(1y - (-1)x) \]
\[ = \mathbf{i}(-z - y) - \mathbf{j}(z - x) + \mathbf{k}(y + x) \]

4. **Calculate \( \mathbf{b} \times \mathbf{a} \)**:
\[ \mathbf{a} = \mathbf{i} - 2\mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} \times \mathbf{a} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -2 & 1 \end{vmatrix} \]
Expanding this determinant:
\[ = \mathbf{i}((-1)(1) - 1(-2)) - \mathbf{j}(1(1) - 1(1)) + \mathbf{k}(1(-2) - (-1)(1)) \]
\[ = \mathbf{i}(-1 + 2) - \mathbf{j}(1 - 1) + \mathbf{k}(-2 + 1) \]
\[ = \mathbf{i}(1) + 0\mathbf{j} - \mathbf{k}(1) = \mathbf{i} - \mathbf{k} \]

5. **Set the two cross products equal**:
\[ \mathbf{b} \times \mathbf{c} = \mathbf{b} \times \mathbf{a} \]
This gives us:
\[ -z - y = 1 \quad \text{(2)} \]
\[ z - x = 0 \quad \text{(3)} \]
\[ y + x = -1 \quad \text{(4)} \]

6. **Solve the equations**:
From equation (3), we have \( z = x \). Substitute \( z \) in equation (2):
\[ -x - y = 1 \implies y = -x - 1 \]
Substitute \( y \) in equation (4):
\[ (-x - 1) + x = -1 \implies -1 = -1 \quad \text{(True)} \]
Now substitute \( y \) back into equation (1):
\[ x - 2(-x - 1) + x = 0 \implies x + 2x + 2 = 0 \implies 4x + 2 = 0 \implies x = -\frac{1}{2} \]
Then, \( z = -\frac{1}{2} \) and \( y = -(-\frac{1}{2}) - 1 = -\frac{1}{2} \).

7. **Find \( c \)**:
\[ \mathbf{c} = -\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k} \]

8. **Calculate \( c \cdot b \)**:
\[ \mathbf{c} \cdot \mathbf{b} = \left(-\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\right) \cdot \left(\mathbf{i} - \mathbf{j} + \mathbf{k}\right) \]
\[ = -\frac{1}{2}(1) - \frac{1}{2}(-1) - \frac{1}{2}(1) = -\frac{1}{2} + \frac{1}{2} - \frac{1}{2} = -\frac{1}{2} \]

### Final Answer:
\[ c \cdot b = -\frac{1}{2} \]