If `a=i-j-k,a*b=1" and "a times b=-j+k`, then k is equal to

Options

A. `i+j-k`
B. `-2j+k`
C. `i`
D. `2j+k`

Answer

To solve the problem, we need to find the value of \( k \) given the vectors \( \mathbf{a} = \mathbf{i} - \mathbf{j} - \mathbf{k} \), \( \mathbf{a} \cdot \mathbf{b} = 1 \), and \( \mathbf{a} \times \mathbf{b} = -\mathbf{j} + \mathbf{k} \).

### Step-by-Step Solution:

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

2. **Use the dot product condition**:
The dot product \( \mathbf{a} \cdot \mathbf{b} = 1 \):
\[ (\mathbf{i} - \mathbf{j} - \mathbf{k}) \cdot (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) = 1 \]
This expands to:
\[ x - y - z = 1 \quad \text{(Equation 1)} \]

3. **Use the cross product condition**:
The cross product \( \mathbf{a} \times \mathbf{b} = -\mathbf{j} + \mathbf{k} \):
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -1 \\ x & y & z \end{vmatrix} \]
This determinant can be calculated as:
\[ = \mathbf{i} \begin{vmatrix} -1 & -1 \\ y & z \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -1 \\ x & z \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1 \\ x & y \end{vmatrix} \]
Calculating each of these determinants:
\[ = \mathbf{i}((-1)z - (-1)y) - \mathbf{j}(1z - (-1)x) + \mathbf{k}(1y - (-1)x) \]
Simplifying gives:
\[ = (y - z)\mathbf{i} - (z + x)\mathbf{j} + (y + x)\mathbf{k} \]
Setting this equal to \( -\mathbf{j} + \mathbf{k} \):
\[ (y - z)\mathbf{i} - (z + x)\mathbf{j} + (y + x)\mathbf{k} = 0\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} \]

4. **Equate coefficients**:
From the above equation, we can equate the coefficients:
\[ y - z = 0 \quad \text{(Equation 2)} \]
\[ - (z + x) = -1 \implies z + x = 1 \quad \text{(Equation 3)} \]
\[ y + x = 1 \quad \text{(Equation 4)} \]

5. **Solve the equations**:
From Equation 2, we have:
\[ y = z \]
Substitute \( y = z \) into Equation 3:
\[ z + x = 1 \implies y + x = 1 \quad \text{(since \( y = z \))} \]
Thus, we have:
\[ x + z = 1 \quad \text{(Equation 3)} \]
Now substituting \( y = z \) into Equation 1:
\[ x - z - z = 1 \implies x - 2z = 1 \quad \text{(Equation 5)} \]

6. **Solve Equations 3 and 5**:
From Equation 3:
\[ x = 1 - z \]
Substitute into Equation 5:
\[ (1 - z) - 2z = 1 \implies 1 - 3z = 1 \implies -3z = 0 \implies z = 0 \]
Therefore, \( y = z = 0 \).

7. **Find \( x \)**:
Substitute \( z = 0 \) into Equation 3:
\[ x + 0 = 1 \implies x = 1 \]

8. **Conclusion**:
We have found \( x = 1 \), \( y = 0 \), and \( z = 0 \). Thus, \( k = z = 0 \).

### Final Answer:
\[ k = 0 \]