If r satisfies the equation `r xx (i+2j+k) = i-k` then for any scalar `lambda,r` is equal to

Options

A. `i+lambda(i+2j+k)`
B. `j+lambda(i+2j+k)`
C. `k+lambda(i+2j+k)`
D. none

Answer

To solve the equation \( \mathbf{r} \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = \mathbf{i} - \mathbf{k} \), we will follow these steps:

### Step 1: Set Up the Cross Product
Let \( \mathbf{r} = r_1 \mathbf{i} + r_2 \mathbf{j} + r_3 \mathbf{k} \). We need to compute the cross product \( \mathbf{r} \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \).

### Step 2: Write the Determinant
The cross product can be calculated using the determinant of a matrix:
\[ \mathbf{r} \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ r_1 & r_2 & r_3 \\ 1 & 2 & 1 \end{vmatrix} \]

### Step 3: Expand the Determinant
Expanding the determinant, we get:
\[ \mathbf{r} \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = \mathbf{i}(r_2 \cdot 1 - r_3 \cdot 2) - \mathbf{j}(r_1 \cdot 1 - r_3 \cdot 1) + \mathbf{k}(r_1 \cdot 2 - r_2 \cdot 1) \]
This simplifies to:
\[ = (r_2 - 2r_3) \mathbf{i} - (r_1 - r_3) \mathbf{j} + (2r_1 - r_2) \mathbf{k} \]

### Step 4: Set the Result Equal to the Right-Hand Side
We know that:
\[ \mathbf{r} \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = \mathbf{i} - \mathbf{k} \]
This gives us the equations:
1. \( r_2 - 2r_3 = 1 \) (coefficient of \( \mathbf{i} \))
2. \( -(r_1 - r_3) = 0 \) (coefficient of \( \mathbf{j} \))
3. \( 2r_1 - r_2 = -1 \) (coefficient of \( \mathbf{k} \))

### Step 5: Solve the System of Equations
From equation 2:
\[ r_1 = r_3 \]
Substituting \( r_1 \) for \( r_3 \) in equation 1:
\[ r_2 - 2r_1 = 1 \quad \Rightarrow \quad r_2 = 2r_1 + 1 \]
Substituting \( r_2 \) in equation 3:
\[ 2r_1 - (2r_1 + 1) = -1 \quad \Rightarrow \quad 2r_1 - 2r_1 - 1 = -1 \]
This is always true, indicating infinite solutions.

### Step 6: Express \( \mathbf{r} \)
Thus, we can express \( \mathbf{r} \) as:
\[ \mathbf{r} = r_1 \mathbf{i} + (2r_1 + 1) \mathbf{j} + r_1 \mathbf{k} \]
Factoring out \( r_1 \):
\[ \mathbf{r} = r_1 (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \mathbf{j} \]

### Step 7: General Solution
For any scalar \( \lambda \), we can let \( r_1 = \lambda \):
\[ \mathbf{r} = \lambda (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \mathbf{j} \]

### Final Result
Thus, the vector \( \mathbf{r} \) can be expressed as:
\[ \mathbf{r} = \lambda \mathbf{i} + (2\lambda + 1) \mathbf{j} + \lambda \mathbf{k} \]