The are three vectors `a = hati+hatj, b = hatj+hatk and hatc = x hata + y hatb`. If the vectors `hati-2hatj+hatk,3hati+2hatj-hatk and c` are coplanar, then `x/y` is equal to

Options

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

Answer

To solve the problem, we need to determine the ratio \( \frac{x}{y} \) given that the vectors \( \mathbf{p_1} = \hat{i} - 2\hat{j} + \hat{k} \), \( \mathbf{p_2} = 3\hat{i} + 2\hat{j} - \hat{k} \), and \( \mathbf{c} = x\hat{a} + y\hat{b} \) are coplanar.

### Step-by-Step Solution

1. **Define the vectors**:
- We have \( \mathbf{a} = \hat{i} + \hat{j} \)
- \( \mathbf{b} = \hat{j} + \hat{k} \)
- The vector \( \mathbf{c} = x\mathbf{a} + y\mathbf{b} = x(\hat{i} + \hat{j}) + y(\hat{j} + \hat{k}) \)

2. **Express vector \( \mathbf{c} \)**:
- Expanding \( \mathbf{c} \):
\[ \mathbf{c} = x\hat{i} + (x + y)\hat{j} + y\hat{k} \]

3. **Set up the matrix for coplanarity**:
- The vectors \( \mathbf{p_1} \), \( \mathbf{p_2} \), and \( \mathbf{c} \) are coplanar if the determinant of the matrix formed by their coefficients is zero:
\[ \begin{vmatrix} 1 & -2 & 1 \\ 3 & 2 & -1 \\ x & x+y & y \end{vmatrix} = 0 \]

4. **Calculate the determinant**:
- Expanding the determinant:
\[ = 1 \begin{vmatrix} 2 & -1 \\ x+y & y \end{vmatrix} - (-2) \begin{vmatrix} 3 & -1 \\ x & y \end{vmatrix} + 1 \begin{vmatrix} 3 & 2 \\ x & x+y \end{vmatrix} \]
- Calculate each of the 2x2 determinants:
- First determinant:
\[ = 2y - (-1)(x+y) = 2y + x + y = x + 3y \]
- Second determinant:
\[ = 3y - (-1)(x) = 3y + x \]
- Third determinant:
\[ = 3(x+y) - 2x = 3x + 3y - 2x = x + 3y \]

5. **Combine the results**:
- The determinant becomes:
\[ 1(x + 3y) + 2(3y + x) + 1(x + 3y) = 0 \]
- Simplifying:
\[ x + 3y + 6y + 2x + x + 3y = 0 \implies 4x + 12y = 0 \]

6. **Solve for \( \frac{x}{y} \)**:
- Rearranging gives:
\[ 4x = -12y \implies \frac{x}{y} = -3 \]

### Final Answer
Thus, the value of \( \frac{x}{y} \) is \( -3 \).