If three points A, B and C have position vectors `hati + x hatj + 3 hatk, 3 hati + 4 hatj + 7 hatk and y hati -2hatj - 5 hatk ` respectively are collinear, then (x, y) =

Options

A. (2, -3)
B. (-2, 3)
C. (-2, -3)
D. (2, 3)

Answer

To determine the values of \(x\) and \(y\) such that the points \(A\), \(B\), and \(C\) are collinear, we start with their position vectors:

- Point \(A\): \(\mathbf{A} = \hat{i} + x \hat{j} + 3 \hat{k}\)
- Point \(B\): \(\mathbf{B} = 3 \hat{i} + 4 \hat{j} + 7 \hat{k}\)
- Point \(C\): \(\mathbf{C} = y \hat{i} - 2 \hat{j} - 5 \hat{k}\)

### Step 1: Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

The vectors can be calculated as follows:

\[ \overrightarrow{AB} = \mathbf{B} - \mathbf{A} = (3 - 1)\hat{i} + (4 - x)\hat{j} + (7 - 3)\hat{k} = 2\hat{i} + (4 - x)\hat{j} + 4\hat{k} \]

\[ \overrightarrow{AC} = \mathbf{C} - \mathbf{A} = (y - 1)\hat{i} + (-2 - x)\hat{j} + (-5 - 3)\hat{k} = (y - 1)\hat{i} + (-2 - x)\hat{j} - 8\hat{k} \]

### Step 2: Set up the condition for collinearity

For the points \(A\), \(B\), and \(C\) to be collinear, the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) must be scalar multiples of each other. This can be expressed using the determinant of the matrix formed by these vectors:

\[ \begin{vmatrix} 2 & 4 - x & 4 \\ y - 1 & -2 - x & -8 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant

Expanding the determinant:

\[ = 2 \begin{vmatrix} -2 - x & -8 \\ 1 & 1 \end{vmatrix} - (4 - x) \begin{vmatrix} y - 1 & -8 \\ 1 & 1 \end{vmatrix} + 4 \begin{vmatrix} y - 1 & -2 - x \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -2 - x & -8 \\ 1 & 1 \end{vmatrix} = (-2 - x) \cdot 1 - (-8) \cdot 1 = -2 - x + 8 = 6 - x \)

2. \( \begin{vmatrix} y - 1 & -8 \\ 1 & 1 \end{vmatrix} = (y - 1) \cdot 1 - (-8) \cdot 1 = y - 1 + 8 = y + 7 \)

3. \( \begin{vmatrix} y - 1 & -2 - x \\ 1 & 1 \end{vmatrix} = (y - 1) \cdot 1 - (-2 - x) \cdot 1 = y - 1 + 2 + x = y + x + 1 \)

Substituting these back into the determinant:

\[ = 2(6 - x) - (4 - x)(y + 7) + 4(y + x + 1) = 0 \]

### Step 4: Expand and simplify

Expanding gives:

\[ = 12 - 2x - (4y + 28 - xy - 7x) + 4y + 4x + 4 = 0 \]

Combining like terms:

\[ = 12 + 4 - 28 + 2x + xy = 0 \implies xy + 2x - 12 = 0 \]

### Step 5: Solve for \(x\) and \(y\)

Rearranging gives:

\[ xy + 2x - 12 = 0 \implies y = \frac{12 - 2x}{x} \]

### Step 6: Substitute possible values for \(x\)

Testing \(x = 2\):

\[ y = \frac{12 - 2(2)}{2} = \frac{12 - 4}{2} = \frac{8}{2} = 4 \]

Testing \(x = -2\):

\[ y = \frac{12 - 2(-2)}{-2} = \frac{12 + 4}{-2} = \frac{16}{-2} = -8 \]

### Conclusion

The values of \(x\) and \(y\) that satisfy the collinearity condition are:

\[ (x, y) = (2, 4) \]