For what value of m are the point with position vectors, `10 hat(i) + 3hat(j), 12 hat(i) - 5hat(j) and m hat(i) + 11 hat(j)` collinear?

Options

A. `-8`
B. 4
C. 8
D. 12

Answer

To find the value of \( m \) for which the points with position vectors \( \mathbf{A} = 10 \hat{i} + 3 \hat{j} \), \( \mathbf{B} = 12 \hat{i} - 5 \hat{j} \), and \( \mathbf{C} = m \hat{i} + 11 \hat{j} \) are collinear, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Position Vectors**:
- Let \( \mathbf{A} = 10 \hat{i} + 3 \hat{j} \)
- Let \( \mathbf{B} = 12 \hat{i} - 5 \hat{j} \)
- Let \( \mathbf{C} = m \hat{i} + 11 \hat{j} \)

2. **Find Vectors \( \mathbf{AB} \) and \( \mathbf{BC} \)**:
- The vector \( \mathbf{AB} = \mathbf{B} - \mathbf{A} \):
\[ \mathbf{AB} = (12 - 10) \hat{i} + (-5 - 3) \hat{j} = 2 \hat{i} - 8 \hat{j} \]
- The vector \( \mathbf{BC} = \mathbf{C} - \mathbf{B} \):
\[ \mathbf{BC} = (m - 12) \hat{i} + (11 + 5) \hat{j} = (m - 12) \hat{i} + 16 \hat{j} \]

3. **Set Up the Condition for Collinearity**:
- The vectors \( \mathbf{AB} \) and \( \mathbf{BC} \) are collinear if their cross product is zero:
\[ \mathbf{AB} \times \mathbf{BC} = 0 \]

4. **Calculate the Cross Product**:
- The cross product in determinant form is:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -8 & 0 \\ m - 12 & 16 & 0 \end{vmatrix} \]
- Expanding this determinant:
\[ \hat{k} \left( 2 \cdot 16 - (-8)(m - 12) \right) = 0 \]
- This simplifies to:
\[ 32 + 8(m - 12) = 0 \]

5. **Solve for \( m \)**:
- Expanding and rearranging:
\[ 32 + 8m - 96 = 0 \]
\[ 8m - 64 = 0 \]
\[ 8m = 64 \]
\[ m = \frac{64}{8} = 8 \]

### Conclusion:
The value of \( m \) for which the points are collinear is \( m = 8 \).