Question

The value of \( m \) for which the points with position vectors \( -\hat{i} - \hat{j} + 2\hat{k} \), \( 2\hat{i} + m\hat{j} + 5\hat{k} \) and \( 3\hat{i} + 11\hat{j} + 6\hat{k} \) are collinear, is

• A. \( 8 \)
• B. \( -8 \)
• C. \( 2 \)
• D. \( \frac{5}{2} \)

Hint:

For collinearity of \( A, B, C \), the area of triangle \( ABC \) must be zero.
Alternatively, checking the ratio of the change in components is often the fastest method.

Answer 1

The Correct Option is A

Concept:
• Three points \( A, B, C \) are collinear if the vectors \( \vec{AB} \) and \( \vec{BC} \) (or \( \vec{AC} \)) are parallel.
• Parallel vectors have proportional components: if \( \vec{u} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \) and \( \vec{v} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k} \), then \( \frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{z_1}{z_2} \).

Step 1: Find the vectors connecting the points
Let the points be \( A(-\hat{i} - \hat{j} + 2\hat{k}) \), \( B(2\hat{i} + m\hat{j} + 5\hat{k}) \), and \( C(3\hat{i} + 11\hat{j} + 6\hat{k}) \). Vector \( \vec{AB} = \vec{OB} - \vec{OA} \): \[ \vec{AB} = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k} \] \[ \vec{AB} = 3\hat{i} + (m + 1)\hat{j} + 3\hat{k} \] Vector \( \vec{BC} = \vec{OC} - \vec{OB} \): \[ \vec{BC} = (3 - 2)\hat{i} + (11 - m)\hat{j} + (6 - 5)\hat{k} \] \[ \vec{BC} = 1\hat{i} + (11 - m)\hat{j} + 1\hat{k} \]

Step 2: Apply the condition for collinearity
Since the points are collinear, \( \vec{AB} = \lambda \vec{BC} \). Equating the ratios of the components: \[ \frac{3}{1} = \frac{m + 1}{11 - m} = \frac{3}{1} \]

Step 3: Solve for \( m \)
From the first equality: \[ 3 = \frac{m + 1}{11 - m} \] Cross-multiply: \[ 3(11 - m) = m + 1 \] \[ 33 - 3m = m + 1 \] Group the terms: \[ 33 - 1 = m + 3m \] \[ 32 = 4m \] \[ m = 8 \]