CBSE Class 12 Mathematics Vector Algebra Q #1851
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
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) $\frac{5}{2}$
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Correct Answer: A

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Detailed Solution

Step 1: Define the position vectors

Let the given points be A, B, and C with position vectors: $$ \vec{a} = -\hat{i} - \hat{j} + 2\hat{k} $$ $$ \vec{b} = 2\hat{i} + m\hat{j} + 5\hat{k} $$ $$ \vec{c} = 3\hat{i} + 11\hat{j} + 6\hat{k} $$

Step 2: Determine vectors AB and BC

For the points to be collinear, the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel. $$ \vec{AB} = \vec{b} - \vec{a} = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k} = 3\hat{i} + (m+1)\hat{j} + 3\hat{k} $$ $$ \vec{BC} = \vec{c} - \vec{b} = (3 - 2)\hat{i} + (11 - m)\hat{j} + (6 - 5)\hat{k} = 1\hat{i} + (11-m)\hat{j} + 1\hat{k} $$

Step 3: Apply the condition for collinearity

Since $\vec{AB}$ and $\vec{BC}$ are parallel, their components must be proportional: $$ \frac{3}{1} = \frac{m+1}{11-m} = \frac{3}{1} $$ This simplifies to: $$ 3 = \frac{m+1}{11-m} $$

Step 4: Solve for m

Multiply both sides by $(11-m)$: $$ 3(11 - m) = m + 1 $$ $$ 33 - 3m = m + 1 $$ $$ 32 = 4m $$ $$ m = 8 $$

Final Answer: 8

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the geometric property of collinearity (parallel vectors) to solve for an unknown variable in a vector equation.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a step-by-step algorithmic approach involving vector subtraction and the application of proportionality constants.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This aligns with the Vector Algebra unit, testing the student's ability to translate geometric conditions into algebraic expressions.

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