Class CBSE Class 12 Mathematics Vector Algebra Q #1727
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
If $(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=198$ and $|\vec{a}|=10|\vec{b}|$, then :
(A) $|\vec{a}|=\sqrt{2}$
(B) $|\vec{b}|=\sqrt{2}$
(C) $|\vec{b}|=10\sqrt{2}$
(D) $|\vec{a}|=\frac{10}{\sqrt{2}}$
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Correct Answer: B

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Detailed Solution

Step 1: Expand the dot product

Given the expression (a + b) . (a - b) = 198. Using the distributive property of the dot product, we get: $$|\vec{a}|^2 - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - |\vec{b}|^2 = 198$$ Since the dot product is commutative, a . b = b . a, the middle terms cancel out: $$|\vec{a}|^2 - |\vec{b}|^2 = 198$$

Step 2: Substitute the given relation

We are given |a| = 10|b|. Substitute this into the equation derived in Step 1: $$(10|\vec{b}|)^2 - |\vec{b}|^2 = 198$$ $$100|\vec{b}|^2 - |\vec{b}|^2 = 198$$ $$99|\vec{b}|^2 = 198$$

Step 3: Solve for |b|

Divide both sides by 99: $$|\vec{b}|^2 = \frac{198}{99}$$ $$|\vec{b}|^2 = 2$$ $$|\vec{b}| = \sqrt{2}$$

Final Answer: |b| = \sqrt{2}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the algebraic properties of vector dot products and substitution to solve for the magnitude of a vector.
Knowledge Dimension: PROCEDURAL
Justification: The problem requires a step-by-step algorithmic approach involving expansion, substitution, and algebraic simplification.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the conceptual understanding of vector algebra properties (dot product) beyond simple rote memorization.
||KEY:B

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