Class CBSE Class 12 Mathematics Vector Algebra Q #1293
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2024 AISSCE(Board Exam) VSA
If $\vec{a}$ and $\vec{b}$ are two non-zero vectors such that $(\vec{a}+\vec{b})\perp\vec{a}$ and $(2\vec{a}+\vec{b})\perp\vec{b}$ , then prove that $|\vec{b}|=\sqrt{2}|\vec{a}|$.
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Detailed Solution

Step 1: Use the perpendicularity condition

Since $(\vec{a}+\vec{b})\perp\vec{a}$, their dot product is zero: $$(\vec{a}+\vec{b})\cdot\vec{a} = 0$$ $$\vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{a} = 0$$ $$|\vec{a}|^2 + \vec{a}\cdot\vec{b} = 0 \hspace{1cm} (1)$$ Similarly, since $(2\vec{a}+\vec{b})\perp\vec{b}$, their dot product is zero: $$(2\vec{a}+\vec{b})\cdot\vec{b} = 0$$ $$2\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{b} = 0$$ $$2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 0 \hspace{1cm} (2)$$

Step 2: Solve for $\vec{a}\cdot\vec{b}$

From equation (1), we have: $$\vec{a}\cdot\vec{b} = -|\vec{a}|^2$$ Substitute this into equation (2): $$2(-|\vec{a}|^2) + |\vec{b}|^2 = 0$$ $$-2|\vec{a}|^2 + |\vec{b}|^2 = 0$$

Step 3: Find the relationship between $|\vec{a}|$ and $|\vec{b}|$

Rearrange the equation to solve for $|\vec{b}|^2$: $$|\vec{b}|^2 = 2|\vec{a}|^2$$ Take the square root of both sides: $$|\vec{b}| = \sqrt{2|\vec{a}|^2}$$ $$|\vec{b}| = \sqrt{2}|\vec{a}|$$

Final Answer: $|\vec{b}| = \sqrt{2}|\vec{a}|$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the properties of dot products and perpendicular vectors to derive the relationship between the magnitudes of the vectors.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concepts of dot product, perpendicularity, and vector magnitudes, and how they relate to each other.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of vector algebra, specifically the dot product and its properties related to perpendicularity.

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