Q1250: Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors. Prove t...

Class CBSE Class 12 Mathematics Vector Algebra Q #1250

KNOWLEDGE BASED

REMEMBER

2 Marks 2024 AISSCE(Board Exam) VSA

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors. Prove that $|\vec{a}\times\vec{b}|\le|\vec{a}||\vec{b}|$ . State the condition under which equality holds, i.e., $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|$

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AI Tutor Explanation

Detailed Solution

Step 1: Recall the formula for the magnitude of the cross product

The magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is given by: $$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$$ where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, and $0 \le \theta \le \pi$.

Step 2: Establish the inequality

Since $0 \le \theta \le \pi$, we know that $0 \le \sin\theta \le 1$. Therefore, $$|\vec{a}||\vec{b}|\sin\theta \le |\vec{a}||\vec{b}|$$ This implies that $$|\vec{a}\times\vec{b}| \le |\vec{a}||\vec{b}|$$

Step 3: Determine the condition for equality

The equality $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|$ holds if and only if $\sin\theta = 1$. This occurs when $\theta = \frac{\pi}{2}$. Therefore, the vectors $\vec{a}$ and $\vec{b}$ must be perpendicular (orthogonal) to each other.

Final Answer: $|\vec{a}\times\vec{b}|\le|\vec{a}||\vec{b}|$. Equality holds when $\vec{a}$ and $\vec{b}$ are perpendicular.

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall the formula for the magnitude of the cross product and the range of the sine function to prove the inequality and state the condition for equality.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concepts of cross product, magnitude of vectors, and the geometric interpretation of the cross product.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of vector algebra, specifically the properties of the cross product.