Class CBSE Class 12 Mathematics Vector Algebra Q #1726
KNOWLEDGE BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
For any two vectors $\vec{a}$ and $\vec{b}$, which of the following statements is always true ?
(A) $\vec{a}.\vec{b}\le|\vec{a}||\vec{b}|$
(B) $|\vec{a}+\vec{b}|\ge|\vec{a}|+|\vec{b}|$
(C) $|\vec{a}-\vec{b}|=|\vec{a}|-|\vec{b}|$
(D) $|\vec{a}\times\vec{b}|\ge|\vec{a}||\vec{b}|$
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Correct Answer: A

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Detailed Solution

Step 1: Analyze Option (A)

The scalar product of two vectors is defined as $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$. Since the range of the cosine function is $[-1, 1]$, it follows that $\cos \theta \le 1$. Therefore, $\vec{a} \cdot \vec{b} \le |\vec{a}||\vec{b}|$. This statement is always true.

Step 2: Analyze Option (B)

The triangle inequality for vectors states that $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$. Option (B) suggests the reverse inequality, which is generally false.

Step 3: Analyze Option (C)

The expression $|\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}|$ is only true under specific conditions (e.g., when vectors are collinear and in the same direction), but it is not true for any two arbitrary vectors.

Step 4: Analyze Option (D)

The magnitude of the cross product is $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta$. Since $|\sin \theta| \le 1$, it follows that $|\vec{a} \times \vec{b}| \le |\vec{a}||\vec{b}|$. Option (D) suggests the reverse, which is false.

Final Answer: (A)

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the definitions of dot product, cross product, and triangle inequality to evaluate the validity of mathematical statements.
Knowledge Dimension: CONCEPTUAL
Justification: The question tests the understanding of fundamental vector properties and inequalities rather than rote memorization of formulas.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly assesses the core concepts covered in the 'Vectors' chapter of the NCERT curriculum.

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