Class CBSE Class 12 Mathematics Vector Algebra Q #552
COMPETENCY BASED
APPLY
1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE
Let \(\vec{a}\) be any vector such that \(|\vec{a}|=a\) The value of \(|\vec{a}\times\hat{i}|^{2}+|\vec{a}\times\hat{j}|^{2}+|\vec{a}\times\hat{k}|^{2}\) is:
(A) \(a^{2}\)
(B) \(2a^{2}\)
(C) \(3a^{2}\)
(D) 0
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Step-by-Step Solution

Let \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\). Then \(|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 = a^2\).

Now, we compute the cross products:

\(\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} = a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) = -a_2\hat{k} + a_3\hat{j}\)

\(|\vec{a} \times \hat{i}|^2 = a_2^2 + a_3^2\)

\(\vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) = a_1\hat{k} - a_3\hat{i}\)

\(|\vec{a} \times \hat{j}|^2 = a_1^2 + a_3^2\)

\(\vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) = -a_1\hat{j} + a_2\hat{i}\)

\(|\vec{a} \times \hat{k}|^2 = a_1^2 + a_2^2\)

Adding these squared magnitudes:

\(|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2) = 2a^2\)

Correct Answer: 2a^{2}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of cross product and magnitude of vectors to solve the problem.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of vector algebra, cross products, and their properties.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply vector algebra concepts to solve a problem, aligning with competency-based assessment.

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