Class CBSE Class 12 Mathematics Vector Algebra Q #1725
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
Vector of magnitude 3 making equal angles with x and y axes and perpendicular to z axis is
(A) $\hat{i}+2\sqrt{2}\hat{j}$
(B) $3\hat{k}$
(C) $\frac{3\sqrt{2}}{2}\hat{i}+\frac{3\sqrt{2}}{2}\hat{j}$
(D) $\sqrt{3}\hat{i}+\sqrt{3}\hat{j}+\sqrt{3}\hat{k}$
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Correct Answer: C

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Detailed Solution

Step 1: Define the vector

Let the vector be v = ai + bj + ck. Since the vector is perpendicular to the z-axis, its component along the z-axis must be zero. Thus, c = 0. The vector is v = ai + bj.

Step 2: Apply the condition of equal angles

The vector makes equal angles with the x and y axes. This implies the components along the x and y axes must be equal in magnitude. Thus, a = b. The vector becomes v = ai + aj.

Step 3: Use the magnitude condition

The magnitude of the vector is given as 3. Therefore: $$ \sqrt{a^2 + a^2} = 3 $$ $$ \sqrt{2a^2} = 3 $$ $$ a\sqrt{2} = 3 $$ $$ a = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} $$

Step 4: Formulate the final vector

Substituting the value of a back into the vector expression: $$ v = \frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j} $$

Final Answer: \frac{3\sqrt{2}}{2}\hat{i}+\frac{3\sqrt{2}}{2}\hat{j}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to translate geometric constraints (perpendicularity, equal angles) into algebraic equations and solve for vector components.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of mathematical steps involving vector algebra and magnitude calculation to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the conceptual understanding of vector components and their relationship with coordinate axes rather than rote memorization.

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