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Question #40847

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u}(\theta) = 2\cos\theta \hat{i} + 2\sin\theta \hat{j}$ for $\theta \in [0, 2\pi)$ . A set $S$ consists of all distinct vectors $\vec{v}$ defined by $\vec{v} = [\vec{u} \cdot \hat{i}] \hat{i} + [\vec{u} \cdot \hat{j}] \hat{j}$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Find the value of $\sum_{\vec{v} \in S} |\vec{v}|^2$ .

Options

A.
48
B.
52
C.
56
D.
60

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Projection vs. Coordinate|Dot Product Projection Sign|Step Function Discontinuities

Why This Question Is Tricky

Common Mistakes

The problem asks for the projection of a vector onto coordinate axes, where the scalar multiplier is defined by a Greatest Integer Function (Step Function).

Students often incorrectly assume that the symmetry of the circle $x^2 + y^2 = 4$ implies symmetric bounds for the floor function, leading them to guess combinations like $(\pm 2, 0)$ and $(0, \pm 2)$ or a full $5 \times 5$ grid of integers.

The core trap lies in ignoring the jump discontinuities and the asymmetry of the floor function for negative values (e.g., $y \in (-1, 0) \implies [y] = -1$ , but $y \in (0, 1) \implies [y] = 0$ ).

By forcing a discrete domain analysis along a continuous circular path, the question differentiates students who rigorously check integer boundaries from those who rely on continuous approximations.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Projection vs. Coordinate
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free