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Question #40395

Difficulty: 8/10 | 🟢 Free

The Question

Let the complex numbers $z_1$ and $z_2$ represent two vectors in the Argand plane, given by: $z_1 = \sin\left(\frac{25\pi}{2} + \theta\right) + i \cos\left(\frac{23\pi}{2} - \theta\right)$ $z_2 = \cos\left(\frac{17\pi}{2} - 2\theta\right) + i \sin\left(\frac{19\pi}{2} + 2\theta\right)$ If the scalar projection of the vector represented by $z_2$ onto the vector represented by $z_1$ is $-\frac{\sqrt{3}}{2}$ and $\theta \in (0, \pi)$ , then the sum of all possible values of $\theta$ is:

Options

A.
$\pi$
B.
$2\pi$
C.
$\frac{5\pi}{9}$
D.
$\frac{4\pi}{3}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Allied Angle Periodicity|Argument and Principal Value|Dot Product Projection Sign

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of complex number geometry, vector projections, and trigonometric allied angles.

The primary trap lies in the 'Sign Confusion' induced by the periodic boundary conditions of the allied angles (e.g., incorrectly evaluating the quadrant of $19\pi/2 + 2\theta$ ).

A single sign error in the imaginary or real components of $z_1$ or $z_2$ flips the sign of the dot product projection from $\sin(3\theta)$ to $-\sin(3\theta)$ .

If a student solves $\sin(3\theta) = \sqrt{3}/2$ instead, they will find four roots in the domain $(0, 3\pi)$ yielding an incorrect sum of $2\pi$ , which is a carefully placed distractor.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Allied Angle Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free