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Question #40918

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(\sec^2 \theta, \tan \theta, \sqrt{2} \sin \theta)$ , $B(-\csc^2 \theta, \cot \theta, \sqrt{2} \cos \theta)$ , and $C(0, 1, 1)$ be three points in $\mathbb{R}^3$ . If the sum of all values of $\theta \in [0, 4\pi]$ for which the points $A, B,$ and $C$ are collinear is $S$ , then the value of $S$ is:

Options

A.
$\frac{5\pi}{2}$
B.
$7\pi$
C.
$3\pi$
D.
$\frac{7\pi}{2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Collinearity and Coplanarity|Ratio of Division (k:1)|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

The question tests the concept of collinearity in 3D space combined with trigonometric equations.

The primary trap is 'Extraneous Solution Rejection'.

When students equate the y-coordinates, they find $\tan \theta = 1$ , yielding four possible angles in the given domain.

To verify the z-coordinate, they often square the equation $\sin \theta + \cos \theta = \sqrt{2}$ to get $\sin 2\theta = 1$ , which mathematically validates all four angles.

However, squaring introduces extraneous solutions; the original linear trigonometric equation requires both $\sin \theta$ and $\cos \theta$ to be positive.

Failing to reject the third-quadrant angles ( $\frac{5\pi}{4}$ and $\frac{13\pi}{4}$ ) leads to the incorrect sum of $7\pi$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Collinearity and Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free