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Question #40881

Difficulty: 8/10 | 🟢 Free

The Question

In $\triangle ABC$ , the angles satisfy the relation $\cos A \cos B + \sin A \sin B \sin C = 1$ . Let $\lambda = \frac{\sin(A+B)}{\cos(A-B)} - 4$ . A point $R$ divides the line segment joining $P(5, -1, 2)$ and $Q(1, 3, -2)$ in the ratio $\lambda : 1$ . If $R$ lies in the $n^{\text{th}}$ octant, find the value of $n^2 + |\vec{OR}|^2$ , where $O$ is the origin.

Options

A.
78
B.
34
C.
58
D.
90

Check Your Answer

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

Octant Sign Convention|Section Formula Internally/Externally|Conditional Identity Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the integration of trigonometric bounding identities with 3D geometry's Octant Sign Convention and section formulas.

The primary trap lies in the conditional identity $\cos A \cos B + \sin A \sin B \sin C = 1$ , which forces a negative ratio ( $\lambda = -3$ ).

A negative ratio mathematically dictates external division.

If a student relies on rote memorization and ignores the negative sign (assuming ratios must be positive), they will incorrectly apply the internal division formula.

This causes the calculated coordinate to 'drift' into the 5th octant $(2, 2, -1)$ instead of the correct 6th octant $(-1, 5, -4)$ , leading to a completely different and incorrect final integer value.

Quick Facts About This Question

Question Information

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