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Question #73284

Difficulty: 8/10 | 🟢 Free

The Question

Let $n \in \mathbb{N}$ . Consider the vectors $\vec{a} = 2n(n+1)\hat{i}$ and $\vec{b} = 2n(n+1)\hat{i} + (2n+1)\hat{j}$ . Let $L$ be the line passing through the origin and parallel to the internal angle bisector of $\vec{a}$ and $\vec{b}$ . If the line $L$ intersects the plane $x = 4n(n+1)$ at the point $P(x_0, y_0, z_0)$ , then the value of $\lfloor y_0 \rfloor$ is (where $\lfloor \cdot \rfloor$ denotes the greatest integer function):

Options

A.
$2n-1$
B.
$2n$
C.
$2n+1$
D.
$2n+2$

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

Angle Bisector Direction|Line-Plane intersection logic|Greatest Integer Function Traps

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the critical distinction between the sum of vectors ( $\vec{a} + \vec{b}$ ) and the true angle bisector direction ( $\hat{a} + \hat{b}$ ).

Students frequently bypass normalization to save time.

If they make this mistake, the direction becomes $(4n^2+4n)\hat{i} + (2n+1)\hat{j}$ , which intersects the plane exactly at $y_0 = 2n+1$ .

The Greatest Integer Function acts as a precision trap: the true normalized direction yields $y_0 = 2n+1 - \frac{1}{2n+1}$ , which falls just below the integer boundary, shifting the correct answer to $2n$ .

The clean algebraic simplification of the wrong path creates a powerful false sense of security.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Angle Bisector Direction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free