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Question #40866

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}$ and $\vec{b}$ be the position vectors of two distinct points $A$ and $B$ . A point $P$ with position vector $\vec{p}$ is uniquely determined by the relation: $8(\lambda^2 - 4)\vec{p} = (\lambda^3 + 3\lambda^2 - 4\lambda - 12)\vec{a} + (5\lambda^2 - \lambda^3 + 4\lambda - 20)\vec{b}$ where $\lambda \in \mathbb{R}$ . If $P$ divides the line segment $AB$ strictly internally, then the set of all possible values of $\lambda$ is:

Options

A.
(-3, 5)
B.
(-3, 5) \setminus {-2, 2}
C.
(-3, 5) \setminus {2}
D.
$(-\infty, -3) \cup (5, \infty)$

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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

Ratio of Division (k:1)|Section Formula Internals|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question targets the 'Zero-Factor Cancellation Trap' in vector section formulas.

Students often factorize and immediately cancel the common term $(\lambda^2 - 4)$ from both sides to find the internal division ratio, leading them to the erroneous interval $(-3, 5)$ .

However, they neglect the degenerate case where the canceled factor equals zero.

If $\lambda = \pm 2$ , the equation collapses to $0\vec{p} = 0$ , meaning $P$ is not uniquely determined.

Recognizing that division by a parameter requires the parameter to be non-zero is a crucial boundary condition frequently tested in JEE Advanced to filter out rote algebraic manipulation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Ratio of Division (k:1)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free