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Question #41103

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all points on the line $L: \frac{x-2}{1} = \frac{y-1}{2} = \frac{z-3}{2}$ . Let $\Pi$ be the plane containing the point $B(1, 0, -1)$ and parallel to the vectors $\vec{p} = \hat{i} + \alpha\hat{j} + \hat{k}$ and $\vec{q} = 2\hat{i} - \hat{j} + \beta\hat{k}$ , where $\alpha, \beta \in \mathbb{Z}^+$ and $\alpha < \beta$ . A relation $R$ on $S$ is defined as: for $X, Y \in S$ , $X R Y$ if and only if the midpoint of the line segment $XY$ lies on the plane $\Pi$ . If $R$ is an empty relation on $S$ , find the value of $(\alpha + \beta)(\alpha \beta)$ .

Options

A.
144
B.
324
C.
420
D.
729

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

Empty and Universal Relation|Line-Plane intersection logic|Geometric Interpretation of Cross Product

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of relations on parameterized sets.

Students often fail to translate 'empty relation on a line's points' to the geometric condition of a line being strictly parallel to a plane.

The lethal collision occurs when students confuse the condition for a line being parallel to a plane ( $\vec{n} \cdot \vec{v} = 0$ ) with the condition for a line lying on a plane.

The trap involving the scalar triple product $[\vec{v}, \vec{p}, \vec{q}] = 0$ tests coplanarity, while the Diophantine equation $(\alpha - 2)(\beta - 4) = 5$ with the constraint $\alpha < \beta$ forces modular error-checking, punishing students who blindly guess roots or ignore the inequality, leading them to the distractor 420 instead of the 324 constant.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Empty and Universal Relation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free