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Question #40915

Difficulty: 8/10 | 🟢 Free

The Question

Points $A(1, 2, 3)$ , $B(2, 4, 5)$ , and $C$ are collinear in 3D space. A variable line $L_1$ passes through $C$ and is parallel to the vector $\hat{i} - \hat{j} + \hat{k}$ . Another line $L_2$ passes through the origin and is parallel to the vector $2\hat{i} + \hat{j} - \hat{k}$ . Let $d$ be the shortest distance between the lines $L_1$ and $L_2$ . If $\vec{AC} = \lambda \vec{AB}$ and $d\sqrt{2} = 7\lambda - 4$ , find the sum of the coordinates of all possible positions of point $C$ .

Options

A.
21
B.
$\frac{292}{11}$
C.
$\frac{61}{11}$
D.
17

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

Collinearity and Coplanarity|Shortest distance between skew lines|Modulus Inequality Logic

Why This Question Is Tricky

Common Mistakes

The problem tests the intersection of 3D geometry (collinearity and skew lines) with algebraic constraints (modulus equations).

The core trap lies in the equation $|4\lambda + 5| = 7\lambda - 4$ .

Students commonly square both sides or split into two cases without checking the domain restriction $7\lambda - 4 \ge 0$ (since distance $d \ge 0$ ).

This leads to an extraneous root $\lambda = -1/11$ , which yields a phantom point $C$ with a coordinate sum of $61/11$ .

Adding this to the valid sum gives the distractor $292/11$ .

This enforces the rigorous checking of algebraic domains derived from geometric realities.

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