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Question #40888

Difficulty: 8/10 | 🟢 Free

The Question

Consider two lines $L_1: \frac{x-1}{1} = \frac{y+2}{2} = \frac{z-3}{2}$ and $L_2: \frac{x-1}{2} = \frac{y+2}{1} = \frac{z+24}{-2}$ . Let $\vec{u} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{v} = 2\hat{i} + \hat{j} - 2\hat{k}$ be the direction vectors of $L_1$ and $L_2$ respectively. Let $P$ be a point on $L_1$ and $Q$ be a point on $L_2$ such that the length of the line segment $PQ$ is $15$ . If $V$ is the volume of the parallelepiped formed by the vectors $\vec{PQ}$ , $\vec{u}$ , and $\vec{v}$ , then the value of $V$ is

Options

A.
81
B.
108
C.
135
D.
9

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

Distance Formula in 3D|Orthogonal projection of a line|Geometric Interpretation of Cross Product

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of the scalar triple product and the properties of skew lines.

The 'Lethal Collision' occurs when students assume the volume depends on the specific length of the segment $PQ$ (given as 15) and attempt to use it directly, such as calculating $V = PQ \times |\vec{u} \times \vec{v}| = 135$ , or by finding the projection of $PQ$ onto the plane ( $12$ ) and calculating $12 \times 9 = 108$ .

The trap exploits the misconception that the scalar triple product varies with the choice of points on the lines, whereas it is an invariant representing the volume of a parallelepiped whose height is the shortest distance and base area is $|\vec{u} \times \vec{v}|$ .

The distance $PQ=15$ is a carefully placed red herring.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Distance Formula in 3D
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free