3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The equations of the lines of shortest distance between the lines $\frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$ and $\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ are:
3(x - 2i) = 3y + 92 = 3z - 32
(x - 62/3)/(1/3) = (y + 31)/(1/3) = (z - 31/3)/(1/3)
(x - 21)/(1/3) = (y + 92/3)/(1/3) = (z - 32/3)/(1/3)
(x - 2)/(1/3) = (y + 3)/(1/3) = (z - 1)/(1/3)
Step-by-Step Solution
Key Concept: The shortest distance between skew lines occurs along their common perpendicular.
Let $P(2t_1 - 3r_1, q_1)$ and $Q(3r_2 + 2, -5r_2 + 1, 2r_2 - 2)$ be points on the given lines. The line $PQ$ is the shortest distance when perpendicular to both lines. Setting up the perpendicularity conditions $2(2t_1 - 3r_2 - 2) - 3(3q_1 + 5r_2 - 1) + (q_1 - 2r_2 + 2) = 0$ and solving gives $r_1 = 31/3$, $r_2 = 19/3$. This yields $P\left(\frac{62}{3}, -31, \frac{31}{3}\right)$, $Q\left(21, \frac{-92}{3}, \frac{32}{3}\right)$, and the distance is $\frac{1}{3}\sqrt{1 + 1 + 1} = \frac{\sqrt{3}}{3}$.
Correct Answer: 2