3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade None
Question:
Shortest distance between the lines $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{1}$ is equal to:
Step-by-Step Solution
Key Concept: The shortest distance between two parallel lines is perpendicular to both lines.
Since the given lines are parallel, the distance between them is found by calculating $BC = \frac{(2-1)·1}{\sqrt{3}} + \frac{(3-1)·1}{\sqrt{3}} + \frac{(4-1)·1}{\sqrt{3}} = \frac{1+2+3}{\sqrt{3}} = 2\sqrt{3}$. Next, $AB = \sqrt{1+4+9} = \sqrt{14}$. The shortest distance between the lines is $AC = \sqrt{14-12} = \sqrt{2}$.
Correct Answer: 3