3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade None

Question:

The shortest distance between any two opposite edges of the tetrahedron is:
$\frac{2}{\sqrt{6}}a$
$\frac{1}{\sqrt{6}}a$
$\frac{1}{\sqrt{3}}a$
None of these

Step-by-Step Solution

Key Concept: The shortest distance between two skew lines equals the absolute value of the scalar triple product divided by the magnitude of the cross product of direction vectors.
From the given equations, $A(0,0,0)$ is a point on line (1) and $B(0,0,a)$ is a point on line (2). The shortest distance is $S.D = |[0-0) + m(0-0) + n(0-a)| = \left|-\frac{1}{\sqrt{6}}(-a)\right| = \frac{2}{\sqrt{6}}a$.
Correct Answer: 1

Master 3D Geometry with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free