JEE Advanced level
A JEE Advanced-level example on Three Dimensional Geometry
Question. Two skew lines are given by L₁: (x−1)/2 = (y−2)/1 = (z−3)/3 and L₂: (x−4)/1 = (y−1)/2 = (z−2)/1. Find the equation of the plane containing L₁ and parallel to L₂. Also find the shortest distance between the two lines.
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- L₁ passes through A(1, 2, 3) with direction d₁ = (2, 1, 3). L₂ passes through B(4, 1, 2) with direction d₂ = (1, 2, 1).
- Normal to the required plane is perpendicular to both directions: n = d₁ × d₂ = (1−6, −(2−3), 4−1) = (−5, 1, 3).
- Plane through A with normal (−5, 1, 3): −5(x−1) + (y−2) + 3(z−3) = 0 which simplifies to 5x − y − 3z + 6 = 0.
- For shortest distance, find AB = B − A = (3, −1, −1).
- Shortest distance = |AB · n| / |n| = |−15 − 1 − 3| / √(25+1+9) = 19/√35 = 19√35/35 units.
Answer: Plane: 5x − y − 3z + 6 = 0; Shortest distance = 19√35/35 units
This example was solved twice independently and both solutions agree.