JEE Main 2023 · 25 Jan · Shift 2 · numerical answer

JEE Main 2023 Mathematics question on Three Dimensional Geometry

If the shortest distance between the line joining the points (1,2,3)(1, 2, 3) and (2,3,4)(2, 3, 4), and the line x−12=y+1−1=z−20\dfrac{x-1}{2} = \dfrac{y+1}{-1} = \dfrac{z-2}{0} is aa, then 28a228a^2 is equal to ______.
Show answer and solution

Answer: 18

Solution

The line through (1,2,3) and (2,3,4) has direction b₁ = (1,1,1).
The second line has direction b₂ = (2,-1,0).

b₁ × b₂ = |i j k; 1 1 1; 2 -1 0| = (0·1−1·(−1), 1·2−0·1, 1·(−1)−1·2) = (1, 2, −3).

|b₁ × b₂| = √(1+4+9) = √14.

Connect a point on line 1 to a point on line 2: take A = (1,2,3) on line 1 and B = (1,−1,2) on line 2.

AB = (1−1, −1−2, 2−3) = (0, −3, −1).

Shortest distance a = |AB · (b₁ × b₂)| / |b₁ × b₂|
= |(0)(1) + (−3)(2) + (−1)(−3)| / √14
= |0 − 6 + 3| / √14
= 3/√14.

So a² = 9/14, and 28a² = 28 × 9/14 = 18.

[JEE Main 2023 · 25 Jan · Shift 2]

Practise Three Dimensional Geometry the way it is asked

A timed set of real JEE Main Three Dimensional Geometry questions, graded as the exam grades. No sign-up.

Start a Three Dimensional Geometry test

More: all JEE Main Three Dimensional Geometry questions · JEE Main 2023 · Mathematics chapter list