Verified Solution • 2025–26 Session moderate

QJEEM25Macd96: JEE Main 2025 Mathematics - Class 12 JEEM25Macd96 Solution: The distance of the line from the point along the line is :

The distance of the line x−22=y−63=z−34\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} from the point (1,4,0)(1,4,0) along the line x1=y−22=z+33\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} is :

Multiple Choice Options

  • A 17\sqrt{17}
  • B 13\sqrt{13}
  • C 15\sqrt{15}
  • D 14\sqrt{14}

Step-by-Step Verified Solution

Let the parallel line is

x−11=y−42=z−03

so their point of intersection is

(λ+1,2λ+43λ)=(2t+2,3t+6,4t+3)λ=2t+1

2λ+4=3t+6⇒t=0

so POI is (2,6,3)(2,6,3)

 so distance =(2−1)2+(6−4)2+(3−0)2=14

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring