JEE Main On 08 April 2018 Question 2
Question: If $ {L_1} $ is the line of intersection of the planes $ 2x-2y+3z-2=0,x-y+z+1=0 $ and $ L _2 $ is the line of intersection of the planes $ x+2y-z-3=0,3x-y+2z-1=0 $ , then the distance of the origin from the plane, containing the lines $ L _1 $ and $ L _2, $ is: [JEE Main Online 08-04-2018]
Options:
A) $ \frac{1}{2\sqrt{2}} $
B) $ \frac{1}{\sqrt{2}} $
C) $ \frac{1}{4\sqrt{2}} $
D) $ \frac{1}{3\sqrt{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ 2x2y+3z2+\lambda (xy+z+1)=0 $ $ x+2yz3+\mu (3xy+2z1)=0 $
$ \Rightarrow \frac{2+\lambda }{1+3\mu }=\frac{-2-\lambda }{2-\mu }=\frac{3+\lambda }{-1+2\mu }=\frac{-2+\lambda }{-3-\mu } $
$ \Rightarrow \frac{1}{1+\mu }=\frac{5}{2+3\mu } $
$ \Rightarrow 5+5\mu =2+3\mu $
$ \Rightarrow \mu =\frac{-3}{2} $ $ 2(x+2yz3)3(3xy+2z1)=0 $
$ \Rightarrow -7x+7y8z3=0 $ Distance $ =\frac{3}{\sqrt{162}}=\frac{3}{9\sqrt{2}}=\frac{1}{3\sqrt{2}} $
