JEE Main 12 Jan 2019 Morning Question 6
Question: The perpendicular distance from the origin to the plane containing the two lines, $ \frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7} $ and $ \frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}, $ is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 11
B) $ 6\sqrt{11} $
C) $ \frac{11}{\sqrt{6}} $
D) $ 11\sqrt{6} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ L _1:\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7} $ and $ L _2:\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7} $
$ \therefore $ Equation of plane containing $ L _1 $ and $ L _2 $ is $ \begin{vmatrix} x-1 & y-4 & z+4 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \\ \end{vmatrix} =0 $
$ \Rightarrow $ $ (x-1)(35-28)-(y-4)(21-7)+(z+4)(12-5)=0 $
$ \Rightarrow $ $ 7x-14y+7z+77=0\Rightarrow x-2y+z+11=0 $
$ \therefore $ Perpendicular distance from the origin to the plane $ \frac{|11|}{\sqrt{1+4+1}}=\frac{11}{\sqrt{6}} $ units
