JEE Main Solved Paper 2017 Question 23
Question: The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1), having normal perpendicular to both the lines $ \frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3} $ and $ \frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}, $ is:- [JEE Main Solved Paper-2017]
Options:
A) $ \frac{10}{\sqrt{74}} $
B) $ \frac{20}{\sqrt{74}} $
C) $ \frac{10}{\sqrt{83}} $
D) $ \frac{5}{\sqrt{83}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Normal vector $ \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \\ \end{vmatrix} =5\hat{i}+7\hat{j}+3\hat{k} $
So plane is $ 5(x-1)+7(y+1)+3(z+1)=0 $
$ \Rightarrow $ $ 5x+7y+3z+5=0 $
Distance $ \frac{5+21-21+5}{\sqrt{25+49+9}}=\frac{10}{\sqrt{83}} $
