JEE Main On 8 April 2017 Question 28
Question: The coordinates of the foot of the perpendicular from the point (1, - 2,1) on the plane containing the lines, $ \frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8} $ and $ \frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}, $ is: [JEE Online 08-04-2017]
Options:
A) (2, - 4, 2)
B) (1, 1, 1)
C) (0, 0, 0)
D) (0, 0, 0)
Show Answer
Answer:
Correct Answer: C
Solution:
$ \overline{n}={{\overline{n}}_1}\times {{\overline{n}}_2} $
$ = \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 6 & 7 & 8 \\ 3 & 5 & 7 \\ \end{vmatrix} $
$ =(9,-18,9) $ $ =(1,-2,1) $ $ 1(x+1)-2(y-1)+(2-3)=0 $
$ = $ foot to z $ \frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-1}{1}=\frac{[1+4+1]}{6} $ $ $ 10,000
