Three Dimensional Geometry Question 65
Question: The point of intersection of the lines $ \frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}, $ $ \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4} $ is
[MP PET 2004]
Options:
A) $ (22, \frac{5}{3},\frac{10}{3}) $
B) $ ( 2, 10, 4) $
C) $ (-3, 3, 6) $
D) $ (5, 7, -2) $
Show Answer
Answer:
Correct Answer: A
Solution:
Given lines are, $ \frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=r_1 $ , (say) and $ \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=r_2 $ , (say)
$ \therefore $ $ x=3r_1+5=-36r_2-3 $ , $ y=-r_1+7=3+2r_2 $ and $ z=r_1-2=4r_2+6 $ On solving, we get $ x=21, y=\frac{5}{3}, z=\frac{10}{3} $ . Trick: Check through options.
BETA
