Three Dimensional Geometry Question 310
Question: If the lines $ \frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2} $ , $ \frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5} $ are at right angles, then k =
[MP PET 1997, 2001]
Options:
A) ?10
B) $ \frac{10}{7} $
C) $ \frac{-10}{7} $
D) $ \frac{-7}{10} $
Show Answer
Answer:
Correct Answer: C
Solution:
Lines are perpendicular if $ a_1a_2+b_1b_2+c_1c_2=0 $ Hence, $ -3,(3k)+2k,(1)+2,(-,5)=0\Rightarrow k=-\frac{10}{7} $ .
BETA
