Three Dimensional Geometry Question 163
Question: The equation of the line passing through (1, 2, 3) and parallel to the planes $ x-y+2z=5 $ and $ 3x+y+z=6 $ , is
[DSSE 1986]
Options:
A) $ \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4} $
B) $ \frac{x-1}{-3}=\frac{y-2}{-5}=\frac{z-1}{4} $
C) $ \frac{x-1}{-3}=\frac{y-2}{-5}=\frac{z-1}{-4} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{x-1}{l}=\frac{y-2}{m}=\frac{z-3}{n} $ or $ l-m+2n=0 $ and $ 3l+m+n=0 $
$ \therefore ,\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4} $ .
BETA
