Three Dimensional Geometry Question 217
Question: The symmetric equation of lines $ 3x+2y+z-5=0 $ and $ x+y-2z-3=0 $ , is
Options:
A) $ \frac{x-1}{5}=\frac{y-4}{7}=\frac{z-0}{1} $
B) $ \frac{x+1}{5}=\frac{y+4}{7}=\frac{z-0}{1} $
C) $ \frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1} $
D) $ \frac{x-1}{-5}=\frac{y-4}{7}=\frac{z-0}{1} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let a, b, c be the d.r.’s of required line
$ \therefore $ $ 3a+2b+c=0 $ and $ a+b-2c=0 $ $ \frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2} $ or $ \frac{a}{-5}=\frac{b}{7}=\frac{c}{1} $ In order to find a point on the required line we put $ z=0 $ in the two given equation to obtain, $ 3x+2y=5 $ and $ x+y=3 $ . Solving these two equations, we obtain $ x=-1,,y=4 $ .
$ \therefore $ Co-ordinates of point on required line are $ (-1,,4,,0) $ . Hence required line is $ \frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1} $ .
BETA
