Three Dimensional Geometry Question 197
Question: The equations of the line passing through the point (1,2,-4) and perpendicular to the two lines $ \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} $ and $ \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5} $ , will be
[AI CBSE 1983]
Options:
A) $ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} $
B) $ \frac{x-1}{-2}=\frac{y-2}{3}=\frac{z+4}{8} $
C) $ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Line passing through the point (1, 2, -4) is $ \frac{x-1}{l}=\frac{y-2}{m}=\frac{z+4}{n} $ Now, according to question, $ 3l-16m+7n=0 $ and $ 3l+8m-5n=0 $ Hence required line is, $ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} $ .
BETA
