Three Dimensional Geometry Question 192
Question: The equation of the plane passing through the line $ \frac{x-1}{5}=\frac{y+2}{6}=\frac{z-3}{4} $ and the point (4, 3, 7) is
[MP PET 2001]
Options:
A) $ 4x+8y+7z=41 $
B) $ 4x-8y+7z=41 $
C) $ 4x-8y-7z=41 $
D) $ 4x-8y+7z=39 $
Show Answer
Answer:
Correct Answer: B
Solution:
Any plane through given line is $ A(x-1)+B(y+2)+C(z-3)=0 $ …..(i) and $ 5A+6B+4C=0 $ ?..(ii) Since, plane (i) passes through (4, 3, 7), we get $ 3A+5B+4C=0 $ …..(iii) Solving (ii) and (iii), we get $ \frac{A}{4}=\frac{B}{-8}=\frac{C}{7} $
$ \therefore $ Equation of required plane is $ 4x-8y+7z=41 $ .
BETA
