Three Dimensional Geometry Question 202
Question: The equation of the plane through the point $ (2,-1,-3) $ and parallel to the lines $ \frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4} $ and $ \frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2} $ is
[Kerala (Engg.) 2005]
Options:
A) $ 8x+14y+13z+37=0 $
B) $ 8x-14y+13z+37=0 $
C) $ 8x+14y-13z+37=0 $
D) $ 8x+14y+13z-37=0 $
E) (e) $ 8x-14y-13z-37=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of plane passing through the point (2, ?1, ?3) is, Also, $ A(x-2)+B(y+1)+C(z+3)=0 $ Also, $ 3A+2B-4C=0 $ and $ 2A-3B+2C=0 $ \ $ \frac{A}{-8}=\frac{B}{-14}=\frac{C}{-13}=k $ , (Let) So, $ A=-8k,B=-14k,C=-13k $ Equation of required plane is, $ -k[8(x-2)+14(y+1)+13(z+3)]=0 $ i.e., $ 8x+14y+13z+37=0 $ .
BETA
