Three Dimensional Geometry Question 410
Question: The equation of the plane containing the line $ 2x-5y+z=3;x+y+4z=5 $ , and parallel to the plane, $ x+3y+6z=1 $ , is:
Options:
A) $ x+3y+6z=7 $
B) $ 2x+6y+12z=-13 $
C) $ 2x+6y+12z=13 $
D) $ x+3y+6z=-7 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Equation of the plane containing the lines $ 2x-5y+z=3 $ and $ x+y+4z=5 $ is $ 2x-5y+z-3+\lambda (x+y+4z-5)=0 $
$ \Rightarrow (2+\lambda )x+(-5+\lambda )y+(1+4\lambda )z+(-3-5\lambda )=0 $ ?.(i) Since the plane (i) parallel to the given plane $ x+3y+6z=1 $
$ \therefore \frac{2+\lambda }{1}=\frac{-5+\lambda }{3}=\frac{1+4\lambda }{6}\Rightarrow \lambda =-\frac{11}{2} $ Hence equation of the required plane is $ ( 2-\frac{11}{2} )x+( -5-\frac{11}{2} )y+( 1-\frac{44}{2} )z+( -3+\frac{55}{2} )=0 $
$ \Rightarrow (4-11)x+(-10-11)y+(2-44)z+(-6+55)=0 $
$ \Rightarrow x+3y+6z=7 $
BETA
