Three Dimensional Geometry Question 289
Question: The equation of the plane passing through (1, 1, 1) and (1, ?1, ?1) and perpendicular to $ 2x-y+z+5=0 $ is
[EAMCET 2003]
Options:
A) $ 2x+5y+z-8=0 $
B) $ x+y-z-1=0 $
C) $ 2x+5y+z+4=0 $
D) $ x-y+z-1=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Any plane passing through (1, 1, 1) is $ a(x-1)+b(y-1)+c(z-1)=0 $ …..(i) Plane (i) is also passing through (1, ?1, ?1)
$ \therefore a,.,0+b(-2)+c(-2)=0 $ or, $ 0.a-2b-2c=0 $ or $ 0.a-b-c=0 $ or, $ 0.a+b+c=0 $ ……(ii) Plane (i) is perpendicular to $ 2x-y+z+5=0 $ So, $ 2a-b+c=0 $ …..(iii) From (ii) and (iii), $ a=b=1,,c=-1 $ Substituting in (i) we have $ x+y-z-1=0 $ .
BETA
