Three Dimensional Geometry Question 2
Question: What is the distance between the planes $ x-2y+z-1=0 $ and $ -3x+6y-3z+2=0 $ ?
Options:
A) 3 unit
B) 1 unit
C) 0
D) None of the above
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given planes are $ x-2y+z=1 $ (i) And $ -3x+6y-3z=-2 $ $ \equiv x-2y+z=\frac{3}{2} $ (ii) Since both planes are parallel and $ a=1,b=-2,c=1 $ and $ d_1=-1,d_2=\frac{-2}{3} $
$ \therefore $ Distance $ =| \frac{d_2-d_1}{\sqrt{a^{2}+b^{2}+c^{2}}} | $ Distance $ =| \frac{1-\frac{2}{3}}{\sqrt{1+4+1}} |=\frac{1}{3\sqrt{6}} $
BETA
