Three Dimensional Geometry Question 8
Question: Which one of the following planes is normal the plane $ 3x+y+z=5? $
Options:
A) $ x+2y+z=6 $
B) $ x-2y+z=6 $
C) $ x+2y-z=6 $
D) $ x-2y-z=6 $
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Answer:
Correct Answer: D
Solution:
[d] Direction cosines of the normal to the plane $ 3x+y+z=5 $ are 3, 1, 1 Direction cosines of the normal to the plane $ x-2y-z=6 $ are 1, -2, -1 Sum of the product of direction cosines $ =3\times 1+1\times (-2)+1\times (-1)=0 $ Hence, normal to the two planes are perpendicular to each other. Therefore two planes are also perpendicular.
BETA
