Let the acute angle bisector of the two planes x – 2y – 2z + 1 = 0 and 2x – 3y – 6z + 1 = 0 be the plane P. Then which of the following points lies on P?
Let the acute angle bisector of the two planes x – 2y – 2z + 1 = 0 and 2x – 3y – 6z + 1 = 0 be the plane P. Then which of the following points lies on P?
Option 1 -
(0, 2, -4)
Option 2 -
(4, 0, -2)
Option 3 -
Option 4 -
Asked by Shiksha User
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1 Answer
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Correct Option - 3
Detailed Solution:Angle bisectors are
->x – 5y + 4z + 4 = 0, (i)
3x – 23y – 32z + 10 = 0 . (ii)
As distance of a point (-1, 0,0) on x – 2y – 2z + 1 = 0
from (i) is greater than that form (ii)
(ii) is the acute angle bisector.
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