Consider the planes 3x−6y+2z+5=0 and 4x−12y+3z=3. The plane 67x−162y+47z+44=0 bisects the angle between the given planes which-
A
Contains origin
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B
Is acute
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C
Is obtuse
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D
None of these
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Solution
The correct options are B Contains origin C Is acute For 3x−6y+2z+5=0 and −4x+12y−3z+3=0 bisector are 3x−6y+2z+5√9+36+4=±−4x+12y−3z+3√16+144+9 The plane which bisects the angle between the plane that contains the origin 13(3x−6y+2z+5)=7(−4x+12y−3z+3)⇒67x−162y+47z+44=0 Further 3×(−4)+(−6)×12+2×(−3)<0 Hence, the origin lies in the acute angle.