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Find the equation of the plane which bisects the angle between the planes3x-4y+12z=26 and x+2y-2z=9

Find the equation of the plane which bisects the angle between the planes3x-4y+12z=26 and x+2y-2z=9

Grade:12th pass

1 Answers

Vikas TU
14149 Points
3 years ago
Dear Student,
ATQ given eqns are 3x-4y+12z-26=0 and x+2y-2z-9=0
plane bisector of angles between them:
(3x-4y+12z-26)/sqrt(9+16+144) = (x+2y-2z-9)/sqrt(1+4+4)
=>3(3x-4y+12z-26)= 13(x+2y-2z-9)
=>4x+38y-62z-39=0 and 22x+14y+10z-195=0
theta be the angle between the second plane and 2x+14y+10z-195=0:
cos theta =sqrt(5/39)   (putting formula for cos theta)
sin theta= sqrt(34/39)
so tan theta=sqrt(34/5) >1 or theta >45 degrees.
so 2x+14y+10z-195=0 is obtuse angle bisector and 4x+38y-62z-39=0 is acute angle bisector. [Ans]
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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