Question

Equation of the plane bisecting the angle between two planes

Answer 1

he equation of the planes bisecting the angles between two given planes a1x + b1y + c1z +d1 = 0 and a2x + b2y + c2z +d2 = 0 is

Notes:

If angle between bisector plane and one of the plane is less than 45° then it is acute angle bisector otherwise it is obtuse angle bisector.If a1a2 + b1b2 + c1c2 is negative, then origin lies in the acute angle between the given planes provided d1 and d2 are of same sign and if a1a2 + b1b2 + c1c2 is positive, then origin lies in the obtuse angle between the given planes.

Illustration 13. Show that the origin lies in the acute angle between the planes x + 2y + 2z = 9 and 4x – 3y + 12z + 13 = 0. Find the planes bisecting the angles between them and point out which bisects the acute angle.

Solution: 
Equation of given planes can be written as

– x – 2y – 2z + 9 = 0 and 4x – 3y + 12z + 13 = 0

The bisecting planes are

⇒ – 13x – 26y – 26z + 117 = ± (12x – 9y + 36z + 39)

So 25x + 17y + 62z – 78 = 0 is the plane bisecting the angle containing the origin, and x + 35y – 10z – 156 = 0 is the other bisecting plane.

Let θ be the angle between x + 2y + 2z = 9

and 25x + 17y + 62z – 78 = 0

Therefore cos θ = 61/68 ⇒ tan θ = √903/61

which is less than 1 , thus θ is less than 450 .

Hence the plane 25x + 17y + 62z – 78 = 0 bisects the acute angle and therefore origin lies in the acute angle.