Question

Find the equation of the plane bisecting the obtuse angle between the planes
3x + 4y - 52+1=0 and 5x +12y-13z=0.​

Answer 1

Answer:

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Answer 2

Answer:

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Explanation:

Plane1 :3x+4y−5z+1=0

Plane2 :5x+12y−12z=0

let us construct a ∣∣gm ABCD with AB & AD in direction of normal to plane ⊥ & plane2 respectively.

AB=3i^+4j^−5k^AD=5i^+12j^−13k^

∴AC will be the acute angle bisector whereas BD will be in direction of obtuse angle bisector to the normals.

AC=AB+AD (by ∣∣gm law of addition )

BD=AB−AD (by △ law of addition)

∴BD=−2i^−8j^+8k^ is the direction of the normal to the plane through obtuse angle bisector plane1 & plane2.

∴ Equation of plane through the line of  intersection of plane1 & plane2

(3x+4y−5z+1)+λ(5x+12y−13z)=0(3+5λ)x+(4+12λ