Question

The equation of the bisector of the obtuse angle between the planes $3x+4y-5z+1=0,5x+12y-13z=0$ is

• A. $11x+4y-3z=0$
• B. $14x-8y+13=0$
• C. $2x+8y-8z-1=0$
• D. $13x-7z+18=0$

Answer 1

The correct option is C $2x+8y-8z-1=0$

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 $\perp$ & plane2 respectively.

$\overrightarrow{AB}=3\hat{i}+4\hat{j}-5\hat{k}\overrightarrow{AD}=5\hat{i}+12\hat{j}-13\hat{k}$

$\therefore \overrightarrow{AC}$ will be the acute angle bisector whereas $\overrightarrow{BD}$ will be in direction of obtuse angle bisector to the normals.

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}$ (by $||$gm law of addition )

$\overrightarrow{BD}=\overrightarrow{AB}-\overrightarrow{AD}$ (by $△$ law of addition)

$\therefore \overrightarrow{BD}=-2\hat{i}-8\hat{j}+8\hat{k}$ is the direction of the normal to the plane through obtuse angle bisector plane1 & plane2.

$\therefore$ Equation of plane through the line of intersection of plane1 & plane2

$(3x+4y-5z+1)+\lambda (5x+12y-13z)=0(3+5\lambda )x+(4+12\lambda )y+(-5-13\lambda )+1=0$

The above plane should be parallel to the plane formed as it is normal.

$\therefore \frac{3+5\lambda }{-2}=\frac{4+12\lambda }{-8}=\frac{-5-13\lambda }{8}\Rightarrow \lambda =-1$

$\therefore$ The required plane is ,

$-2x-8y+8z+1=0\Rightarrow 2x+8y-8z-1=0$