Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Write the planes with positive constants: $3x-6y+2z+5=0$ and $-4x+12y-3z+$ $3=0$. Bisector: $\frac{3x-6y+2z+5}{7}=\frac{-4x+12y-3z+3}{13}$, which gives $67x-162y+47z+44=0$. This bisector lies in the region where both expressions are positive, so it bisects the angle containing the origin. The dot product of the normals is $3(-4)+(-6)(12)+2(-3)=-90<0$, so the angle containing the origin is acute.