Find the equation of the angles between the both planes below $$2x-y+2z+3=0$$ $$3x-2y+6z+8=0$$ and specify the plane which bisects the acute angle and the plane which bisects the obtuse angle.
I am facing problems in determining the plane which bisects the acute angle and the one that bisects the obtuse angle.
On
$u=(2,-1,2)\\ v=(3,-2,6)$
If we normalize $u,v$
$\hat u = (\frac 23, -\frac 13, \frac 23)$
$\hat v = (\frac 37, -\frac 27, \frac 67)$
$\hat u + \hat v$ and $\hat u - \hat v$ will be normal vectors to the bisecting planes we seek.
Find a point on the line of intersection and describing the planes should be easy after that.
Let : $$A_1=2,\:B_1=-1,\:C_1=2,\:D_1=3$$ $$A_2=3,\:B_2=-2,\:C_2=6,\:D_2=8$$
Use the formula below : $$\cos \left(\theta \right)=\frac{\left|\left(A_1\cdot \:A_2\right)+\left(B_1\cdot \:B_2\right)+\left(C_1\cdot \:C_2\right)\right|}{\sqrt{\left(A_1\right)^2+\left(B_1\right)^2+\left(C_1\right)^2}\cdot \sqrt{\left(A_2\right)^2+\left(B_2\right)^2+\left(C_2\right)^2}}$$ $$\cos \left(\theta \right)=\frac{\left|6+2+12\right|}{\sqrt{4+1+4}\cdot \sqrt{9+4+36}}$$ $$\cos \left(\theta \right)=\frac{20}{\sqrt{9}\cdot \sqrt{49}}$$ $$\cos \left(\theta \right)=\frac{20}{3\cdot 7}$$ $$\cos \left(\theta \right)=\frac{20}{21}$$ $$\theta \:=\cos ^{-1}\left(\frac{20}{21}\right)$$ $$\theta \:=17.752790161946674°$$