Given planes P1: $4x + y − 8z − 1 = 0$ and P2: $x − 2y + 2z + + 3 = 0$. Find the equation of a plane P3 such that P2 halves the dihedral angle between P1 and P3.
I tried to find P3 using normals, vector sums, determined the angle between P1 and P2, but the answer is still at odds with the answer: $64x − 47y − 16z + 75 = 0$.
The system is rectangular.
Any plane containing the line of intersection of the two planes $P_1$ and $P_2$ can be written in the form $$\lambda(4x+y-8z-1)+\mu(x-2y+2z+3)=0$$
The angle $\theta$ between $P_1$ and $P_2$ is given by $$\cos\theta=\hat{n_1}\cdot\hat{n_2}$$
So using the values given in the question, we get $\cos\theta=-\frac{14}{27}$.
This must be the same as the angle between $P_2$ and $P_3$.
So using the same formula, we have the following equation:
$$\frac{\left(\begin{matrix}1\\-2\\2\end{matrix}\right)\cdot\left(\begin{matrix}4\lambda+\mu\\\lambda-2\mu\\-8\lambda+2\mu\end{matrix}\right)}{3\sqrt{(4\lambda+\mu)^2+(\lambda-2\mu)^2+(-8\lambda+2\mu)^2}}=-\frac{14}{27}$$
This can be simplified in a few lines to become $$9\mu=28\lambda$$
So we can choose $\lambda=9$ and $\mu=28$, and this leads directly to the expected answer, namely $$64x-47y-16z+75=0$$