Angle between planes with x, y and z variables Problem: Find the angle between the planes $3x-y+z-5=0$ and $x+2y+2z+2=0$.
We have been thought 1 formula for solving angles which is : Angle $= Arctan(\frac{m1-m2}{1+m1m2})$ but that is for parallel lines only and for equation with x and y only. 
How to answer this kind of problem? Thank you.

First find the normal of each plane and then use the dot product
Let us consider normal of $3x-y+z-5=0$ as $P_1$ and
$x+2y+2z+2=0$ as $P_2$, then we have
$$\ P_1=(3,-1,1),\ P_2=(1,2,2)$$
The formula for finding the angle is $$P_1\cdot P_2=|P_1|\cdot|P_2|\cdot\cos(\theta)$$
$$\cos(\theta)=\frac{P_1\cdot P_2}{|P_1|\cdot|P_2|}$$
$$|P_1|=\sqrt{3^2+(-1)^2+1^2}=\sqrt{9+1+1}=\sqrt{11}$$
$$|P_2|=\sqrt{1^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3$$
$$P_1\cdot P_2=3(1)+(-1)(2)+1(2)=3-2+2=3$$
Then,
$$\cos(\theta)=\frac{3}{3\cdot\sqrt{11}}$$
$$\cos(\theta)=\frac{1}{\sqrt{11}}$$
$$\theta=\cos^{-1}\frac{1}{\sqrt{11}}$$
$$\theta=72.5^o$$ 
