<s>[INST] 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.
 [/INST] Finding the angle between two planes is similar to finding the angle between two vector (normal of the planes). The first plane is $3x-y+z-5=0$ has normal  $\overrightarrow n_0 =\langle 3, -1, 1 \rangle$. The second plane is $x+2y+2z+2=0$ has normal  $\overrightarrow n_1 =\langle 1, 2, 2 \rangle$.
To find the angle between two vectors we use the formula
$$cos(\theta)= \frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}$$
Insert A as $\overrightarrow n_0$ and B as $\overrightarrow n_1$.
$$\overrightarrow n_0 \bullet \overrightarrow n_1=3*1+(-1)*(2)+1(2) = 3$$
$$|\overrightarrow n_0| = \sqrt{3^2+(-1)^2+1^2}=\sqrt{11}$$
$$|\overrightarrow n_1| = \sqrt{1^2+2^2+2^2}=3$$
So,
$$cos(\theta)= \frac{3}{\sqrt{11}*3}=\frac{1}{\sqrt{11}}$$
Therefore
$$\theta= \arccos(\frac{1}{\sqrt{11}}) \approx 1.26451896 rad$$
Relevant Find the angle between two planes using their normal vectors
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