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, $$\cos(\theta)=\frac{3}{3\cdot\sqrt{11}}$$
$$\cos(\theta)=\frac{1}{\sqrt{11}}$$ $$\theta=\cos^{-1}\frac{1}{\sqrt{11}}$$ $$\theta=72.5^o$$
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
