The problem:
Given a pyramid with $P_0=(0,0,0)$, $P_1=(1,1,1)$, $P_2=(2,-1,2)$, $P_3=(3,0,1)$, find the angle between the $P_1P_2P_3$ face the $P_0P_1P_2$ face.
My idea for the solution is to find the proper plane for the proper face, $\underline u$ and $\underline v$, and then using the direction vector use - $$\cos{\theta}=\frac{u\cdot v}{||u||\cdot||v||}$$ To find the angle between these two planes (faces).
I found the normal for the $P_1P_2P_3$ face to be $\underline u =<1, 2, 1>$, and the normal for the $P_0P_1P_2$ face -
$$\overrightarrow{P_0P_1}X\overrightarrow{P_0P_2}=3i +3k$$
So the normal for the $P_0P_1P_2$ face is $\underline v=<3, 0, 3>$.
So $u\cdot v=6$ and $||u|| = \sqrt{6}, ||v||=3\cdot \sqrt{2}$
So - $$\cos{\theta}=\frac{6}{\sqrt{6}\cdot 3\cdot \sqrt{2}} \Rightarrow \theta=54.73$$
Is this correct?
Thanks!
The equation of a plane looks like this: $$n\cdot(r-r_0)= 0\quad \text{ or }\quad n\cdot r = n\cdot r_0$$
You need a vector $n$ normal to the plane and a point $r_0$ of the plane. You can get $n$ by using the cross product $n=(p_2-p_1)\times (p_3-p_1)$
But for your problem you will don't need to build the plane equations, you can work with the vectors alone. The angle between those two planes is the same as the angle between a vector, which is normal to plane 1, and a vector, which is normal to plane 2.
There are a few things to look for though: If the angle between those two normal vectors is smaller than 90°, then this result is correct. Although if the result is greater than 90° and smaller than 180°, you have to subtract your result from 180° (This can happen, if the normal vectors look in different directions). Also, if your angle is greater than 180°, subtract it from 360° first, and then look, which case you get.