I have a square based pyramid with vertices at $O(0,0,0)$ $A(1,0,0)$, $B(1,1,0)$, $C(0,1,0)$, and $D(0,0,1)$.
I need to find the angle between the faces $ABD$ and $BCD$ $(1,0,1)$ and $(0,1,1)$.
When I use the angle formula I get a solution of $\theta =60°$.
However the solution says that it should be $120°$ why is the obtuse angle correct in this case?
Best Answer
Note that it depends upon the orientation you have used for the vectors, indeed
$$60°=180°-120°$$
then your calculation is almost correct but you need to fix orientations according to the convention adopted.
If we choose as direction of the normal vectors the outer direction, your calculation is correct, notably we have
normal to face $ADB$: $v=DA\times DB=(1,0,-1)\times(1,1,-1)=(1,0,1)$
normal to face $BCD$: $w=DB\times DC=(1,1,-1)\times(1,1,-1)=(0,1,1)$
then
$$\cos \theta =\frac{v\cdot w}{|v||w|}=\frac{1}{2}\implies \theta=60°$$
As observed by Jyrki Lahtonen, if we consider the dihedral angle between the two half-planes bordered by the shared edge, it can be easily shown that this angle is exactly $180°-60°=120°$.