I have 3 vectors
$$A\langle0,-1,-1\rangle$$
$$B\langle-1,0,-1\rangle$$
$$C \langle-1,-1,-2\rangle$$
Where C is $A+B$
Simply put I wanted to find the angle between $A$ and $B$
Using dot product I found that $A\cdot B = 1$ and then dividing by the magnitudes $||A||=\sqrt{2}$, $||B|| = \sqrt{2}$, and $||C||=\sqrt{6}$
Giving me an angle of $60^\circ$
However when I find the angle between $A$ and $C$ I find an angle of $30^\circ$ and this is the same answer I get for the angle between $B$ and $C$.
$A\cdot C=3$ then $cos\theta=\frac{3}{\sqrt{2}\sqrt{6}}$ where $\theta=30^\circ$
This is the same for $B\cdot C$
Adding all of these angles together I only get a grand total of $120^\circ$, how is this possible as I have to have $180^\circ$
Can anyone help me figure out what I am missing or where I went wrong?
Best Answer
The problem is that there are two possible angles for any dot product, $\theta$ and $\pi - \theta$ (Draw the intersecting lines and you will see the adjacent supplementary angles.)
Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees. That will work.
You could also try 180 degrees - 30 degrees = 150 degrees but that would be too large.