Angle between diagonals, vectors So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis $\vec a$  , $\vec b$, and i want to find out the angle between its diagonals starting only from the fact that $$\Vert \vec a \Vert= 3$$ and $$\Vert\vec b \Vert =2$$ and the angle between them pi/3 , how would i go about it?
I know the diagonals are $\vec a - \vec b$ and $\vec a+ \vec b$ respectively, but im not sure how I shoul apply the formula for sum and difference of vectors.
 A: You have\begin{align}\require{cancel}\left\|\vec a-\vec b\right\|^2&=\left\|\vec a\right\|^2-2\left\langle\vec a,\vec b\right\rangle+\left\|\vec b\right\|^2\\&=9-\cancel2\times3\times2\times\cancel{\cos\left(\frac\pi3\right)}+4\\&=7.\end{align}Can you deal with $\left\|\vec a+\vec b\right\|$ now?
A: $$\vec a = ||a|| dir(a) =(||a||\cos{\theta}, ||a||\sin{\theta}) = (3\cos{\theta},3\sin{\theta})$$
$$\vec b = ||b||  dir(b) =(||b||\cos{(\theta+\frac{\pi}{3})}, ||a||\sin{(\theta+\frac{\pi}{3})}) = (2\cos{(\theta+\frac{\pi}{3})},2\sin{(\theta+\frac{\pi}{3})}) $$
Now, you are looking for $\alpha = \arccos\displaystyle\frac{(\vec a +\vec b) \cdot(\vec a -\vec b)}{||\vec a + \vec b ||||\vec a - \vec b ||}$.
You can easily calculate the parts, indeed
$$(\vec a +\vec b) \cdot(\vec a -\vec b) = \vec a^2-\vec b^2  = 9-4 = 5 $$
Besides,
$$||\vec a - \vec b || = \sqrt{(3\cos\theta-2\cos(\theta+\frac{\pi}{3}))^2 + (3\sin \theta -2\sin(\theta+\frac{\pi}{3}))^2} = \sqrt{9+4-12(\cos\theta\cos(\theta+\frac{\pi}{3})+\sin\theta\sin(\theta+\frac{\pi}{3}))} = \sqrt{7}  $$
$$||\vec a + \vec b || = \sqrt{(3\cos\theta+2\cos(\theta+\frac{\pi}{3}))^2 + (3\sin \theta +2\sin(\theta+\frac{\pi}{3}))^2} = \sqrt{9+4+12(\cos\theta\cos(\theta+\frac{\pi}{3})+\sin\theta\sin(\theta+\frac{\pi}{3}))} = \sqrt{19}  $$
Here, I've used the fact that
$$\cos\theta\cos(\theta+\frac{\pi}{3})+\sin\theta\sin(\theta+\frac{\pi}{3}) = \frac12$$
Therefore, $\alpha =\displaystyle \arccos \frac{5}{\sqrt{133}} .$
