Finding angle between two vectors . I was studying vectors when i read a line stating that
" Angle bewtween two vectors is obtained by their dot product ( not from cross product ) ie. $ \theta = \cos^{-1}(\frac {A\cdot B}  {AB}) $ $\;$.
It is not always $ \sin^{-1}{ \frac {| A×B |}{AB} } $"
I cant seem to understand the line. can somebody pleaese explain it to me.
 A: The angle between any 2 vectors has the range $[0,\pi]$.
Now, the principal range of $\cos^{-1}$  is $[0,\pi]$, while the range of $\sin^{-1}$  is from $[-\frac{\pi}{2},\frac{\pi}{2}]$. 
So, $\sin^{-1}$ cannot describe angles from $[\frac{\pi}{2},\pi]$ properly, while $\cos^{-1}$ can.
For example, if $A×B = 0$, then $\sin^{-1} (0)$ can take the values $0$ and $\pi$. Thus, you are not able to know whether $A$ and $B$ are parallel or anti-parallel just from the cross product.
A: The problem is that $\sin{\theta}=\sin(\pi-\theta)$ so that $\arcsin$ can't distinguish between these possibilities.  If the angle between the vectors in $105^\circ$, the $\arcsin$ function will report that the angle is $75^\circ$, since it takes values between ${-\pi\over2}$ and ${\pi\over2}$.
A: The first equation is a property of the dot product in $\mathbb{R}^2$. I.e. we always have that
$$ \vec{A}.\vec{B} = (a_1, a_2) . (b_1, b_2) = a_1b_1 + a_2b_2 = |\vec{A}||\vec{B}|.\cos(\theta)$$
from which the required formula quickly follows. You could derive this using the Pythagorean theorem.
A similar property holds for the cross product, although the given expression $\frac{|A \times B|}{AB}$ only takes values between $0$ and $1$, from which it should be obvious that it cannot always be correct (in particular when the angle is such that $\sin(\theta) < 0$).
