Finding angle between two vectors .

Question

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.

Answer 1

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.

Answer 2

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 $\sin(\theta) < 0$).

Answer 3

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 ${\pi\over2}$.