<s>[INST] Angle between vectors by using cross product Given $A=(0,10,2)$ and $B=(0,-4,0.5)$, find the angle between $A$ and $B$ using the cross product.
My attempt by using dot product:
$$\cos (\theta)=\frac{A\cdot B}{|A||B|}=\frac{(0)(0)+10(-4)+2(0.5)}{\sqrt{0^2+10^2+2^2}\sqrt{0^2+(-4)^2+(0.5)^2}}=\frac{-39}{41.1}$$
Then $\theta=\arccos(-0.948)= 161.44$
Now using the cross product:
$$\sin(\theta)=\frac{|A\times B|}{|A||B|}=\frac{ \begin{vmatrix}
    i & j & k \\
    0 & 10 & 2 \\
    0 & -4 & 0.5 \\
    \end{vmatrix}}{\sqrt{0^2+10^2+2^2} \sqrt{0^2+(-4)^2+(0.5)^2}}=
\frac{|10(0.5)i+0(-4)k+0(2)j-0(10)k-2(-4)i-(0.5)(0)j|}{\sqrt{1690}}=\frac{|13i|}{41.1}=\frac{13}{41.1}=0.316$$
but then I get $\theta=\arcsin(0.316)=18.42$
What am I doing wrong?
 [/INST] The cross product isn’t a conclusive means to determine the angle $\theta$ between two vectors, since within $\theta$’s range $[0,180^\circ],$ $$\sin\theta=x\implies\\\theta=\arcsin x\text{ or }180^\circ-\arcsin x.$$
(On the other hand, for $\theta\in[0,180^\circ],$ $$\cos\theta=x\implies\theta=\arccos x.)$$
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