Question: The primitive lattice vectors for a body-centered cubic lattice are given by $\vec a_1 =\frac{a}{2}(1,1,-1)$ , $\vec a_2 =\frac{a}{2}(-1,1,1)$ , $ \vec a_3 =\frac{a}{2}(1,-1,1)$. Calculate the angle between two of the primitive lattice vectors in bcc?
When calculating the angle, I have used the formula $\vec a_1 \cdot \vec a_2 = |a_1||a_2|cos\theta$ which gives me the correct answer $109.47° $.
My question is, why don't I get the correct answer when using the cross product instead of the dot product? When I use this formula $|a_1 \times a_2|=|a_1||a_2|sin\theta$ I get the incorrect angle $70.53°$.
I also noticed that when I take $180°-70.53°$, I get the correct answer. Why is this?
Hint
$\cos \theta$ can be negative if $\theta \in \left(\frac {\pi}{2},\pi\right)$
But in case of sine, it is positive in this interval too(i.e $\sin(\pi-\theta) =\sin \theta)$) thus the angle isn't precisely correct