I am trying to find the bond angle between two vectors in a tetrahedron that is, the angle between BE and DE
however seems like the vectors are wrong.
Those vectors we find subtracting points $$E = (1/2,1/2,1/2) $$ $$B=(0,1,0)$$ $$D = (1,1,1)$$
$$BE = (1/2,-1/2,1/2)$$ $$ED = (1/2,1/2,1/2)$$
However, solving for $\theta$ in the dot product gives me the wrong angle
$$\cos\theta = \frac{BE \cdot DE}{|BE||DE|}$$ $$\theta = 1,91.$$
Continuing your work, we have that $\cos(\theta)=\frac{BE \cdot DE}{|BE||DE|}=\frac{\frac{-1}{4}}{\frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}}=\frac{\frac{-1}{4}}{\frac{3}{4}}=\frac{-1}{3}.$
So, the angle $\theta$ between $BE$ and $DE$ is given by $\cos^{-1}\big(\frac{-1}{3}\big) \approx1.9106^c.$