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Suppose $\alpha < \pi/2$ is an angle between vectors $\boldsymbol{u}$ and $\boldsymbol{v}$.

I am writing this way

$$tan(\alpha)$$

$$= \frac{sin(\alpha)}{cos(\alpha)}$$

$$= \frac{\frac{|\boldsymbol{u}\times\boldsymbol{v}|}{|u|\cdot|v|}}{\frac{\boldsymbol{u}\cdot\boldsymbol{v}}{|u|\cdot|v|}}$$

$$= \frac{|\boldsymbol{u}\times\boldsymbol{v}|}{\boldsymbol{u}\cdot\boldsymbol{v}}$$

Is this correct?

Now

$$(tan(\alpha))^2=$$

$$= \frac{(\boldsymbol{u}\times\boldsymbol{v})^2}{(\boldsymbol{u}\cdot\boldsymbol{v})^2}$$

$$= \frac{(u_2v_3-u_3v_2)^2 +(u_3v_1-u_1v_3)^2 + (u_1v_2-u_2v_1)^2}{(u_1v_1)^2 + (u_2v_2)^2 + (u_3v_3)^2}$$

is this also correct?

Is this really independent of vectors lengths?

I was trying to use this to perform geodesic calculations (vectors of lenth of 6 millions of meters) and was hoping it will produce smaller errors but I am obtainins something completely wrong, and only if I normalize vectors, I achieve somethong reasonable.

Where did I lost normalization or something?

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    If $\alpha = \pi/2$ your calculation doesn't work. Furthermore it is reasonable that the angle $\alpha$ (and hence $\tan(\alpha)$ between two vectors does not depend on their length. – 2017-02-14
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    $(u \cdot v)^2=(u_1v_1+u_2v_2+u_3v_3)^2\neq (u_1v_1)^2+(u_2v_2)^2+(u_3v_3)^2$ in general – 2017-02-14
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    Observe that for $u \cdot v \neq 0$ we have $$\tan \alpha = \frac{|u \times v|}{u \cdot v} = \frac{|u| |v|}{|u| |v|} \frac{|u \times v|}{u \cdot v} = \frac{| \frac{u}{|u|} \times \frac{v}{|v| } |}{\frac{u}{|u|} \cdot \frac{v}{|v|}}.$$ – 2017-02-14
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    @user160738 oh God, thanks a lot, I was stupid :)))) – 2017-02-14

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