Assume $x$ and $y$ are two vectors in $\mathbb{R}^3$ and we want to compute the acute angle $\alpha\in(0,\pi/2]$ between these two (noncolinear) vectors. There are (at least) two possibilities:

  1. In the naive approach, we compute the absolute value of the dot product of the normalized vectors $x$ and $y$ $$\frac{x^Ty}{\|x\|\|y\|}$$ and take the inverse cosine of the result.

  2. The less naive approach is based on the fact that $$\|x\times y\|=\|x\|\|y\|\sin\alpha\quad\text{and}\quad|x^Ty|=\|x\|\|y\|\cos\alpha$$ so $\alpha$ is equal to an angle in a right triangle with legs of the length $\|x\times y\|$ and $|x^Ty|$ (for convenience, one can use a variant of the inverse tangent implemented in the atan2 function which is available in most programming languages; the function takes the side lengths of the legs as two arguments).

Now assume that $x$ is given and $y=x+\Delta x$ where $\|\Delta x\|\leq\tau\|x\|$, $\tau\ll 1$, that is, the vectors are almost colinear (for simplicity also of almost same norms). Assume that $\tilde\alpha_1$ and $\tilde\alpha_2$ are, respectively, the angles computed by the naive and the less naive approaches. Recently, I've run several tests which suggest that $$\tag{1} \frac{|\alpha-\tilde\alpha_1|}{\alpha}\leq \epsilon\mathcal{O}(\tau^{-2}) \quad\text{and}\quad \frac{|\alpha-\tilde\alpha_2|}{\alpha}\leq \epsilon\mathcal{O}(\tau^{-1}), $$ where $\epsilon$ is the machine precision. I understand that both algorithms suffer from a certain inaccuracy when $y\approx x$; in particular, both computing the dot and cross products. I suppose the inverse trigonometric functions are not an issue as these are usually implemented to give very accurate results.

I'm not asking anybody for performing any kind of analysis. I was just wondering whether there is a known reference where the accuracy of the two approaches is considered, if possible revealing why (1) (probably) holds. Thanks a lot in advance.


2 Answers 2


It's easy to see why this is: $\cos(\alpha) \sim 1 - \alpha^2/2$ for $\alpha$ near $0$, so an error of $\delta$ in $\cos(\alpha)$ can produce an error of about $\sqrt{2\delta}$ in $\alpha$ as computed using $\arccos(\cos(\alpha))$.

  • $\begingroup$ Thank you for the input. Although it still a bit puzzles me because the cosine is also involved in the second approach. $\endgroup$ Feb 12, 2015 at 22:13
  • $\begingroup$ Ok I understand it now (at least from the "qualitative" point of view). As $\cos(\alpha)\approx 1$ the error of the inverse tangent formula is essentially determined by the error of the norm of the cross product which is of the order $\delta$. $\endgroup$ Feb 12, 2015 at 22:25

For some background on these sort of issues, this might be interesting: R.W. Sinnott, "Virtues of the Haversine", Sky and Telescope, vol. 68, no. 2, 1984, p. 159

  • $\begingroup$ Thank you for the interesting reference. I'll try to look it up. $\endgroup$ Feb 12, 2015 at 22:18

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