I am currently trying to understand the derivation of Theorem 3.3 of this paper (it's on pdf-page 42, or book-page 705).
Equations 3.8 and 3.9 are fine, but I am having troubles with understanding the third transformation of 3.10.
In particular, they seem to say
$$
x_{1j}\cdot x_{1k}=\frac{1}{2}\left(d_{1j}^2+d_{1k}^2-d_{jk}^2\right)^2
$$
where $x_{ij}$ is the vector from point $i$ to $j$ and $d_{ij}$ is the distance between these two points.
I can get a similar result by transforming the law of cosine: \begin{align*} & &x_{jk}^2 &= x_{1j}^2+x_{1k}^2-2x_{1j}x_{1k}\cos\left(\theta\right) \\ \Leftrightarrow & &d_{jk}^2 &= d_{1j}^2+d_{1k}^2 - 2x_{1j}x_{1k}\cos\left(\theta\right) \\ \Leftrightarrow & &2x_{1j}x_{1k}\cos\left(\theta\right) &= d_{1j}^2+d_{1k}^2 - d_{jk}^2 \\ \Leftrightarrow & &x_{1j}x_{1k} &= \frac{1}{\cos\left(\theta\right)} \frac{1}{2} \left(d_{1j}^2+d_{1k}^2 - d_{jk}^2\right) \\ \end{align*} Unfortunately, I am then still missing the square around the brackets and have a superfluous $\frac{1}{\cos\left(\theta\right)}$ sticking around. The fraction would of course vanish for parallel vectors, but I don't see why that would be a given.
I could expand the RHS with $\frac{\left(d_{1j}^2+d_{1k}^2 - d_{jk}^2\right)}{\left(d_{1j}^2+d_{1k}^2 - d_{jk}^2\right)}$, yielding $$ x_{1j}x_{1k} = \frac{1}{2} \left(d_{1j}^2+d_{1k}^2 - d_{jk}^2\right)^2 \cdot \frac{1}{\cos\left(\theta\right)\cdot\left(d_{1j}^2+d_{1k}^2 - d_{jk}^2\right)} $$ but why would $\cos\left(\theta\right)\cdot\left(d_{1j}^2+d_{1k}^2 - d_{jk}^2\right) = 1$ hold?
Am I on the right track, or should I maybe try a completely different approach?
The square is a mistake in their paper. Consider that
$$ \mathbf{x}_{1j}\cdot\mathbf{x}_{1k} = \|\mathbf{x}_{1j}\|\|\mathbf{x}_{1k}\|\cos\theta = d_{1j}d_{1k} \cos\theta $$
Thus by the cosine law it follows that
$$ \mathbf{x}_{1j}\cdot\mathbf{x}_{1k} = d_{1j}d_{1k} \cos\theta = \frac{1}{2}(d_{1j}^2 + d_{1k}^2 - d_{jk}^2) $$
If you look at the next equality after this equation in the paper, then it is clear that they did not take the square in their derivation.