Let $r: B^n \to S^{n-1}$ be definied by: $$r\left(x\right):=\frac{x-f\left(x\right)}{\left\lVert x-f\left(x\right)\right\rVert_2}\in S^{n-1}$$ ($r\left(x\right)$ is the normed direction vector from $f\left(x\right)$ to $x$.)
Let $\lambda: B^n\to \mathbb{R}^+_0$ be defined by: $$\lambda\left(x\right) := - \left\langle x, r\left(x\right) \right\rangle + \sqrt{1-\left\lVert x\right\rVert_2^2 + \left\langle x, r\left(x\right) \right\rangle^2}\in\mathbb{R}^+_0$$
Finally, let $F:B^n\to S^{n-1}$ be defined by: $$F\left(x\right)=x+\lambda\left(x\right)r\left(x\right)$$ ($F\left(x\right)$ is the intersection of the half-ray from $f\left(x\right)$ to $x$ with the sphere $S^{n-1}$ and $\lambda\left(x\right)$ is the corresponding scalar.)
My construction should follow the proof from Wikipedia:
I now want to check that $F\left(x\right)$ actually lies in $S^{n-1}$. \begin{align} \left\lVert F\left(x\right) \right\rVert_2^2 = &\left\langle F\left(x\right), F\left(x\right) \right\rangle\newline = &\left\langle x+\lambda\left(x\right)r\left(x\right), x+\lambda\left(x\right)r\left(x\right) \right\rangle\newline = &\left\langle x, x+\lambda\left(x\right)r\left(x\right) \right\rangle\newline &+ \left\langle \lambda\left(x\right)r\left(x\right), x+\lambda\left(x\right)r\left(x\right) \right\rangle\newline =&\left\langle x, x\right\rangle + \left\langle x, \lambda\left(x\right)r\left(x\right) \right\rangle\newline &+ \left\langle \lambda\left(x\right)r\left(x\right), x \right\rangle + \left\langle \lambda\left(x\right)r\left(x\right), \lambda\left(x\right)r\left(x\right) \right\rangle\newline =&\left\langle x, x\right\rangle + \lambda\left(x\right) \left\langle x, r\left(x\right) \right\rangle\newline &+ \lambda\left(x\right) \left\langle r\left(x\right), x \right\rangle + \lambda\left(x\right)^2\left\langle r\left(x\right), r\left(x\right) \right\rangle\newline =&\left\lVert x\right\rVert_2^2 + 2\lambda\left(x\right) \left\langle x, r\left(x\right) \right\rangle + \lambda\left(x\right)^2 \left\lVert r\left(x\right)\right\rVert_2^2 \overset{!}{=} 1\newline \Leftrightarrow \lambda_{1,2}\left(x\right) =&\frac{-2\left\langle x, r\left(x\right) \right\rangle \pm \sqrt{\left(2\left\langle x, r\left(x\right) \right\rangle\right)^2 - 4\left\lVert r\left(x\right)\right\rVert_2^2 \left\lVert x\right\rVert_2^2}}{2\left\lVert r\left(x\right)\right\rVert_2^2}\newline =&-\left\langle x, r\left(x\right) \right\rangle \pm \sqrt{\left\langle x, r\left(x\right) \right\rangle^2 - \left\lVert x\right\rVert_2^2} \end{align}
However, I am missing the 1 in the discriminant. Where did I go wrong?
Thanks to Izaak van Dongen.
\begin{align*} \left\lVert F\left(x\right)\right\rVert_2=\ &\left\lVert x\right\rVert_2^2 + 2\lambda\left(x\right) \left\langle x, r\left(x\right) \right\rangle + \lambda\left(x\right)^2 \left\lVert r\left(x\right)\right\rVert_2^2 \overset{!}{=} 1\\ \Leftrightarrow\ &\lambda\left(x\right)^2 + \left(2\left\langle x, r\left(x\right)\right\rangle\right) \lambda\left(x\right) + \left(\left\lVert x\right\rVert_2^2 - 1\right) = 0\\ \Leftrightarrow \lambda_{1,2}\left(x\right) =&\frac{-2\left\langle x, r\left(x\right) \right\rangle \pm \sqrt{\left(2\left\langle x, r\left(x\right) \right\rangle\right)^2 - 4 \left(\left\lVert x\right\rVert_2^2-1\right)}}{2}\\ =&-\left\langle x, r\left(x\right) \right\rangle \pm \sqrt{\left\langle x, r\left(x\right) \right\rangle^2 + 1 - \left\lVert x\right\rVert_2^2} \end{align*}