Taking the results of (Determine distance of point to straight), where the straight is given by:

%$g: \vec{x} = \vec{p} + \mu \vec{u_{0}} $%

let the point R be an element of a second straight h.

%$ h: R = \vec{r}=\vec{a}+\lambda\vec{b_0},\; |\vec{b_0}|=1$%

So the distance d becomes a function of λ

For convenience I now treat d² and will come back to d later. Now calculate the minium of

%$ d(\lambda) $%:

i.e.

%$\frac{\displaystyle d}{\displaystyle d\lambda}d \stackrel{!}{=} 0 \mbox{ and } \left.\frac{\displaystyle d^2}{\displaystyle d\lambda^2}d\;\right|_{\lambda=\lambda_{min}} > 0$%

%$\frac{\displaystyle d}{\displaystyle d\lambda}d = \frac{\displaystyle \lambda A + B}{\displaystyle \sqrt{\lambda^2 \cdot A + 2\cdot \lambda \cdot B + C}} = \frac{\displaystyle \lambda A + B}{\displaystyle d}$%

%$\frac{\displaystyle {d^2}}{\displaystyle {d \lambda^2}} d = \frac{\displaystyle d}{\displaystyle d\lambda} \left( \frac{\displaystyle \lambda A + B}{\displaystyle d} \right) = \frac{\displaystyle {A \cdot d - (\lambda A + B) \cdot \left( \frac{\displaystyle d}{\displaystyle d\lambda}d \right)}}{\displaystyle {d^2}}$%

%$\to \frac{\displaystyle {d^2}}{\displaystyle {d \lambda^2}} d = \frac{\displaystyle A}{\displaystyle d} - \frac{\displaystyle 1}{\displaystyle d} \cdot \left(\frac{\displaystyle d}{\displaystyle d\lambda} d \right)^2$%

%$\mbox{i.e. with: } \left.\frac{\displaystyle d}{\displaystyle d\lambda}\right|_{\lambda_{min}}d = 0 $%

%$\to \left.\frac{\displaystyle {d^2}}{\displaystyle {d \lambda^2}} d\right|_{\lambda=\lambda_{min}} = \frac{\displaystyle A}{\displaystyle d}$%

with

%$ A = b_{0x}^2 (1 - u_{0x}^2) + b_{0y}^2 (1 - u_{0y}^2) + b_{0z}^2 (1 - u_{0z}^2) $% %$\mbox { and } |\vec{u_0}| = 1,\; 0 \le u_{0i} \le 1, |\vec{b_0}| = 1,\; 0 \le b_{0i} \le 1 $% %$\mbox { and } \vec{u_0} \neq \vec{b_0} \mbox { and } \vec{u_0} \times \vec{u_0} \neq \vec{0} \mbox{ i.e. }\vec{u_0} \not\| \vec{b_0}$%

%$ \to A > 0 $%

%$ \mbox{ i.e. if } \frac{\displaystyle d}{\displaystyle d\lambda}d = 0 \mbox{ then it is a minimum.} $%

%$ \mbox{solving: } \frac{\displaystyle d}{\displaystyle d\lambda}d = 0 $% %$ \mbox{, given } d \neq 0 $% %$ \mbox{, gives: } \frac{\displaystyle \lambda A + B}{\displaystyle d} = 0 \to \lambda_{min}=\frac{\displaystyle B}{\displaystyle A} $%

%$ \to \lambda_{min} = \frac{\displaystyle{\Big[ (a_x b_{0x} - b_{0x} p_x)(1-u_{0x}) + (a_y b_{0y} - b_{0y} p_y)(1-u_{0y}) + (a_z b_{0z} - b_{0z} p_z)(1-u_{0z}) \Big]}}{\displaystyle{\Big[ b_{0x}^2 (1 - u_{0x}^2) + b_{0y}^2 (1 - u_{0y}^2) + b_{0z}^2 (1 - u_{0z}^2) \Big]}}$%

%$ \to \vec{r}_{min}=\vec{a}+\frac{\displaystyle{\Big[ (a_x b_{0x} - b_{0x} p_x)(1-u_{0x}) + (a_y b_{0y} - b_{0y} p_y)(1-u_{0y}) + (a_z b_{0z} - b_{0z} p_z)(1-u_{0z}) \Big]}}{\displaystyle{\Big[ b_{0x}^2 (1 - u_{0x}^2) + b_{0y}^2 (1 - u_{0y}^2) + b_{0z}^2 (1 - u_{0z}^2) \Big]}}\vec{b_0} $%

analogue to this calculation you can derive the equivalent point of straight g.

%$ \mu_{min} = \frac{\displaystyle B'}{\displaystyle A'} $%

%$ \to \mu_{min} = \frac{\displaystyle{\Big[ (p_x u_{0x} - u_{0x} a_x)(1-b_{0x}) + (p_y u_{0y} - u_{0y} a_y)(1-b_{0y}) + (p_z u_{0z} - u_{0z} a_z)(1-b_{0z}) \Big]}}{\displaystyle{\Big[ u_{0x}^2 (1 - b_{0x}^2) + u_{0y}^2 (1 - b_{0y}^2) + u_{0z}^2 (1 - b_{0z}^2) \Big]}}$%

%$ \to \vec{x}_{min}=\vec{p}+\mu_{min}\vec{u_0} $%

%$ \to \vec{x}_{min}=\vec{p}+\frac{\displaystyle{\Big[ (p_x u_{0x} - u_{0x} a_x)(1-b_{0x}) + (p_y u_{0y} - u_{0y} a_y)(1-b_{0y}) + (p_z u_{0z} - u_{0z} a_z)(1-b_{0z}) \Big]}}{\displaystyle{\Big[ u_{0x}^2 (1 - b_{0x}^2) + u_{0y}^2 (1 - b_{0y}^2) + u_{0z}^2 (1 - b_{0z}^2) \Big]}}\vec{u_0} $%

%$ \mbox{Vertex }\vec{v} = \frac{\displaystyle{1}}{\displaystyle{2}}(\vec{r}_{min} - \vec{x}_{min})$%

	%$ \vec{v}$%	%$=$%	%$\frac{\displaystyle{1}}{\displaystyle{2}}\Bigg(\vec{a} - \vec{p}$%
			%$+\frac{\displaystyle{\Big[ (a_x b_{0x} - b_{0x} p_x)(1-u_{0x}) + (a_y b_{0y} - b_{0y} p_y)(1-u_{0y}) + (a_z b_{0z} - b_{0z} p_z)(1-u_{0z}) \Big]}}{\displaystyle{\Big[ b_{0x}^2 (1 - u_{0x}^2) + b_{0y}^2 (1 - u_{0y}^2) + b_{0z}^2 (1 - u_{0z}^2) \Big]}}\vec{b_0}$%
			%$\left.- \frac{\displaystyle{\Big[ (p_x u_{0x} - u_{0x} a_x)(1-b_{0x}) + (p_y u_{0y} - u_{0y} a_y)(1-b_{0y}) + (p_z u_{0z} - u_{0z} a_z)(1-b_{0z}) \Big]}}{\displaystyle{\Big[ u_{0x}^2 (1 - b_{0x}^2) + u_{0y}^2 (1 - b_{0y}^2) + u_{0z}^2 (1 - b_{0z}^2) \Big]}}\vec{u_0}\right)$%

...Back to Vertex Finding

-- PeterZumbruch - 23 Feb 2004