Dec 12, 2016

$d = \sqrt{\frac{2}{3}} - \frac{1}{\sqrt{6}} = 0.408248$

#### Explanation:

${L}_{a} \to \left(x + 1 = 2 {t}_{1} , y + 1 = 3 {t}_{1} , z + 1 = 4 {t}_{1}\right)$
${L}_{b} \to \left(x + 1 = 3 {t}_{2} , y = 4 {t}_{2} , z = 5 {t}_{2}\right)$

or

${L}_{a} \to p = {p}_{a} + {t}_{1} {\vec{v}}_{a}$
${L}_{b} \to p = {p}_{b} + {t}_{2} {\vec{v}}_{b}$

with

${p}_{a} = \left(- 1 , - 1 , - 1\right)$ and ${\vec{v}}_{a} = \left(2 , 3 , 4\right)$
${p}_{b} = \left(- 1 , 0 , 0\right)$ and ${\vec{v}}_{b} = \left(3 , 4 , 5\right)$

calling now ${\hat{v}}_{c} = \frac{{\vec{v}}_{a} \times {\vec{v}}_{b}}{\left\lVert {\vec{v}}_{a} \times {\vec{v}}_{b} \right\rVert}$ and also making

$\Delta p = {p}_{a} + {t}_{1} {\vec{v}}_{a} - \left({p}_{b} + {t}_{2} {\vec{v}}_{b}\right) = {p}_{a} - {p}_{b} + {t}_{1} {\vec{v}}_{a} - {t}_{2} {\vec{v}}_{b}$

making the scalar product by ${\hat{v}}_{c}$ we have

$\left\langle\Delta p , {\hat{v}}_{c}\right\rangle = \left\langle{p}_{a} - {p}_{b} , {\hat{v}}_{c}\right\rangle$ because $\left\langle{\hat{v}}_{c} , {\vec{v}}_{a}\right\rangle = \left\langle{\hat{v}}_{c} , {\vec{v}}_{b}\right\rangle = 0$ and finally

$d = \left\mid \left\langle\Delta p , {\hat{v}}_{c}\right\rangle \right\mid = \left\mid \left\langle{p}_{a} - {p}_{b} , {\hat{v}}_{c}\right\rangle \right\mid$

In our case study we have

${p}_{a} - {p}_{b} = \left(0 , 1 , 1\right)$

${\vec{v}}_{a} \times {\vec{v}}_{b} = \left(- 1 , 2 , - 1\right)$ and

${\hat{v}}_{c} = \left(- \frac{1}{\sqrt{6}} , \sqrt{\frac{2}{3}} , - \frac{1}{\sqrt{6}}\right)$ and

$d = \sqrt{\frac{2}{3}} - \frac{1}{\sqrt{6}} = 0.408248$