Find the shortest distance between two lines and their coordinates.

Question

$$g_1: \vec x = \vec b_1 +s \vec r_1, s, \in \mathbb{R}$$ $$g_2: \vec x = \vec b_2 +t \vec r_2, t, \in \mathbb{R}$$

Given this information calculate the values of $s$ and $t$. Using that find the coordinates where the distance is the shortest and then calculate the actual shortest distance between $g_1$ and $g_2$.

Here is my attempt.

$$F_{g_1}=(1+2s \ | \ 6 \ | \ 1+s)$$ $$F_{g_2}=(6+9t \ | \ 8+6t \ | \ 9+9t)$$

$$\vec {F_{g_1}F_{g_2}}= \vec f_{g_2}-\vec f_{g_1}= \begin{bmatrix}5+9t-2s\\2+6t\\8+9t-s\end{bmatrix}$$

1. $$\vec {F_{g_1}F_{g_2}} \cdot \vec r_1= \vec 0 \implies 27t-5s= -18$$
2. $$\vec {F_{g_1}F_{g_2}} \cdot \vec r_2= \vec 0 \implies 198t-27s= -129$$

Using Gaussian Elimination I get that $t= \frac {-53}{87}$ and $\frac {9}{29}$, which I then put back into the equation and I get the coordinates:

$$F_{g_1}=(\frac{47}{29} \ | \ 6 \ | \ \frac{38}{29})$$ $$F_{g_2}=(\frac{15}{29} \ | \ \frac{126}{29} \ | \ \frac{102}{29})$$

$$\vec {F_{g_1}F_{g_2}}= \vec f_{g_2}-\vec f_{g_1}=\begin{bmatrix}\frac{-32}{29}\\\frac{-48}{29}\\\frac{64}{29}\end{bmatrix}$$

Therefore $d(g_1,g_2) = |\vec {F_{g_1}F_{g_2}}|= \frac{16\sqrt{29}}{29} \approx 3.0$

Im confused about two things. Firstly, if this is correct and if it is correct then how do we know that this is actually the shortest distance between the two line?

Answer 1

Let's choose a point on $g_1$ and a point on $g_2$. The relative position of one point with respect to the other is $$\vec l(s,t) =(\vec b_2+t\vec r_2)-(\vec b_1+s\vec r_1)$$ We say that the distance between lines is the minimum of $|\vec l(s,t)|$. To simplify calculations, this will also be the minimum of $|\vec l(s,t)|^2=\vec l(s,t)\cdot \vec l(s,t)$. Now just take the derivatives with respect to $s$ and $t$ and set them to $0$: $$\begin{align}\frac{d}{ds}(\vec l(s,t)\cdot \vec l(s,t))&=-2\vec r_1\cdot \vec l(s,t)&=0\\\frac{d}{dt}(\vec l(s,t)\cdot \vec l(s,t))&=2\vec r_2\cdot \vec l(s,t)&=0\end{align}$$ These is how you get your equations 1 and 2.

Answer 2

Decompose $\vec{F_{g_1}F_{g_2}}=\lambda_1\vec{r_1}+\lambda_2\vec{r_2}+\vec{v}$ where $\vec{v}\in\operatorname{span}\{\vec{r_1},\vec{r_2}\}^{\perp}$ then is it easy to see if $\vec{v}\neq \vec{0}$ that: $$\vec{v}=\pm\langle \vec{b_2-b_1},\frac{\vec{v}}{||\vec{v}||}\rangle \frac{\vec{v}}{||\vec{v}||}$$ is indenpendant from $\lambda_1$ and $\lambda_2$ and clearly $$|\vec{F_{g_1}F_{g_2}}|^2=\langle\lambda_1\vec{r_1}+\lambda_2\vec{r_2},\lambda_1\vec{r_1}+\lambda_2\vec{r_2}\rangle +\langle \vec{v},\vec{v}\rangle$$ is minimal if and only if $\lambda_1=\lambda_2=0$.

Answer 3

The line joining the two points $\vec x_1=\vec b_1 +s \vec r_1$ and $\vec x_2=\vec b_2 +t \vec r_2$ is parallel to

$$\vec x_1-\vec x_2=\vec b_1-\vec b_2+s\vec r_1-t\vec r_2$$

And if this is the shortest line between $g_1$ and $g_2$, it must be perpendicular to both $\vec r_1$ and $\vec r_2$. You can evaluate all the products in the two equations $$(\vec b_1-\vec b_2+s\vec r_1-t\vec r_2).\vec r_1=0$$ $$(\vec b_1-\vec b_2+s\vec r_1-t\vec r_2).\vec r_2=0$$

and get two equations in $s$ and $t$ of the form $$as+bt=e$$ $$cs+dt=f$$ which I expect you know how to solve.