Question:
Find the shortest distance between GT and BA.
- Parametric equations for GT and BA:
$GT\leftrightarrow\begin{cases}x=r\\y=2-r\\z=2+2r\end{cases}\rightarrow P(r,2-r,2+2r)$, $AB\leftrightarrow\begin{cases}x=2\\y=s\\z=0\end{cases}\rightarrow Q(2,s,0)$ - Find $co(\vec{PQ})=co(\vec{P}-\vec{Q})=(2-r,r+s-2,-2-2r)$
- $PQ \perp GT\Leftrightarrow \vec{PQ}.\vec{GT}=0\Leftrightarrow -6r-s=0$
- $PQ \perp AB\Leftrightarrow \vec{PQ}.\vec{AB}=0\Leftrightarrow r+s-2=0$
- $\begin{cases}-6r-s=0\\r+s-2=0\end{cases}\Leftrightarrow r=-\frac{2}{5}, s=\frac{12}{5}$
- $P(-\frac{2}{5},\frac{12}{5},\frac{6}{5}),Q(2,\frac{12}{5},0)$
- $|PQ|=\frac{6\sqrt{5}}{5}$
This is the drawing associated to this exercise inside my book:
I don't understand how there could exist a line between GT and AB that could be perpendicular to both. Wouldn't that be impossible based on the drawing?
Also, when looking on the internet, I don't seem to find a lot of resources on analytic geometry (if that's how it is called), would you have some links to share?
Best Answer
First, imagine a plane perpendicular to $AB$ through some point $Q$ (not necessarily the $Q$ in your picture) and a plane perpendicular to $GT$ through some point $P$ (not necessarily the $P$ in your picture.) Since the lines are skew, those planes have to intersect. The line of intersection may not intersect $AB$ or $GT$, but you have the freedom to move both planes (which means you have the freedom to move $P$ and $Q$.) So slide $P$ and $Q$ cleverly until the line of intersection of the planes intersects $AB$ and $GT$. Since every line in the first plane through $Q$ is perpendicular to $AB$ and every line in the second plane through $P$ is perpendicular to $GT$, the intersection is perpendicular to both lines.