# Shortest Distance Between Skew Lines with Basic Geometry

Cube ABCD.EFGH have side length s cm, calculate the shortest distance of segment line AH and DG without involving vector (or even partial differentiation) and only using basic geometry (such as pythagorean theorem, trigonometry, etc).

My work: Find the plane which is perpendicular to DG which is ADEH (Because ADEH is perpendicular to DCGH which DG lies on), ADEH intersect DG at line AD on point D. So i thought the distance between those two skew lines is perpendicular distance from D to AH . Im using $$base.height = base.height$$ to find perpendicular distance from D to AH,which is $$\frac{S}{\sqrt{2}}$$. But when i checked using vector my answer is wrong and the correct answer is $$\frac{S}{\sqrt{3}}$$.

What's going on ? which line segment is the correct shortest distance ? and how does skew lines distance are calculated without vector ? please show me how did you do it, pretty please.

Since $$AH||BG$$ and $$HF||DB$$, we obtain: $$(AHF)||(DBG)$$.

Id est, our distance it's just the distance between planes: $$(AHF)$$ and $$(DBG).$$

But $$\overrightarrow{EC}$$ it's a normal to both planes.

Thus, the needed distance is equal to $$\frac{1}{3}EC=\frac{s\sqrt3}{3}=\frac{s}{\sqrt3}.$$

• Well that's all, i just want to know where is my mistake Jul 26, 2019 at 11:39
• This is nice! technically, @Michael Rozenberg translated the distance between the two lines to a distance between two planes. The questions are why do the planes $AHF$ and $DBG$ cut the line $EC$ in three equal lengths and why does this translation guarantee the smallest distance? Jul 26, 2019 at 12:02
• @Ahmed Hossam Take a parallelogram $EGCA$, $M$ is a midpoint of $EG$ and $N$ is a midpoint of $AC$. Also, let $EC\cap AM=\{P\}$ and $EC\cap GN=\{Q\}$. Thus, $EP=PQ=QC$ because $P$ and $Q$ are gravitation centers of $\Delta AEG$ and $\Delta AGC$ respectively. Jul 26, 2019 at 12:08
• @Michael Rozenberg So, $EP + PQ + QC = \sqrt{3}s$ and $EP = PQ = QC$. Three questions: How do we know that $P$ and $Q$ are the gravitational centers of the triangles $\Delta AEG$ and $\Delta AGC$? Why is $EC$ normal to $AM$ and to $NG$? Does this translation of distance between the two lines $AH$ and $DG$ to the distance between the planes $AHF$ and $DBG$ guarantee the smallest distance, because $PQ$ is normal to both planes? Like $PQ$ is the distance? Jul 26, 2019 at 12:27
• @Ahmed Hossam Let $K$ be a midpoint of $EC$. Thus, $AM$ and $EK$ are medians of $\Delta AEG$. For $Q$ it's a similar. Also, $EC\perp(AHF)$ because $EC\perp HF$ and $EC\perp AH$. For example, $HF\perp EG$ and $HF\perp AE,$ which says $HF\perp(AEGC),$ which gives $HF\perp EC.$ Jul 26, 2019 at 12:41

I don't think there is some easier way than differentiation, because questions like this are asked when the topic's name is differentiation. So here is the way to solve:

Imagine two opposite vertices of the cube are moving towards each other. Then these would be the vertices of a smaller cube inside of the original cube. The diagonal of the smaller cube is a distance between the two red lines and this distance or cube diagonal $$D$$ can be calculated with a geometry formula (3D pythagoras)

\begin{align*}D(h) &= \sqrt{(s-h)^2 + (s-h)^2 + (s-2h)^2} \\&= \sqrt{2(s-h)^2 + (s-2h)^2}\\&= \sqrt{6h^2-8hs+3s^2} \end{align*}

Now we ask for which $$0 \leq h \leq \frac{s}{2}$$ does $$D(h)$$ become the smallest possible $$D(h)$$. We need the first derivative of $$D(h)$$. The first derivative of a function $$D(h)$$ that looks like this $$\sqrt{f(h)}$$ is something that looks like this $$\dfrac{f'(h)}{2\sqrt{f(h)}}$$. Here we have $$f(h) =6h^2-8hs+3s^2$$ and $$f'(h)= 12h-8s$$. In total we have

$$D'(h) = \dfrac{12h-8s}{2\sqrt{6h^2-8hs+3s^2}}$$

We then set $$D'(h)=0$$ to find the values of $$h$$ at which $$D(h)$$ has a max or a min value. Set $$D'(h)=0$$ and get $$h=\frac{2}{3}s$$. This is the only value for $$h$$, that makes $$D'(h)=0$$. So, the solution for the minimal distance is $$D\left(h=\frac{2}{3}s\right)= \sqrt{6\left(\dfrac{2}{3}s\right)^2-8\left(\dfrac{2}{3}s\right)s+3s^2}=\dfrac{s}{\sqrt{3}}$$

• Seems like basic geometry (with pythagorean) for 3D distances isnt really taught much in other countries, maybe it's even already forgotten. But people from my country was really able to use basic geometry to do this somehow Jul 26, 2019 at 11:26