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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.

Cube

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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 $$\frac{1}{3}EC=\frac{s\sqrt3}{3}=\frac{s}{\sqrt3}.$$

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  • $\begingroup$ Yes, this is what i need to see, so i thought we would need to make a perpendicular plane, but instead it need to be parallel $\endgroup$ – Tobi123 Jul 26 '19 at 11:31
  • $\begingroup$ @Tobi123 It was a hint only. I am waiting for your questions. $\endgroup$ – Michael Rozenberg Jul 26 '19 at 11:34
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    $\begingroup$ Well that's all, i just want to know where is my mistake $\endgroup$ – Tobi123 Jul 26 '19 at 11:39
  • $\begingroup$ Ahmed has also given me a hint on how the distance between skew lines change (with visualisation) $\endgroup$ – Tobi123 Jul 26 '19 at 11:41
  • $\begingroup$ 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? $\endgroup$ – Ahmed Hossam Jul 26 '19 at 12:02
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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:

Cube

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 $$D\left(h=\frac{2}{3}s\right)= \sqrt{6\left(\dfrac{2}{3}s\right)^2-8\left(\dfrac{2}{3}s\right)s+3s^2}=\dfrac{s}{\sqrt{3}}$$

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    $\begingroup$ 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 $\endgroup$ – Tobi123 Jul 26 '19 at 11:26

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