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