Distance between diagonals in rectangular parallelipipedon

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Distance between diagonals in rectangular parallelipipedon

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#1 h Dec 4, 2015, 9:33 AM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Extension. Let $ABCDA'B'C'D'$ be a rectangular parallelipipedon with the base $[ABCD]$ and the vertical edges $[AA']$ , $[BB']$ , $[CC']$ , $[DD']$ so that

$\left\{\begin{array}{ccc} AB & = & a\\BC & = & b\\\ AA' & = & c\end{array}\right\|$ . Prove that the distance $d\equiv d(c)$ between the diagonals $AC$ si $D'B$ is given by the relation $\boxed{d(c)=\frac {abc}{\sqrt{4a^2b^2+c^2\cdot \left(a^2+b^2\right)}}}\ .$

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#2 h Dec 4, 2015, 5:52 PM • 2 Y

Y by AndrewTom, Adventure10

Solution, using analytic geometry.
Let $A(a,0,0), B(0,0,0), C(0,b,0), D(a,b,0), D'(a,b,c), B'(0,0,c)$ .

Point $P \in AC : P(\frac{ab^{2}c^{2}+2a^{3}b^{2}}{b^{2}c^{2}+a^{2}c^{2}+4a^{2}b^{2}},\frac{a^{2}bc^{2}+2a^{2}b^{3}}{b^{2}c^{2}+a^{2}c^{2}+4a^{2}b^{2}},0)$ .

Point $Q \in BD' : Q(\frac{2a^{3}b^{2}}{b^{2}c^{2}+a^{2}c^{2}+4a^{2}b^{2}},\frac{2a^{2}b^{3}}{b^{2}c^{2}+a^{2}c^{2}+4a^{2}b^{2}},\frac{2a^{2}b^{2}c}{b^{2}c^{2}+a^{2}c^{2}+4a^{2}b^{2}})$ .

$PQ \bot AC$ and $PQ \bot BD'$ .
$PQ = \frac{abc}{\sqrt{b^{2}c^{2}+a^{2}c^{2}+4a^{2}b^{2}}}$ .

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