Question Number 24778 by ajfour last updated on 25/Nov/17
$${Show}\:{that}\:{the}\:{shortest}\:{distance} \\ $$$${between}\:{two}\:{opposite}\:{edges}\:\boldsymbol{{a}},\boldsymbol{{d}}\: \\ $$$${of}\:{a}\:{tetrahedron}\:{is}\:\mathrm{6}\boldsymbol{{V}}/\boldsymbol{{ad}}\mathrm{sin}\:\boldsymbol{\theta}, \\ $$$${where}\:\theta\:{is}\:{the}\:{angle}\:{between}\:{the} \\ $$$${edges}\:{and}\:{V}\:{is}\:{the}\:{volume}\:{of}\:{the} \\ $$$${tetrahedron}. \\ $$
Commented by ajfour last updated on 25/Nov/17
Commented by ajfour last updated on 27/Nov/17
$${please}\:{someone}\:{attempt}\:{this}\:? \\ $$
Commented by jota+ last updated on 27/Nov/17
![b=BC a=AD B(0 0 0) C(b 0 0) A(a_x a_y 0) D=(d_x d_y d_z ) H=((6V)/(absinθ)) 6V=Habsinθ (New tesis) Case 1/2) d_x =a_x in this case θ=90. 6V=(H=a_y (d_z /a))ab=ba_y d_z . True Case 2/2) Sea A^∗ (d_x a_y 0) auxiliar point ⇒H=a_y (d_z /([A^∗ D])) and sinθ=(([A^∗ D])/a) Luego 6V=Habsinθ=ba_y d_z . True](Q24863.png)
$$\boldsymbol{{b}}=\boldsymbol{{B}}{C} \\ $$$$\boldsymbol{{a}}=\boldsymbol{{A}}{D} \\ $$$${B}\left(\mathrm{0}\:\:\mathrm{0}\:\:\mathrm{0}\right)\:\:\:{C}\left({b}\:\:\mathrm{0}\:\:\mathrm{0}\right)\:\:{A}\left({a}_{{x}} \:\:{a}_{{y}} \:\:\mathrm{0}\right) \\ $$$${D}=\left({d}_{{x}} \:\:{d}_{{y}} \:\:{d}_{{z}} \right) \\ $$$${H}=\frac{\mathrm{6}{V}}{{absin}\theta} \\ $$$$\mathrm{6}{V}={Habsin}\theta\:\:\:\left({New}\:{tesis}\right) \\ $$$$\left.{Case}\:\mathrm{1}/\mathrm{2}\right)\:\:{d}_{{x}} ={a}_{{x}} \:{in}\:{this}\:{case}\: \\ $$$$\theta=\mathrm{90}. \\ $$$$\:\mathrm{6}{V}=\left({H}={a}_{{y}} \frac{{d}_{{z}} }{{a}}\right){ab}={ba}_{{y}} {d}_{{z}} .\:\:{True} \\ $$$$\left.{Case}\:\mathrm{2}/\mathrm{2}\right)\:{Sea}\:{A}^{\ast} \left({d}_{{x}} \:\:{a}_{{y}} \:\:\mathrm{0}\right)\:{auxiliar}\: \\ $$$$\:{point} \\ $$$$\:\Rightarrow{H}={a}_{{y}} \:\frac{{d}_{{z}} }{\left[{A}^{\ast} {D}\right]}\:\:\:\:{and} \\ $$$$\:\:\:\:\:\:{sin}\theta=\frac{\left[{A}^{\ast} {D}\right]}{{a}} \\ $$$${Luego} \\ $$$$\mathrm{6}{V}={Habsin}\theta={ba}_{{y}} {d}_{{z}} .\:\:{True} \\ $$$$ \\ $$
Commented by ajfour last updated on 28/Nov/17
$${thanks}\:{a}\:{lot},\:{but}\:{let}\:{me}\:{see}\:{if}\:{i} \\ $$$${can}\:{follow}\:{it}\:{or}\:{not}\:! \\ $$