Show that the shortest distance between two opposite edges a,d of a tetrahedron is 6V/adsin 𝛉, where θ is the angle between the edges and V is the volume of the tetrahedron.


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Question Number 24778 by ajfour last updated on 25/Nov/17

$S h o w t h a t t h e s h o r t e s t d i s t a n c e$ $b e t w e e n t w o o p p o s i t e e d g e s a, d$ $o f a t e t r a h e d r o n i s 6 V / a d \sin θ,$ $w h e r e θ i s t h e a n g l e b e t w e e n t h e$ $e d g e s a n d V i s t h e v o l u m e o f t h e$ $t e t r a h e d r o n .$
Commented by ajfour last updated on 25/Nov/17

Commented by ajfour last updated on 27/Nov/17

$p l e a s e s o m e o n e a t t e m p t t h i s ?$
Commented by jota+ last updated on 27/Nov/17

$b = B C$ $a = A D$ $B (0 0 0) C (b 0 0) A (a_{x} a_{y} 0)$ $D = (d_{x} d_{y} d_{z})$ $H = \frac{6 V}{a b s i n θ}$ $6 V = H a b s i n θ (N e w t e s i s)$ $C a s e 1 / 2) d_{x} = a_{x} i n t h i s c a s e$ $θ = 90 .$ $6 V = (H = a_{y} \frac{d_{z}}{a}) a b = {b a}_{y} d_{z} . T r u e$ $C a s e 2 / 2) S e a A^{} (d_{x} a_{y} 0) a u x i l i a r$ $p o i n t$ $\Rightarrow H = a_{y} \frac{d_{z}}{[A^{} D]} a n d$ $s i n θ = \frac{[A^{*} D]}{a}$ $L u e g o$ $6 V = H a b s i n θ = {b a}_{y} d_{z} . T r u e$
Commented by ajfour last updated on 28/Nov/17

$t h a n k s a l o t, b u t l e t m e s e e i f i$ $c a n f o l l o w i t o r n o t!$
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