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

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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