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Question

If volume of regular tetrahedron of edge length k is V and shortest distance between any pair of opposite edges of same regular tetrahedron is d, then the value of d3V is

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Solution

Length of edge of regular tetrahedron =k
Consider a tetrahedron ABCD
and let A(0,0,0), B(k,0,0), C(k2,3k2,0)
Projection of point D on plane containing points A,B and C is the centroid of equilateral triangle ABC
Projection of D is (k2,3k6,0)
Let coordinates of D be (k2,3k6,h)

|AD|=k
k24+k212+h2=k2
h2=k2k23=2k23
h=23k
D is (k2,3k6,2k3)

Since, distance between any two skew lines of tetrahedron ABCD=d,
ACCD×ABCD×AB=d
d=(k2^i+3k2^j)(23k6^j+2k3^k)×k^i∣ ∣(236^j+2k3^k)×k^i∣ ∣
d=(k2^i+3k2^j)(23k26^k+2k23^j)k2 (13)2+(23)2
=3k2.2k23k2
d=12k

Area of equilateral ΔABC=34k2
Volume of tetrahedron, V=13×(ar(ΔABC))×h
=13×34k2×23k=2k312

d3V=(k2)3(2k312)
=k322.122k2=124=3

Alternate Solution :

We first draw a cube of side length d and draw a tetrahedron in it.
By Pythagoras theorem,
d2+d2=DC2=k2
d=k2

Volume of regular tetrahedron with edge length k is V=k362
d3V=3

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