Question

If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular, then show that the edges in the third pair are also perpendicular.

Answer 1

Let ABCD is a tetrahedron and AD,

BC and AC, BD are two pairs of opposite edges which are perpendicular to each other

I.e Ad.Bd = 0 amd AC.Bd and OD = d

Now AD.BC = 0

$\Rightarrow (d-a)(c-b)=0$ ...(1)

and Ac.Bd = 0

$\Rightarrow (c-a)(d-b)=0$ ...(2)

Now equation (1) and (2) we get

(d-a) (c-d) - (c-a) (d-b) = 0

$\Rightarrow (dc-db-ac+ab)-(cd-cb+cd+ab)=0$

$cd-db-ac+ab-cd+cb-ad-ab=0$

$\Rightarrow cd-ad-ab-ac=0$

$\Rightarrow cb-db-ad-ac=0$

$\Rightarrow b(c-d)+a(d-c)=0$

$\Rightarrow b(c-d)-a(c-d)=0$

$\Rightarrow (c-d)(b-a)=0$

$\Rightarrow CD.AB=0$

Hence, the third pair of opposite edges AB,CD are also perpendicular to each other.