Let the equation to one of the pair of opposite edges OA and BC be

y + z = 0, x + z = 0 .....(1) and x + y = 0, x + y + z = a .....(2)

equation (1) and (2) can be expressed in symmetrical form as

..... (3) and, .......(4)

d. r. of OA and BC are respectively (1, 1, – 1) and (1, – 1, 0).

Let PQ be the shortest distance between OA and BC having direction cosines (  , m, n)

∴ PQ is perpendicular to both OA and BC.

∴  + m – n = 0 and  – m = 0

Solving (5) and (6), we get, = k (say)

also,  2 + m 2 + n 2 = 1

∴ k 2 + k 2 + 4k 2 = 1 ⇒ k = ±

∴  = ± , m = ± , n = ±

Shortest distance between OA and BC

i.e. PQ = The length of projection of OC on PQ

= | (x 2 – x 1 )  + (y 2 – y 1 ) m + (z 2 – z 1 ) n |

= = a.

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