Question

Prove that the shortest distance between any two opposite edges of a tetrahedron formed by the planes y + z = 0, x + z = 0, x + y = 0, x + y + z = √3 a is √2 a

Answer 1

Given planes are y + z = 0 ..... (i) 

x + z = 0 ..... (ii) 

x + y = 0 ..... (iii) 

x + y + z = √3 a ..... (iv)

Clearly planes (i), (ii) and (iii) meet at O(0, 0, 0) 

Let the tetrahedron be OABC O (0, 0, 0) (0, 0, 3a) 

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

y + z = 0, x + z = 0 ..... (1)

x + y = 0, x + y + z = √3 a ..... (2)

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

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

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

∴ PQ is perpendicular to both OA and BC. 

∴ l + m – n = 0 and l – m = 0 Solving (5) and (6), we get,