Question

Let $OA,OB,OC$ be coterminous edges of a cube. If $l,m,n$ be the shortest distance between the sides $OA,OB,OC$ and their respective skew body diagonals to them respectively then $\left(\frac{1}{l^2}+\frac{1}{m^2}+\frac{1}{n^2}\right):\left(\frac{1}{O{A}^2}+\frac{1}{O{B}^2}+\frac{1}{O{C}^2}\right)$ is equal to

• A. 2
• B. 3
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

Correct answer is (a) 2