3D Geometry
Shortest Distance
MMTS_Full_Test_12
Grade 12
Question:
Let $Q$ be the cube with vertices $\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1,x_2,x_3\in\{0,1\}\}$. Let $F$ be the set of all 12 lines containing face diagonals and $S$ be the set of 4 main diagonals. For lines $l_1\in F$ and $l_2\in S$, let $d(l_1,l_2)$ denote shortest distance. Maximum of $d(l_1,l_2)$ is $\lambda$. Find $\lambda^{-2}$.
Step-by-Step Solution
Key Concept: Compute shortest distance between a face diagonal and a body diagonal
The maximum shortest distance between face diagonal and body diagonal in unit cube: $\lambda=1/\sqrt{6}$... After careful computation: $\lambda^{-2}=6$.
Correct Answer: 6