Skew Lines and Distance
DAILY_CHALLENGE
Grade None
Question:
Let $Q$ be the cube with the set of 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 twelve lines containing the diagonals of the six faces of the cube $Q$. Let $S$ be the set of all four lines containing the main diagonals of the cube $Q$; for instance, the line passing through the vertices $(0,0,0)$ and $(1,1,1)$ is in $S$. For lines $\ell_1$ and $\ell_2$, let $d(\ell_1,\ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1,\ell_2)$, as $\ell_1$ varies over $F$ and $\ell_2$ varies over $S$, is
$\dfrac{1}{\sqrt{6}}$
$\dfrac{1}{\sqrt{8}}$
$\dfrac{1}{\sqrt{3}}$
$\dfrac{1}{\sqrt{12}}$
Step-by-Step Solution
Key Concept: Distance between skew lines = |( P₂−P₁)·(d₁×d₂)|/|d₁×d₂|; face diagonal parallel to main diagonal gives d=0
Take $\ell_1\in F$: face diagonal on face $z=0$ from $(1,0,0)$ to $(0,1,0)$, direction $\vec{d_1}=(-1,1,0)/\sqrt{2}$.
Take $\ell_2\in S$: main diagonal from $(0,0,0)$, direction $\vec{d_2}=(1,1,1)/\sqrt{3}$.
Point on $\ell_1$: $(1,0,0)$. Point on $\ell_2$: $(0,0,0)$. Connecting vector: $\vec{v}=(1,0,0)$.
$\vec{d_1}\times\vec{d_2}=\dfrac{1}{\sqrt{6}}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&0\\1&1&1\end{vmatrix}=\dfrac{1}{\sqrt{6}}(1,1,-2)$.
$|\vec{d_1}\times\vec{d_2}|=\sqrt{(1+1+4)/6}=\sqrt{6/6}=1$.
$d=|\vec{v}\cdot(\vec{d_1}\times\vec{d_2})|/|\vec{d_1}\times\vec{d_2}|=|(1,0,0)\cdot(1,1,-2)/\sqrt{6}|/1=1/\sqrt{6}$.
This is the maximum over all pairs.
Correct Answer: A