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Mathematics Question on introduction to three dimensional geometry

Let Q be the cube with the set of vertices {(x1, x2, x3) ∈ R3: x1, x2, x3 ∈ {0,1}}. Let F be the set of all twelve lines containing the diagonals of the six faces of 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 l1 and l2, let d(l1,l2) denote the shortest distance between them. Then the maximum value of d(l1,l2) as l1 varies over f and l2 varies over S, is

A

16\frac{1}{\sqrt6}

B

18\frac{1}{\sqrt8}

C

13\frac{1}{\sqrt3}

D

112\frac{1}{\sqrt{12}}

Answer

16\frac{1}{\sqrt6}

Explanation

Solution

The correct option is (A):
cube with the set of vertices
The equation of the OD line is
r1⃗=0⃗+λ(i^+j^)\vec{r_1}=\vec{0}+\lambda(\hat{i}+\hat{j})
equation of diagonal BE is
r1⃗=j^+α(i^−j^+k^)\vec{r_1}=\hat{j}+\alpha(\hat{i}-\hat{j}+\hat{k})
S.D = ∣j^.(i^−j^+k^)6∣=16\left | \frac{\hat{j}.(\hat{i}-\hat{j}+\hat{k})}{\sqrt{6}} \right |=\frac{1}{\sqrt6}