Problem Statement:

We are given:

  • A cube Q with vertices at all points where x1,x2,x3{0,1}.

  • F: The set of all 12 lines containing the face diagonals of the cube (2 diagonals per face × 6 faces).

  • S: The set of all 4 lines containing the main (space) diagonals of the cube (e.g., the line through (0,0,0) and (1,1,1)).

We need to find the maximum value of the shortest distance d(1,2) where 1F (a face diagonal) and 2S (a main diagonal). The options are given in terms of 1n.


Step 1: Understand the Geometry

Cube Q:

  • Vertices: All combinations of (x1,x2,x3) where each xi is 0 or 1.

  • Edge length: 1 (since adjacent vertices differ by 1 in one coordinate).

Main Diagonals (S):

  • There are 4 main diagonals, each passing through two opposite vertices of the cube.

  • Example: The line through (0,0,0) and (1,1,1).

Face Diagonals (F):

  • Each face of the cube has 2 diagonals.

  • There are 6 faces, so F=12.

  • Example: The line through (0,0,0) and (0,1,1) (a diagonal of the face x1=0).


Step 2: Parametrize the Lines

Main Diagonal (2):

  • Let’s take 2 as the line through (0,0,0) and (1,1,1).

  • Parametric equations:

    2:r2(t)=(t,t,t),tR.

Face Diagonal (1):

  • Consider the face diagonal on the face x1=0, from (0,0,0) to (0,1,1).

  • Parametric equations:

    1:r1(s)=(0,s,s),sR.

Step 3: Compute the Shortest Distance Between 1 and 2

The shortest distance d(1,2) between two skew lines r1(s)=a1+sd1 and r2(t)=a2+td2 is given by:

d=(a2a1)(d1×d2)d1×d2.

For our lines:

  • a1=(0,0,0)d1=(0,1,1).

  • a2=(0,0,0)d2=(1,1,1).

Compute d1×d2:

d1×d2=ijk011111=(1111)i(0111)j+(0111)k=(0,1,1).

Compute d1×d2:

d1×d2=02+12+(1)2=2.

Since a2a1=(0,0,0), the numerator is 0. This suggests the lines intersect, which is true (they meet at (0,0,0)). Thus, d=0.

This is not the maximum distance. We need to choose non-intersecting lines.


Step 4: Choose Non-Intersecting Lines

Let’s take:

  • 1: The face diagonal from (0,0,1) to (0,1,0) (on the face x1=0).

    r1(s)=(0,s,1s),d1=(0,1,1).
  • 2: The main diagonal from (0,0,0) to (1,1,1).

    r2(t)=(t,t,t),d2=(1,1,1).

Compute a2a1=(0,0,0)(0,0,1)=(0,0,1).

Compute d1×d2:

d1×d2=ijk011111=(2,1,1).

Compute d1×d2:

d1×d2=4+1+1=6.

Compute the numerator:

(a2a1)(d1×d2)=(0,0,1)(2,1,1)=1.

Thus:

d=16.

Step 5: Verify for Other Pairs

We must check if this is indeed the maximum possible distance between any 1F and 2S. By symmetry, other non-intersecting pairs will yield the same or smaller distances. For example:

  • If 1 is the diagonal from (1,0,0) to (1,1,1) (a different face diagonal), the distance to 2 (the main diagonal) will be smaller or zero.

Thus, the maximum distance is 16.


Final Answer:

The maximum value of d(1,2) is 16. Therefore, the correct option is (A).

A

Comments