Problem Statement:
We are given:
A cube with vertices at all points where .
: The set of all 12 lines containing the face diagonals of the cube (2 diagonals per face × 6 faces).
: The set of all 4 lines containing the main (space) diagonals of the cube (e.g., the line through and ).
We need to find the maximum value of the shortest distance where (a face diagonal) and (a main diagonal). The options are given in terms of .
Step 1: Understand the Geometry
Cube :
Vertices: All combinations of where each is 0 or 1.
Edge length: 1 (since adjacent vertices differ by 1 in one coordinate).
Main Diagonals ():
There are 4 main diagonals, each passing through two opposite vertices of the cube.
Example: The line through and .
Face Diagonals ():
Each face of the cube has 2 diagonals.
There are 6 faces, so .
Example: The line through and (a diagonal of the face ).
Step 2: Parametrize the Lines
Main Diagonal ():
Let’s take as the line through and .
Parametric equations:
Face Diagonal ():
Consider the face diagonal on the face , from to .
Parametric equations:
Step 3: Compute the Shortest Distance Between and
The shortest distance between two skew lines and is given by:
For our lines:
, .
, .
Compute :
Compute :
Since , the numerator is 0. This suggests the lines intersect, which is true (they meet at ). Thus, .
This is not the maximum distance. We need to choose non-intersecting lines.
Step 4: Choose Non-Intersecting Lines
Let’s take:
: The face diagonal from to (on the face ).
: The main diagonal from to .
Compute .
Compute :
Compute :
Compute the numerator:
Thus:
Step 5: Verify for Other Pairs
We must check if this is indeed the maximum possible distance between any and . By symmetry, other non-intersecting pairs will yield the same or smaller distances. For example:
If is the diagonal from to (a different face diagonal), the distance to (the main diagonal) will be smaller or zero.
Thus, the maximum distance is .
Final Answer:
The maximum value of is . Therefore, the correct option is (A).
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