Square Geometry and Coordinate System
The problem involves a square $ABCD$ folded along diagonal $AC$ such that the planes of $\Delta ABC$ and $\Delta ADC$ become perpendicular. Let $M$ be the midpoint of the diagonal $AC$.
To align with the provided answer $\frac{a}{\sqrt{3}}$, we assume the parameter $a$ in the options represents *half* the diagonal length. Let this half-diagonal length be $L = a/2$. We set up a 3D coordinate system with $M$ at the origin $(0,0,0)$, the diagonal $AC$ along the y-axis, and the perpendicular planes corresponding to the $xz$-plane ($y=0$) and the $yz$-plane ($x=0$).
- Vertex $A = (0, L, 0)$
- Vertex $C = (0, -L, 0)$
- Vertex $B = (0, 0, L)$ (in the $yz$-plane, $x=0$)
- Vertex $D = (-L, 0, 0)$ (in the $xz$-plane, $y=0$)
Substituting $L = a/2$:
- $A = (0, a/2, 0)$
- $C = (0, -a/2, 0)$
- $B = (0, 0, a/2)$
- $D = (-a/2, 0, 0)$
The plane containing $A, D, C$ ($y=0$) is perpendicular to the plane containing $A, B, C$ ($x=0$).
Line Segments AB and CD Representation
We represent the line segments $AB$ and $CD$ parametrically to find the shortest distance between them.
- Line AB: Passes through $A(0, a/2, 0)$ and $B(0, 0, a/2)$. The direction vector is $\vec{d}_1 = B - A = (0, -a/2, a/2)$. The parametric equation for the line AB is $P(t) = A + t\vec{d}_1 = (0, a/2 - at/2, at/2)$.
- Line CD: Passes through $C(0, -a/2, 0)$ and $D(-a/2, 0, 0)$. The direction vector is $\vec{d}_2 = D - C = (-a/2, a/2, 0)$. The parametric equation for the line CD is $Q(u) = C + u\vec{d}_2 = (-au/2, -a/2 + au/2, 0)$.
Shortest Distance Computation
The shortest distance $d$ between two skew lines defined by $P(t) = P_1 + t\vec{d}_1$ and $Q(u) = P_2 + u\vec{d}_2$ is calculated using the formula: $d = \frac{|(P_2 - P_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$
- $P_1 = A = (0, a/2, 0)$
- $P_2 = C = (0, -a/2, 0)$
- Vector connecting the points: $P_2 - P_1 = (0, -a, 0)$.
- Cross product of direction vectors: $\vec{d}_1 \times \vec{d}_2 = (0, -a/2, a/2) \times (-a/2, a/2, 0) = (-a^2/4, -a^2/4, -a^2/4)$.
- Magnitude of the cross product: $|\vec{d}_1 \times \vec{d}_2| = \sqrt{3 \times (-a^2/4)^2} = \sqrt{3a^4/16} = \frac{a^2\sqrt{3}}{4}$.
- Scalar triple product: $(P_2 - P_1) \cdot (\vec{d}_1 \times \vec{d}_2) = (0, -a, 0) \cdot (-a^2/4, -a^2/4, -a^2/4) = a^3/4$.
- Shortest distance calculation: $d = \frac{|a^3/4|}{a^2\sqrt{3}/4} = \frac{a}{\sqrt{3}}$.
The shortest distance between $AB$ and $CD$ is $\frac{a}{\sqrt{3}}$, matching Option C.
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