A square ABCD of diagonal 2a is folded along | vedclass

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(B) Let the diagonal $AC$ lie along the $x$-axis with the origin at the midpoint of $AC$. The coordinates are $A(-a, 0, 0)$ and $C(a, 0, 0)$.Initially

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$2a/\sqrt{3}$

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(B) Let the diagonal $AC$ lie along the $x$-axis with the origin at the midpoint of $AC$. The coordinates are $A(-a, 0, 0)$ and $C(a, 0, 0)$.Initially, $D$ and $B$ are at $(0, a, 0)$ and $(0, -a, 0)$ respectively.When folded such that planes $DAC$ and $BAC$ are at a right angle, $D$ moves to $(0, 0, a)$ and $B$ moves to $(0, -a/\sqrt{2}, -a/\sqrt{2})$ is incorrect; let's use a standard frame: $A(-a, 0, 0)$, $C(a, 0, 0)$, $D(0, 0, a)$, $B(0, -a, 0)$ is not right-angled. Correct coordinates: $A(-a, 0, 0)$, $C(a, 0, 0)$, $D(0, a/\sqrt{2}, a/\sqrt{2})$, $B(0, -a/\sqrt{2}, a/\sqrt{2})$.The shortest distance $d$ between two skew lines is given by $|(\vec{r_2} - \vec{r_1}) \cdot (\vec{v_1} 	imes \vec{v_2})| / |\vec{v_1} 	imes \vec{v_2}|$.For lines $DC$ and $AB$, the shortest distance is calculated as $\frac{2a}{\sqrt{3}}$.

$A$ square $ABCD$ of diagonal $2a$ is folded along the diagonal $AC$ so that the planes $DAC$ and $BAC$ are at a right angle. The shortest distance between $DC$ and $AB$ is

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