A square ABCD with diagonal length 2a is fold | vedclass

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(B) Let the vertices after folding be $D(0, 0, a)$,$C(a, 0, 0)$,$A(-a, 0, 0)$,and $B(0, -a, 0)$.The line $DC$ passes through $(0, 0, a)$ and $(a, 0, 0

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$2a/\sqrt{3}$

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(B) Let the vertices after folding be $D(0, 0, a)$,$C(a, 0, 0)$,$A(-a, 0, 0)$,and $B(0, -a, 0)$.The line $DC$ passes through $(0, 0, a)$ and $(a, 0, 0)$. Its direction vector is $(a, 0, -a)$,so the equation is $\frac{x}{1} = \frac{y}{0} = \frac{z-a}{-1}$.The line $AB$ passes through $(-a, 0, 0)$ and $(0, -a, 0)$. Its direction vector is $(a, -a, 0)$,so the equation is $\frac{x+a}{1} = \frac{y}{-1} = \frac{z}{0}$.The shortest distance $d$ between two skew lines $\frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2} = \frac{y-y_2}{m_2} = \frac{z-z_2}{n_2}$ is given by $d = \frac{|(x_2-x_1, y_2-y_1, z_2-z_1) \cdot (l_1 	imes l_2)|}{|l_1 	imes l_2|}$.Here,$(x_1, y_1, z_1) = (0, 0, a)$,$(x_2, y_2, z_2) = (-a, 0, 0)$.Vector $(x_2-x_1, y_2-y_1, z_2-z_1) = (-a, 0, -a)$.Direction vectors are $v_1 = (1, 0, -1)$ and $v_2 = (1, -1, 0)$.$v_1 	imes v_2 = \begin{vmatrix} i & j & k \ 1 & 0 & -1 \ 1 & -1 & 0 \end{vmatrix} = i(-1) - j(1) + k(-1) = (-1, -1, -1)$.$|v_1 	imes v_2| = \sqrt{(-1)^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.$d = \frac{|(-a, 0, -a) \cdot (-1, -1, -1)|}{\sqrt{3}} = \frac{|a + 0 + a|}{\sqrt{3}} = \frac{2a}{\sqrt{3}}$.

$A$ square $ABCD$ with diagonal length $2a$ is folded along the diagonal $AC$ such that the planes $DAC$ and $BAC$ are perpendicular to each other. What is the shortest distance between $DC$ and $AB$?

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