The edge of a cube is of length a. Then the s | vedclass

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(D) Let the vertices of the cube be $O(0,0,0)$,$A(a,0,0)$,$B(0,a,0)$,$C(0,0,a)$,$E(a,a,0)$,$F(0,a,a)$,$G(a,0,a)$,and $D(a,a,a)$.Consider the diagonal

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$\frac{a}{\sqrt{2}}$

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(D) Let the vertices of the cube be $O(0,0,0)$,$A(a,0,0)$,$B(0,a,0)$,$C(0,0,a)$,$E(a,a,0)$,$F(0,a,a)$,$G(a,0,a)$,and $D(a,a,a)$.Consider the diagonal $OD$ connecting $(0,0,0)$ and $(a,a,a)$. The equation of the line $OD$ is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$.Consider an edge skew to it,for example,the edge $AE$ connecting $(a,0,0)$ and $(a,a,0)$. The equation of the line $AE$ is $x=a, z=0$.The shortest distance between two skew lines is given by the projection of the vector joining a point on each line onto the common perpendicular.Alternatively,using the coordinates of the midpoint $K$ of the diagonal $OD$ as $(\frac{a}{2}, \frac{a}{2}, \frac{a}{2})$ and the midpoint $L$ of the edge $AE$ as $(a, \frac{a}{2}, 0)$,the distance $KL$ is:$KL = \sqrt{(a - \frac{a}{2})^2 + (\frac{a}{2} - \frac{a}{2})^2 + (0 - \frac{a}{2})^2}$$KL = \sqrt{(\frac{a}{2})^2 + 0^2 + (-\frac{a}{2})^2}$$KL = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{2a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$.

The edge of a cube is of length $a$. Then the shortest distance between the diagonal of a cube and an edge skew to it is:

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