In a cube $ABCDA'B'C'D' $two points are considered, $M \in (CD')$ and $N \in (DA')$. Show that the $MN$ is common perpendicular to the lines $CD'$ and $DA'$ if and only if $$\frac{D'M}{D'C}=\frac{DN}{DA'} =\frac{1}{3}.$$
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Tags: 3D geometry, geometry, cube, perpendicular, ratio
In a cube $ABCDA'B'C'D' $two points are considered, $M \in (CD')$ and $N \in (DA')$. Show that the $MN$ is common perpendicular to the lines $CD'$ and $DA'$ if and only if $$\frac{D'M}{D'C}=\frac{DN}{DA'} =\frac{1}{3}.$$