Let $\mathrm{A}\left(\overrightarrow{\mathrm{a}}\right)$ and $\mathrm{B}\left(\overrightarrow{\mathrm{b}}\right)$ be points on two skew lines $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\mathrm{u}\overrightarrow{\mathrm{q}}$ and the shortest distance between the skew lines is l, where $\overrightarrow{\mathrm{p}} \mathrm{and} \overrightarrow{\mathrm{q}}$ are unit vectors forming adjacent sides of a parallelogram enclosing an area of ! units. If an angle between l8 and the line of shortest distance is 60^o, then AB =

• A

  $\frac{1}{2}$

• B

  2

• C

  1

• D

  $\mathrm{\lambda }\in \mathrm{R}-\left\{0\right\}$