1. Tardigrade
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  4. Let A( veca) and B( vecb) be points on two skew line vecr= veca+ vecλ and vecr= vecb+u vecq and the shortest distance between the skew line is 1 , where vecp and vecq are unit vectors forming adjacent sides of a parallelogram enclosing an area of (1/2) units. If an angle between A B and the line of shortest distance is 60°, then A B=

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Q. Let $A(\vec{a})$ and $B(\vec{b})$ be points on two skew line $\vec{r}=\vec{a}+\vec{\lambda}$ and $\vec{r}=\vec{b}+u \vec{q}$ and the shortest distance between the skew line is 1 , where $\vec{p}$ and $\vec{q}$ are unit vectors forming adjacent sides of a parallelogram enclosing an area of $\frac{1}{2}$ units. If an angle between $A B$ and the line of shortest distance is $60^{\circ}$, then $A B=$

Three Dimensional Geometry

Solution:

$1=\left|(\vec{b}-\vec{a}) \cdot \frac{(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}\right| $
$\Rightarrow |\vec{a}-\vec{b}| \cos 60^{\circ}=1 $
$\Rightarrow A B=2$