Q. Let $A\left( \overrightarrow{a} \right)$ and $B\left( \overrightarrow{b} \right)$ be points on two skew lines $\overrightarrow{r} = \overrightarrow{a} + \lambda\ \overrightarrow{p}$ and $\overrightarrow{r} = \overrightarrow{b} + u\overrightarrow{q}$ and the shortest distance between the skew lines is 1, where $\overrightarrow{p}$ and $\overrightarrow{q}$ are unit vectors forming adjacent sides of a parallelogram enclosing an area of $\frac{1}{2}$ units. If an angle between $AB$ and the line of shortest distance is $60{^\circ}$, then $AB =$
A) $\frac{1}{2}$
B) 2
C) 1
D) $\lambda \in R - \{ 0\}$
Explanation
(b)
$1 = \left| \left( \overrightarrow{b} - \overrightarrow{a} \right) \bullet \frac{(\overrightarrow{p} \times \overrightarrow{q})}{\left| \overrightarrow{p} \times \overrightarrow{q} \right|} \right|$
$\Rightarrow \ \left| \overrightarrow{b} - \overrightarrow{a} \right|\cos{60{^\circ}} = 1\ \Rightarrow \ AB = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.