Three Dimensional Geometry Question 368
Question: 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 AB and the line of shortest distance is $ 60{}^\circ $ , then $ AB= $
Options:
A) $ \frac{1}{2} $
B) $ 2 $
C) $ 1 $
D) $ \lambda \in R-{0} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ 1=| (\vec{b}-\vec{a}).\frac{(\vec{p}\times \vec{q})}{| \vec{p}\times \vec{q} |} |\Rightarrow | \vec{a}-\vec{b} |\cos 60{}^\circ =1 $ $ AB=2 $
BETA
