Three Dimensional Geometry Question 360
Question: The shortest distance between the skew lines $ l_1:\vec{r}={{\vec{a}}_1}+\lambda {{\vec{b}}_1}l_2:\vec{r}={{\vec{a}}_2}+\mu {{\vec{b}}_2} $ is
Options:
A) $ \frac{|({{{\vec{a}}}_2}-{{{\vec{a}}}_1}).{{{\vec{b}}}_1}\times {{{\vec{b}}}_2}|}{|{{{\vec{b}}}_1}\times {{{\vec{b}}}_2}|} $
B) $ \frac{| ({{{\vec{a}}}_2}-{{{\vec{a}}}_1}).{{{\vec{a}}}_2}\times {{{\vec{b}}}_2} |}{| {{{\vec{b}}}_1}\times {{{\vec{b}}}_2} |} $
C) $ \frac{| ({{{\vec{a}}}_2}-{{{\vec{b}}}_2}).{{{\vec{a}}}_1}\times {{{\vec{b}}}_1} |}{| {{{\vec{b}}}_1}\times {{{\vec{b}}}_2} |} $
D) $ \frac{| ({{{\vec{a}}}_1}-{{{\vec{b}}}_2}).{{{\vec{b}}}_1}\times {{{\vec{a}}}_2} |}{| {{{\vec{b}}}_1}\times {{{\vec{a}}}_2} |} $
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Answer:
Correct Answer: A
Solution:
[a] Let PQ be the shortest distance vector between $ l_1 $ and $ l_2 $ . Now, $ l_1 $ passes through $ A_1({{\vec{a}}_1}) $ and is parallel to $ {{\vec{b}}_1} $ and $ l_2 $ passes through $ A_2({{\vec{a}}_2}) $ and is parallel to $ {{\vec{b}}_2} $ . Since, PQ is perpendicular to both $ l_1 $ and $ l_2 $ is is parallel to $ {{\vec{b}}_1}\times {{\vec{b}}_2}. $ Let $ \hat{n} $ be the unit vector along PQ. Then, $ \hat{n}=\frac{{{{\vec{b}}}_1}\times {{{\vec{b}}}_2}}{| {{{\vec{b}}}_1}\times {{{\vec{b}}}_2} |} $ Let d be the shortest distance between the given lines $ l_1 $ and $ l_2 $ . $ | \overrightarrow{PQ} |=d $ and $ \overrightarrow{PQ}=d,\hat{n}. $ Next PQ being the line of shortest distance between $ l_1 $ and $ l_2 $ is the projection of the line joining the points $ A_1({{\vec{a}}_1}) $ and $ A_2({{\vec{a}}_2}) $ on $ \hat{n} $ . $ | \overrightarrow{PQ} |=| {{{\vec{A}}}_1}{{{\vec{A}}}_2}.\hat{n} |\Rightarrow d=| \frac{({{{\vec{a}}}_2}-{{{\vec{a}}}_1}).{{{\vec{b}}}_1}\times {{{\vec{b}}}_2}}{| {{{\vec{b}}}_1}\times {{{\vec{b}}}_2} |} | $
BETA
