A) \[\frac{|({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}).{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}|}{|{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}|}\]
B) \[\frac{\left| ({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}).{{{\vec{a}}}_{2}}\times {{{\vec{b}}}_{2}} \right|}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|}\]
C) \[\frac{\left| ({{{\vec{a}}}_{2}}-{{{\vec{b}}}_{2}}).{{{\vec{a}}}_{1}}\times {{{\vec{b}}}_{1}} \right|}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|}\]
D) \[\frac{\left| ({{{\vec{a}}}_{1}}-{{{\vec{b}}}_{2}}).{{{\vec{b}}}_{1}}\times {{{\vec{a}}}_{2}} \right|}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{a}}}_{2}} \right|}\]
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, \[\hat{n}=\frac{{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|}\] Let d be the shortest distance between the given lines \[{{l}_{1}}\] and\[{{l}_{2}}\]. \[\left| \overrightarrow{PQ} \right|=d\] 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\[\hat{n}\]. \[\left| \overrightarrow{PQ} \right|=\left| {{{\vec{A}}}_{1}}{{{\vec{A}}}_{2}}.\hat{n} \right|\Rightarrow d=\left| \frac{({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}).{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|\]
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