JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    The shortest distance between the lines \[\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\] and \[\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{2}\] is [JEE MAIN Held On 08-01-2020 Morning]

    A) \[2\sqrt{30}\]    

    B) \[\frac{7}{2}\sqrt{30}\]

    C) \[3\sqrt{30}\]

    D) 3

    Correct Answer: C

    Solution :

    [c]             Shortest distance \[=\left| \frac{[{{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}{{{\vec{b}}}_{1}}{{{\vec{b}}}_{2}}]}{\left| {{{\vec{b}}}_{1}}\times {{b}_{2}} \right|} \right|\]             \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    3 & -2 & 1  \\    -3 & 2 & 4  \\ \end{matrix} \right|=-6\hat{i}-15\hat{j}+3\hat{k}\]             \[\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{36+225+9}=\sqrt{270}\]             and \[\left[ {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}{{{\vec{b}}}_{1}}{{{\vec{b}}}_{2}} \right]=-36-225-9=-270\] \[\Rightarrow \] Shortest distance \[=\sqrt{270}=3\sqrt{30}\]          


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