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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Line

  • question_answer
    The shortest distance between the lines \[\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\] and \[\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\] is [RPET 2001; MP PET 2002]

    A)                                                               \[\sqrt{30}\]

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

    C)                                                               \[5\sqrt{30}\]

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

    Correct Answer: D

    Solution :

                    S.D.\[=\]\[\frac{\left| \begin{matrix}    6 & 15 & -3  \\    3 & -1 & 1  \\    -3 & 2 & 4  \\ \end{matrix} \right|}{\sqrt{{{(-4-2)}^{2}}+{{(12+3)}^{2}}+{{(6-3)}^{2}}}}\]               \[=\,\,\frac{270}{\sqrt{270}}=\sqrt{270}\] = \[3\sqrt{30}\].


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