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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    The equations of the line passing through the point  (1,2,-4) and perpendicular to the two lines \[\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\] and \[\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\], will be [AI CBSE 1983]

    A) \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\]

    B) \[\frac{x-1}{-2}=\frac{y-2}{3}=\frac{z+4}{8}\]

    C) \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\]

    D) None of these

    Correct Answer: A

    Solution :

    • Line passing through the point (1, 2, -4) is \[\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+4}{n}\] Now, according to question, \[3l-16m+7n=0\] and \[3l+8m-5n=0\]
    • Hence required line is, \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\].


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