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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • question_answer
    The line, \[\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}\] intersects the curve \[xy={{c}^{2}},z=0\] if c is equal to

    A) \[\pm 1\]

    B) \[\pm \frac{1}{3}\]

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

    D) None

    Correct Answer: C

    Solution :

    [c] We have, z = 0 for the point where the line intersects the curve. Therefore, \[\frac{x-2}{3}=\frac{y+1}{2}=\frac{0-1}{-1}\] \[\Rightarrow \frac{x-2}{3}=1\] and \[\frac{y+1}{2}=1\] \[\Rightarrow x=5\] and y=1 Put these value in \[xy={{c}^{2}},\] we get \[5={{c}^{2}}\] \[\Rightarrow c=\pm \sqrt{5}\]


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