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

  • question_answer
    The distance of the point (1, -2, 3) from the plane \[x-y+z=5\] measured parallel to the line \[\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}\] is

    A) 1

    B) 2

    C) 4

    D) \[2\sqrt{3}\]

    Correct Answer: A

    Solution :

    [a] Equation of the line through (1, -2, 3) parallel to the line \[\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}\] is \[\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{-6}=r(say)\]       ?..(1) Then any point on (1) is \[(2r+1,3r-2,-6r+3)\]if this point lies on the plane \[x-y+z=5\] then \[(2r+1)-(3r-2)+(-6r+3)=5\Rightarrow r=\frac{1}{7}\] Hence the point is \[\left( \frac{9}{7},-\frac{11}{7},\frac{15}{7} \right)\] Distance between \[(1,-2,3)\] and \[\left( \frac{9}{7},-\frac{11}{7},\frac{15}{7} \right)\] \[=\sqrt{\left( \frac{4}{49}+\frac{9}{49}+\frac{36}{49} \right)}=\sqrt{\left( \frac{49}{59} \right)}=1\]


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