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

  • question_answer
    The distance of the point (1, -2, 3) from the plane \[x-y+z=5\] measured parallel to the line \[\frac{x}{2}=\frac{y}{3}=\frac{z}{-6},\]is [AI CBSE 1984]

    A) 1

    B) 6/7

    C) 7/6

    D) None of these

    Correct Answer: A

    Solution :

    • Direction cosines of line \[=\left( \frac{2}{7},\frac{3}{7},\frac{-6}{7} \right)\] Now, \[{x}'=1+\frac{2r}{7},\,\,{y}'=-2+\frac{3r}{7}\] and \[{z}'=3-\frac{6r}{7}\]   
    • \[\therefore \,\,\,\left( 1+\frac{2r}{7} \right)-\left( -2+\frac{3r}{7} \right)+\left( 3-\frac{6r}{7} \right)=5\,\,\Rightarrow \,\,r=1.\]


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