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

  • question_answer
    The distance of the point of intersection of the line \[\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\]and the plane \[x+y+z=17\] from the point (3, 4, 5) is given by

    A) 3

    B) 3/2

    C) \[\sqrt{3}\]

    D) None of these

    Correct Answer: A

    Solution :

    • Any point on the line \[\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\] is \[(r+3,\,\,2r+4,\,\,2r+5)\] which satisfies the plane. So, \[r+3+2r+4+2r+5=\,\,\Rightarrow \,\,r=1.\]           
    • \[\therefore \] The point is (4, 6, 7).                   
    • Hence required distance is \[\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}=3.\]


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